Chapter Test – Continuity and Differentiability | ML Aggarwal Class 12 Solutions

Solution: We observe that the function \( f(x)=x^{2}-\sin x+5 \) is composed of the following parts: \[ x^2,\quad -\sin x,\quad 5. \] The polynomial \( x^2 \) and the constant \(5\) are continuous for all \( x \) and the sine function \( \sin x \) is also continuous for all \( x \). Since the sum of continuous functions is continuous, it follows that \( f(x) \) is continuous at every point, including at \( x=\pi \).

Solution: For \( f(x) \) to be continuous at \( x=0 \), we require \[ \lim_{x\to0^-} f(x)=f(0). \] Given that for \( x\geq0 \) \[ f(0)=8-3\cdot 0=8. \] Now, consider the left-hand limit: \[ \lim_{x\to0^-}\frac{\sin kx}{x}. \] We rewrite the expression as: \[ \frac{\sin kx}{x}=k\frac{\sin kx}{kx}. \] As \( x\to0 \), note that \[ \lim_{x\to0}\frac{\sin kx}{kx}=1. \] Hence, \[ \lim_{x\to0^-}\frac{\sin kx}{x}=k. \] Setting this equal to \( f(0) \) for continuity: \[ k=8. \]

Solution: For continuity at \( x=0 \), we require that \[ \lim_{x\to 0} \frac{1-\cos 2x}{2x^{2}}=f(0)=k. \] To evaluate the limit, we use the standard result: \[ \lim_{x\to 0}\frac{1-\cos ax}{x^{2}}=\frac{a^{2}}{2}. \] Here, \( a=2 \). Therefore, \[ \lim_{x\to 0}\frac{1-\cos2x}{x^{2}}=\frac{2^{2}}{2}=\frac{4}{2}=2. \] Since the original expression has a denominator of \( 2x^{2} \), the limit becomes: \[ \lim_{x\to 0}\frac{1-\cos2x}{2x^{2}}=\frac{1}{2}\lim_{x\to 0}\frac{1-\cos2x}{x^{2}}=\frac{1}{2}\times 2=1. \] Equating this with \( k \) for continuity: \[ k=1. \]

Solution: For continuity at \( x=\frac{\pi}{4} \), we require \[ \lim_{x\to \frac{\pi}{4}} f(x)=f\left(\frac{\pi}{4}\right)=k. \] Substitute \( x=\frac{\pi}{4}+h \). Then, when \( h\to 0 \), we have: \[ f\left(\frac{\pi}{4}+h\right)=\frac{1-\tan\left(\frac{\pi}{4}+h\right)}{4\left(\frac{\pi}{4}+h\right)-\pi}. \] Simplify the denominator: \[ 4\left(\frac{\pi}{4}+h\right)-\pi=\pi+4h-\pi=4h. \] Thus, the function becomes: \[ f\left(\frac{\pi}{4}+h\right)=\frac{1-\tan\left(\frac{\pi}{4}+h\right)}{4h}. \] Now, expand \( \tan\left(\frac{\pi}{4}+h\right) \) using the Taylor series expansion around \( h=0 \): \[ \tan\left(\frac{\pi}{4}+h\right)=\tan\frac{\pi}{4}+\sec^2\frac{\pi}{4}\cdot h+O(h^2)=1+2h+O(h^2), \] since \( \tan\frac{\pi}{4}=1 \) and \( \sec^2\frac{\pi}{4}=2 \). Substitute this expansion into the numerator: \[ 1-\tan\left(\frac{\pi}{4}+h\right)\approx 1-(1+2h)= -2h. \] Therefore, the limit becomes: \[ \lim_{h\to 0}\frac{-2h}{4h}=\lim_{h\to 0}\frac{-2}{4}=-\frac{1}{2}. \] For continuity, \( k \) must equal this limit.

Solution: For \( x<0 \), the function is given by \[ f(x)=\frac{x}{|x|}. \] When \( x<0 \), note that \(|x|=-x\). Thus, \[ f(x)=\frac{x}{-x}=-1. \] For \( x\geq0 \), the function is explicitly defined as \[ f(x)=-1. \] Hence, for all \( x\neq 0 \) (in the domain of the first piece) and at \( x=0 \) (the second piece), the function takes the constant value \(-1\). Therefore, \( f(x) \) is continuous everywhere on its domain.

Solution: The function involves absolute value functions which are continuous for all real \( x \). More precisely, both \( |x| \) and \( |x+1| \) are continuous everywhere. The difference of two continuous functions is also continuous. Therefore, \( f(x)=|x|-|x+1| \) is continuous for all \( x \) in \( \mathbb{R} \).

Solution: We first note that the absolute value function is continuous for all \( x \). Since \( f(x) \) is a sum of absolute value functions, it is continuous everywhere, including at \( x = 1 \) and \( x = 3 \).

For \( x < 1 \):
\[ |x – 1| = 1 – x, \quad |x – 3| = 3 – x, \] so \[ f(x) = (1 – x) + (3 – x) = 4 – 2x. \] The derivative here is \[ f'(x) = -2. \]
For \( 1 < x < 3 \):
\[ |x – 1| = x – 1, \quad |x – 3| = 3 – x, \] thus \[ f(x) = (x – 1) + (3 – x) = 2, \] a constant, so the derivative is \[ f'(x) = 0. \]
At \( x = 1 \):
\[ f'(1^-) = -2, \quad f'(1^+) = 0. \] Since the one-sided derivatives differ, \( f(x) \) is not differentiable at \( x = 1 \).

For \( x > 3 \):
\[ |x – 1| = x – 1, \quad |x – 3| = x – 3, \] so \[ f(x) = (x – 1) + (x – 3) = 2x – 4. \] The derivative for \( x > 3 \) is \[ f'(x) = 2. \]
For \( 1 < x < 3 \) we already found \( f'(x) = 0 \). Thus, at \( x = 3 \):
\[ f'(3^-) = 0, \quad f'(3^+) = 2. \] Since these are not equal, \( f(x) \) is not differentiable at \( x = 3 \).

In summary, \( f(x) \) is continuous for all \( x \) but is not differentiable at \( x = 1 \) and \( x = 3 \).

Solution: Write \[ u(x)=\sqrt{a}+\sqrt{x} \quad \text{and} \quad v(x)=\sqrt{a}-\sqrt{x}. \] Then the derivative \( f'(x) \) using the quotient rule is: \[ f'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{\left(v(x)\right)^{2}}. \] First, compute the derivatives: \[ u'(x)=\frac{d}{dx}(\sqrt{a}+\sqrt{x})=0+\frac{1}{2\sqrt{x}}=\frac{1}{2\sqrt{x}}, \] \[ v'(x)=\frac{d}{dx}(\sqrt{a}-\sqrt{x})=0-\frac{1}{2\sqrt{x}}=-\frac{1}{2\sqrt{x}}. \] Next, substitute into the quotient rule: \[ f'(x)=\frac{\frac{1}{2\sqrt{x}}\left(\sqrt{a}-\sqrt{x}\right)-\left(\sqrt{a}+\sqrt{x}\right)\left(-\frac{1}{2\sqrt{x}}\right)}{\left(\sqrt{a}-\sqrt{x}\right)^{2}}. \] Simplify the numerator: \[ \frac{1}{2\sqrt{x}}\left(\sqrt{a}-\sqrt{x}\right)+\frac{1}{2\sqrt{x}}\left(\sqrt{a}+\sqrt{x}\right) =\frac{1}{2\sqrt{x}}\left[ \left(\sqrt{a}-\sqrt{x}\right)+\left(\sqrt{a}+\sqrt{x}\right)\right] =\frac{1}{2\sqrt{x}}(2\sqrt{a}) =\frac{\sqrt{a}}{\sqrt{x}}. \] Thus, the derivative is: \[ f'(x)=\frac{\sqrt{a}}{\sqrt{x}\,\left(\sqrt{a}-\sqrt{x}\right)^{2}}. \]

Solution: We start by writing \[ u(x)=1-\tan x,\quad v(x)=1+\tan x. \] Then by the quotient rule, \[ f'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{\left(v(x)\right)^{2}}. \] First, compute the derivatives: \[ u'(x)=-\sec^{2}x,\quad v'(x)=\sec^{2}x. \] Then, \[ \begin{aligned} f'(x) &= \frac{-\sec^{2}x\,(1+\tan x) – (1-\tan x)\,\sec^{2}x}{\left(1+\tan x\right)^{2}}\\[1mm] &= \frac{-\sec^{2}x\left[(1+\tan x)+(1-\tan x)\right]}{\left(1+\tan x\right)^{2}}\\[1mm] &= \frac{-\sec^{2}x\,(2)}{\left(1+\tan x\right)^{2}}\\[1mm] &= -\frac{2\,\sec^{2}x}{\left(1+\tan x\right)^{2}}. \end{aligned} \] Multiply both the numerator and the denominator by \(\cos^{2}x\) (since \(\sec^{2}x=\frac{1}{\cos^{2}x}\)): \[ f'(x)=-\frac{2}{\left(1+\tan x\right)^{2}\cos^{2}x}. \] Notice that \[ (1+\tan x)^{2}\cos^{2}x=\left(\cos x+\sin x\right)^{2}. \] Also, using the identity \[ (\cos x+\sin x)^{2}=\cos^{2}x+2\sin x\cos x+\sin^{2}x=1+2\sin x\cos x=1+\sin 2x, \] we obtain: \[ f'(x)=-\frac{2}{1+\sin 2x}. \]

Solution: We start with \[ f(x)=\sin \left(\sqrt{\sin \sqrt{x}}\right). \] Let \[ u=\sqrt{\sin \sqrt{x}}, \] so that \[ f(x)=\sin (u). \] Differentiating \( f(x) \) using the chain rule yields: \[ f'(x)=\cos (u)\cdot \frac{du}{dx}. \] Since \[ u=\left(\sin \sqrt{x}\right)^{\frac{1}{2}}, \] we differentiate \( u \) with respect to \( x \): \[ \frac{du}{dx}=\frac{1}{2}\left(\sin \sqrt{x}\right)^{-\frac{1}{2}} \cdot \frac{d}{dx}\left(\sin \sqrt{x}\right). \] Now, differentiate \(\sin \sqrt{x}\) by applying the chain rule: \[ \frac{d}{dx}\left(\sin \sqrt{x}\right)=\cos \sqrt{x}\cdot \frac{d}{dx}\left(\sqrt{x}\right)=\cos \sqrt{x}\cdot \frac{1}{2\sqrt{x}}. \] Thus, \[ \frac{du}{dx}=\frac{1}{2}\left(\sin \sqrt{x}\right)^{-\frac{1}{2}} \cdot \cos \sqrt{x}\cdot \frac{1}{2\sqrt{x}}=\frac{\cos \sqrt{x}}{4\sqrt{x}\sqrt{\sin \sqrt{x}}}. \] Substituting back into the expression for \( f'(x) \): \[ f'(x)=\cos \left(\sqrt{\sin \sqrt{x}}\right)\cdot \frac{\cos \sqrt{x}}{4\sqrt{x}\sqrt{\sin \sqrt{x}}}. \]

Solution: Let \[ f(x)=\frac{\tan 2x}{1-\cot 2x}. \] Define \[ u(x)=\tan 2x \quad \text{and} \quad v(x)=1-\cot 2x. \] Then, \[ u'(x)=\frac{d}{dx}(\tan 2x)=2\sec^2 2x, \] and \[ v'(x)=\frac{d}{dx}(1-\cot 2x)=2\,\mathrm{cosec}^2 2x. \] Using the quotient rule, \[ f'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{\left[v(x)\right]^2}=\frac{2\sec^2 2x\,(1-\cot 2x)-\tan 2x\,(2\,\mathrm{cosec}^2 2x)}{(1-\cot 2x)^2}. \] Factor out \( 2 \) in the numerator: \[ f'(x)=\frac{2\left[\sec^2 2x\,(1-\cot 2x)-\tan 2x\,\mathrm{cosec}^2 2x\right]}{(1-\cot 2x)^2}. \] Notice that \[ \tan 2x\,\mathrm{cosec}^2 2x=\frac{\sin 2x}{\cos 2x}\cdot \frac{1}{\sin^2 2x}=\frac{1}{\cos 2x\,\sin 2x}=\frac{2}{2\cos 2x\,\sin 2x}=\frac{2}{\sin 4x}=2\,\mathrm{cosec}4x. \] Therefore, the derivative simplifies to: \[ f'(x)=\frac{2\left[\sec^2 2x\,(1-\cot 2x)-2\,\mathrm{cosec}4x\right]}{(1-\cot 2x)^2}. \]

Solution: Let \[ f(x)=\sin\left(\frac{1+x^{2}}{1-x^{2}}\right). \] Define \[ \theta(x)=\frac{1+x^{2}}{1-x^{2}}. \] Then, by the chain rule, \[ f'(x)=\cos\left(\theta(x)\right)\cdot\theta'(x). \] To compute \(\theta'(x)\), write \[ \theta(x)=\frac{u(x)}{v(x)}, \] where \[ u(x)=1+x^{2} \quad \text{and} \quad v(x)=1-x^{2}. \] Differentiating, we obtain: \[ u'(x)=2x, \quad v'(x)=-2x. \] Applying the quotient rule, \[ \theta'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{\left[v(x)\right]^2} = \frac{2x(1-x^{2})- (1+x^{2})(-2x)}{(1-x^{2})^2}. \] Simplify the numerator: \[ 2x(1-x^{2})+2x(1+x^{2})=2x\Big[(1-x^{2})+(1+x^{2})\Big]=2x(2)=4x. \] Hence, \[ \theta'(x)=\frac{4x}{(1-x^{2})^2}. \] Substituting back into the derivative: \[ f'(x)=\cos\left(\frac{1+x^{2}}{1-x^{2}}\right)\cdot \frac{4x}{(1-x^{2})^2}. \]

Solution: We start with \[ f(x)=\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}. \] To simplify \( f(x) \), multiply the numerator and denominator by the conjugate of the denominator: \[ f(x)=\frac{\left(\sqrt{1+\sin x}+\sqrt{1-\sin x}\right)^2}{\left(\sqrt{1+\sin x}\right)^2-\left(\sqrt{1-\sin x}\right)^2}. \] Since \[ \left(\sqrt{1+\sin x}\right)^2=1+\sin x \quad \text{and} \quad \left(\sqrt{1-\sin x}\right)^2=1-\sin x, \] the denominator becomes: \[ (1+\sin x)-(1-\sin x)=2\sin x. \] Next, expanding the numerator: \[ \left(\sqrt{1+\sin x}+\sqrt{1-\sin x}\right)^2 = (1+\sin x)+(1-\sin x)+2\sqrt{(1+\sin x)(1-\sin x)}. \] Notice that \[ (1+\sin x)+(1-\sin x)=2 \quad \text{and} \quad \sqrt{(1+\sin x)(1-\sin x)}=\sqrt{1-\sin^2 x}=\lvert\cos x\rvert. \] Assuming the domain where \(\cos x\ge0\) (for instance, \(x\) in \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\)), we have: \[ \sqrt{(1+\sin x)(1-\sin x)}=\cos x. \] Thus, the numerator simplifies to: \[ 2+2\cos x=2(1+\cos x). \] Therefore, \[ f(x)=\frac{2(1+\cos x)}{2\sin x}=\frac{1+\cos x}{\sin x}. \] Using the half-angle identity, \[ \cot\frac{x}{2}=\frac{1+\cos x}{\sin x}, \] we write \[ f(x)=\cot\frac{x}{2}. \] Now, differentiating \( f(x)=\cot\frac{x}{2} \) with respect to \( x \): \[ f'(x)=\frac{d}{dx}\left(\cot\frac{x}{2}\right). \] Recall that: \[ \frac{d}{dx}\left(\cot u\right)=-\mathrm{cosec}^{2}u\cdot\frac{du}{dx}, \] where here \( u=\frac{x}{2} \) and \(\frac{du}{dx}=\frac{1}{2}\). Thus, \[ f'(x)=-\mathrm{cosec}^{2}\frac{x}{2}\cdot\frac{1}{2}=-\frac{1}{2}\,\mathrm{cosec}^{2}\frac{x}{2}. \]

Solution: We begin by rewriting the function \( y \) as the difference of two logarithms: \[ y=\log\!\left(a+b \tan \frac{x}{2}\right)-\log\!\left(a-b \tan \frac{x}{2}\right). \] Differentiating term by term, we have \[ \frac{d}{dx}\left[\log\!\left(a+b \tan \frac{x}{2}\right)\right] = \frac{1}{a+b \tan \frac{x}{2}}\cdot \frac{d}{dx}\left(a+b \tan \frac{x}{2}\right), \] and \[ \frac{d}{dx}\left[\log\!\left(a-b \tan \frac{x}{2}\right)\right] = \frac{1}{a-b \tan \frac{x}{2}}\cdot \frac{d}{dx}\left(a-b \tan \frac{x}{2}\right). \] Since \[ \frac{d}{dx}\left(\tan \frac{x}{2}\right) = \sec^2\frac{x}{2}\cdot \frac{1}{2}, \] we obtain: \[ \frac{d}{dx}\left(a+b \tan \frac{x}{2}\right) = b\cdot \frac{1}{2}\sec^2\frac{x}{2} = \frac{b}{2}\sec^2\frac{x}{2}, \] and similarly, \[ \frac{d}{dx}\left(a-b \tan \frac{x}{2}\right) = -\frac{b}{2}\sec^2\frac{x}{2}. \] Therefore, the derivative of \( y \) is \[ \frac{dy}{dx} = \frac{\frac{b}{2}\sec^2\frac{x}{2}}{a+b \tan \frac{x}{2}} – \left(\frac{-\frac{b}{2}\sec^2\frac{x}{2}}{a-b \tan \frac{x}{2}}\right) = \frac{b}{2}\sec^2\frac{x}{2}\left[\frac{1}{a+b \tan \frac{x}{2}} +\frac{1}{a-b \tan \frac{x}{2}}\right]. \] Combine the two fractions: \[ \frac{1}{a+b \tan \frac{x}{2}}+\frac{1}{a-b \tan \frac{x}{2}} = \frac{\Bigl(a-b \tan \frac{x}{2}\Bigr)+\Bigl(a+b \tan \frac{x}{2}\Bigr)} {\Bigl(a+b \tan \frac{x}{2}\Bigr)\Bigl(a-b \tan \frac{x}{2}\Bigr)} = \frac{2a}{a^{2}-b^{2}\tan^{2}\frac{x}{2}}. \] Substituting back, we get \[ \frac{dy}{dx}=\frac{b}{2}\sec^2\frac{x}{2}\cdot\frac{2a}{a^{2}-b^{2}\tan^{2}\frac{x}{2}} =\frac{ab\,\sec^2\frac{x}{2}}{a^{2}-b^{2}\tan^{2}\frac{x}{2}}. \] Express the denominator in terms of cosine using \[ \tan^{2}\frac{x}{2}=\frac{\sin^{2}\frac{x}{2}}{\cos^{2}\frac{x}{2}}, \] so that \[ a^{2}-b^{2}\tan^{2}\frac{x}{2} = a^{2}-b^{2}\frac{\sin^{2}\frac{x}{2}}{\cos^{2}\frac{x}{2}} = \frac{a^{2}\cos^{2}\frac{x}{2}-b^{2}\sin^{2}\frac{x}{2}}{\cos^{2}\frac{x}{2}}. \] Thus, the derivative becomes \[ \frac{dy}{dx}=\frac{ab\,\sec^2\frac{x}{2}}{\displaystyle \frac{a^{2}\cos^{2}\frac{x}{2}-b^{2}\sin^{2}\frac{x}{2}}{\cos^{2}\frac{x}{2}}} =\frac{ab\,\sec^2\frac{x}{2}\cos^{2}\frac{x}{2}}{a^{2}\cos^{2}\frac{x}{2}-b^{2}\sin^{2}\frac{x}{2}}. \] Since \( \sec^2\frac{x}{2}\cos^{2}\frac{x}{2}=1 \), we obtain the desired result: \[ \frac{dy}{dx}=\frac{ab}{a^{2}\cos^{2}\frac{x}{2}-b^{2}\sin^{2}\frac{x}{2}}. \]

Solution: We start with \[ 2y=x\sqrt{x^{2}-a^{2}}-a^{2}\log\!\left(x+\sqrt{x^{2}-a^{2}}\right). \] Differentiating both sides with respect to \( x \), we obtain: \[ \frac{d}{dx}(2y)=\frac{d}{dx}\left[x\sqrt{x^{2}-a^{2}}\right]-a^{2}\frac{d}{dx}\left[\log\!\left(x+\sqrt{x^{2}-a^{2}}\right)\right]. \] That is, \[ 2\frac{dy}{dx}=\frac{d}{dx}\left[x\sqrt{x^{2}-a^{2}}\right]-a^{2}\frac{d}{dx}\left[\log\!\left(x+\sqrt{x^{2}-a^{2}}\right)\right]. \] **Step 1. Differentiate \( x\sqrt{x^{2}-a^{2}} \):** Let \[ F(x)=x\sqrt{x^{2}-a^{2}}. \] Using the product rule, \[ F'(x)=\sqrt{x^{2}-a^{2}}+x\cdot\frac{d}{dx}\left(\sqrt{x^{2}-a^{2}}\right). \] Since \[ \sqrt{x^{2}-a^{2}}=(x^{2}-a^{2})^{\frac{1}{2}}, \] we have \[ \frac{d}{dx}\left(\sqrt{x^{2}-a^{2}}\right) =\frac{1}{2}(x^{2}-a^{2})^{-\frac{1}{2}}\cdot2x =\frac{x}{\sqrt{x^{2}-a^{2}}}. \] Therefore, \[ F'(x)=\sqrt{x^{2}-a^{2}}+x\cdot\frac{x}{\sqrt{x^{2}-a^{2}}} =\sqrt{x^{2}-a^{2}}+\frac{x^{2}}{\sqrt{x^{2}-a^{2}}}. \] Combining the terms over a common denominator yields: \[ F'(x)=\frac{x^{2}-a^{2}+x^{2}}{\sqrt{x^{2}-a^{2}}} =\frac{2x^{2}-a^{2}}{\sqrt{x^{2}-a^{2}}}. \] **Step 2. Differentiate \( \log\!\left(x+\sqrt{x^{2}-a^{2}}\right) \):** Let \[ G(x)=\log\!\left(x+\sqrt{x^{2}-a^{2}}\right). \] Its derivative is \[ G'(x)=\frac{1}{x+\sqrt{x^{2}-a^{2}}}\cdot\frac{d}{dx}\left(x+\sqrt{x^{2}-a^{2}}\right). \] We have \[ \frac{d}{dx}\left(x+\sqrt{x^{2}-a^{2}}\right) =1+\frac{x}{\sqrt{x^{2}-a^{2}}}. \] Thus, \[ G'(x)=\frac{1}{x+\sqrt{x^{2}-a^{2}}}\left(1+\frac{x}{\sqrt{x^{2}-a^{2}}}\right) =\frac{\sqrt{x^{2}-a^{2}}+x}{\sqrt{x^{2}-a^{2}}\left(x+\sqrt{x^{2}-a^{2}}\right)}. \] Notice that the numerator and denominator have the common factor \( x+\sqrt{x^{2}-a^{2}} \), so it simplifies to: \[ G'(x)=\frac{1}{\sqrt{x^{2}-a^{2}}}. \] **Step 3. Combine the derivatives:** Substituting the derivatives of \( F(x) \) and \( G(x) \) into the differentiation of the given equation, we have: \[ 2\frac{dy}{dx}=\frac{2x^{2}-a^{2}}{\sqrt{x^{2}-a^{2}}}-a^{2}\cdot\frac{1}{\sqrt{x^{2}-a^{2}}} =\frac{2x^{2}-a^{2}-a^{2}}{\sqrt{x^{2}-a^{2}}} =\frac{2x^{2}-2a^{2}}{\sqrt{x^{2}-a^{2}}}. \] Simplify further: \[ 2\frac{dy}{dx}=\frac{2\left(x^{2}-a^{2}\right)}{\sqrt{x^{2}-a^{2}}} =2\sqrt{x^{2}-a^{2}}. \] Dividing both sides by 2, \[ \frac{dy}{dx}=\sqrt{x^{2}-a^{2}}. \] Hence, we have proved that \[ \frac{dy}{dx}=\sqrt{x^{2}-a^{2}}. \]

Solution: We begin with \[ x=2a\sin^{-1}\sqrt{\frac{y}{2a}}-\sqrt{2ay-y^2}. \] Define \[ I=2a\sin^{-1}\sqrt{\frac{y}{2a}}, \quad II=\sqrt{2ay-y^2}. \]

Step 1: Differentiate \( I \) with respect to \( y \).

Let \[ u=\sqrt{\frac{y}{2a}}, \quad \text{so that} \quad I=2a\sin^{-1}(u). \] Differentiating using the chain rule: \[ \frac{d}{dy}\left[\sin^{-1}(u)\right]=\frac{1}{\sqrt{1-u^2}}\cdot\frac{du}{dy}. \] Since \[ u=\left(\frac{y}{2a}\right)^{\frac{1}{2}}, \quad \text{we have} \quad \frac{du}{dy}=\frac{1}{2\sqrt{2a}}\frac{1}{\sqrt{y}}. \] Note that \[ u^2=\frac{y}{2a} \quad \Rightarrow \quad 1-u^2=1-\frac{y}{2a}=\frac{2a-y}{2a}, \] and hence \[ \sqrt{1-u^2}=\sqrt{\frac{2a-y}{2a}}. \] Therefore, \[ \frac{d}{dy}\left[\sin^{-1}\sqrt{\frac{y}{2a}}\right]=\frac{1}{\sqrt{\frac{2a-y}{2a}}}\cdot\frac{1}{2\sqrt{2a}}\frac{1}{\sqrt{y}} =\sqrt{\frac{2a}{2a-y}}\cdot\frac{1}{2\sqrt{2a}}\frac{1}{\sqrt{y}}. \] Multiplying by \( 2a \) gives \[ I_y=2a\cdot \sqrt{\frac{2a}{2a-y}}\cdot\frac{1}{2\sqrt{2a}}\frac{1}{\sqrt{y}} =\frac{a}{\sqrt{y(2a-y)}}. \]

Step 2: Differentiate \( II \) with respect to \( y \).

Write \[ II=\sqrt{2ay-y^2}=\left(2ay-y^2\right)^{\frac{1}{2}}. \] Differentiating using the chain rule, \[ II_y=\frac{1}{2}(2ay-y^2)^{-\frac{1}{2}}\cdot (2a-2y) =\frac{2a-2y}{2\sqrt{2ay-y^2}}. \] Recognizing that \[ \sqrt{2ay-y^2}=\sqrt{y(2a-y)}, \] we have \[ II_y=\frac{a-y}{\sqrt{y(2a-y)}}. \]

Step 3: Compute \(\frac{dx}{dy}\) and then \(\frac{dy}{dx}\).

Since \[ x= I- II, \] it follows that \[ \frac{dx}{dy}=I_y- II_y =\frac{a}{\sqrt{y(2a-y)}}-\frac{a-y}{\sqrt{y(2a-y)}} =\frac{a-(a-y)}{\sqrt{y(2a-y)}} =\frac{y}{\sqrt{y(2a-y)}}. \] Thus, \[ \frac{dx}{dy}=\sqrt{\frac{y}{2a-y}}. \] Taking the reciprocal, we conclude that \[ \frac{dy}{dx}=\sqrt{\frac{2a-y}{y}}. \]

Solution: Let \[ u(x)=\frac{a+b\cos x}{b+a\cos x}. \] Then \[ y=\sin^{-1}\left(u(x)\right) \] and by the chain rule, \[ \frac{dy}{dx}=\frac{u'(x)}{\sqrt{1-u(x)^2}}. \] Step 1. Compute \(u'(x)\): Write \[ u(x)=\frac{N(x)}{D(x)},\quad \text{with } N(x)=a+b\cos x,\quad D(x)=b+a\cos x. \] Differentiating gives: \[ N'(x)=-b\sin x,\quad D'(x)=-a\sin x. \] Applying the quotient rule, \[ u'(x)=\frac{N'(x)D(x)-N(x)D'(x)}{[D(x)]^2} =\frac{(-b\sin x)(b+a\cos x)- (a+b\cos x)(-a\sin x)}{(b+a\cos x)^2}. \] Simplify the numerator: \[ \begin{aligned} &(-b\sin x)(b+a\cos x)+(a+b\cos x)(a\sin x)\\[2mm] &= -b\sin x\,(b+a\cos x)+a\sin x\,(a+b\cos x)\\[2mm] &=\sin x\Bigl[a(a+b\cos x)-b(b+a\cos x)\Bigr]\\[2mm] &=\sin x\Bigl[a^2+ab\cos x-b^2-ab\cos x\Bigr]\\[2mm] &=\sin x\,(a^2-b^2). \end{aligned} \] Hence, \[ u'(x)=\frac{(a^2-b^2)\sin x}{(b+a\cos x)^2}. \] Step 2. Compute \(1-u(x)^2\): Since \[ u(x)^2=\frac{(a+b\cos x)^2}{(b+a\cos x)^2}, \] it follows that \[ 1-u(x)^2=\frac{(b+a\cos x)^2-(a+b\cos x)^2}{(b+a\cos x)^2}. \] Expand the numerator: \[ \begin{aligned} (b+a\cos x)^2&=b^2+2ab\cos x+a^2\cos^2 x,\\[2mm] (a+b\cos x)^2&=a^2+2ab\cos x+b^2\cos^2 x. \end{aligned} \] Their difference gives: \[ \begin{aligned} (b+a\cos x)^2-(a+b\cos x)^2 &=\left[b^2+2ab\cos x+a^2\cos^2 x\right]-\left[a^2+2ab\cos x+b^2\cos^2 x\right]\\[2mm] &= (b^2-a^2)+(a^2\cos^2 x-b^2\cos^2 x)\\[2mm] &= (b^2-a^2)(1-\cos^2 x)\\[2mm] &= (b^2-a^2)\sin^2 x. \end{aligned} \] Thus, \[ 1-u(x)^2=\frac{(b^2-a^2)\sin^2 x}{(b+a\cos x)^2}. \] Taking the square root (assuming \(b+a\cos x>0\) and a suitable domain for \(x\)), \[ \sqrt{1-u(x)^2}=\frac{\sqrt{b^2-a^2}\,|\sin x|}{b+a\cos x}. \] Within a domain where \(\sin x\) is nonnegative, this simplifies to \[ \sqrt{1-u(x)^2}=\frac{\sqrt{b^2-a^2}\,\sin x}{b+a\cos x}. \] Step 3. Combine the results: Substitute \(u'(x)\) and \(\sqrt{1-u(x)^2}\) into the chain rule expression: \[ \frac{dy}{dx}=\frac{\displaystyle \frac{(a^2-b^2)\sin x}{(b+a\cos x)^2}}{\displaystyle \frac{\sqrt{b^2-a^2}\,\sin x}{b+a\cos x}} =\frac{(a^2-b^2)\sin x}{(b+a\cos x)^2} \cdot \frac{b+a\cos x}{\sqrt{b^2-a^2}\,\sin x}. \] Cancel \(\sin x\) and one factor of \((b+a\cos x)\): \[ \frac{dy}{dx}=\frac{a^2-b^2}{(b+a\cos x)\sqrt{b^2-a^2}}. \] Since \(a^2-b^2=-(b^2-a^2)\), we have \[ \frac{dy}{dx}=-\frac{\sqrt{b^2-a^2}}{b+a\cos x}. \]

Solution: Let \[ u(x)=\frac{\sqrt{1+x^{2}}-1}{x}. \] Then \[ y=\cot^{-1}(u(x)) \quad\text{and}\quad \frac{dy}{dx}=-\frac{u'(x)}{1+u(x)^2}. \] We first differentiate \( u(x) \) using the quotient rule. Writing \[ u(x)=\frac{N(x)}{D(x)} \quad \text{with} \quad N(x)=\sqrt{1+x^{2}}-1 \quad\text{and}\quad D(x)=x, \] we have \[ N'(x)=\frac{x}{\sqrt{1+x^{2}}}, \quad D'(x)=1. \] Thus, \[ u'(x)=\frac{D(x)N'(x)-N(x)D'(x)}{D(x)^2} =\frac{x\left(\frac{x}{\sqrt{1+x^{2}}}\right)-\Bigl(\sqrt{1+x^{2}}-1\Bigr)}{x^2}. \] Simplify the numerator: \[ \frac{x^2}{\sqrt{1+x^{2}}}-\sqrt{1+x^{2}}+1. \] Writing all terms over the common denominator \(\sqrt{1+x^{2}}\), \[ u'(x)=\frac{\frac{x^2-\Bigl(1+x^{2}\Bigr)+\sqrt{1+x^{2}}}{\sqrt{1+x^{2}}}}{x^2} =\frac{\sqrt{1+x^{2}}-1}{x^2\sqrt{1+x^{2}}}. \] Next, we compute \(1+u(x)^2\). Since \[ u(x)^2=\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)^2=\frac{\Bigl(\sqrt{1+x^{2}}-1\Bigr)^2}{x^2}, \] we have \[ 1+u(x)^2=\frac{x^2+\Bigl(\sqrt{1+x^{2}}-1\Bigr)^2}{x^2}. \] Note that \[ \Bigl(\sqrt{1+x^{2}}-1\Bigr)^2=1+x^{2}-2\sqrt{1+x^{2}}+1=x^{2}+2-2\sqrt{1+x^{2}}, \] so that \[ 1+u(x)^2=\frac{x^2+(x^{2}+2-2\sqrt{1+x^{2}})}{x^2} =\frac{2x^2+2-2\sqrt{1+x^{2}}}{x^2}. \] Factor out 2: \[ 1+u(x)^2=\frac{2\Bigl(x^2+1-\sqrt{1+x^{2}}\Bigr)}{x^2}. \] Notice that \[ x^2+1=\left(\sqrt{1+x^{2}}\right)^2, \] so that \[ x^2+1-\sqrt{1+x^{2}}=\sqrt{1+x^{2}}\Bigl(\sqrt{1+x^{2}}-1\Bigr). \] Therefore, \[ 1+u(x)^2=\frac{2\sqrt{1+x^{2}}\Bigl(\sqrt{1+x^{2}}-1\Bigr)}{x^2}. \] Now, substituting \( u'(x) \) and \( 1+u(x)^2 \) into the derivative formula: \[ \frac{dy}{dx}=-\frac{u'(x)}{1+u(x)^2} =-\frac{\displaystyle \frac{\sqrt{1+x^{2}}-1}{x^2\sqrt{1+x^{2}}}}{\displaystyle \frac{2\sqrt{1+x^{2}}\,(\sqrt{1+x^{2}}-1)}{x^2}} =-\frac{\sqrt{1+x^{2}}-1}{x^2\sqrt{1+x^{2}}}\cdot\frac{x^2}{2\sqrt{1+x^{2}}(\sqrt{1+x^{2}}-1)}. \] Cancel the common factors \( x^2 \) and \(\sqrt{1+x^{2}}-1\) (provided \( x\neq 0 \)): \[ \frac{dy}{dx}=-\frac{1}{2\sqrt{1+x^{2}}\cdot\sqrt{1+x^{2}}} =-\frac{1}{2(1+x^{2})}. \]

Solution: Begin with \[ y=\tan^{-1}\sqrt{\frac{a-x}{a+x}}. \] Define \[ u=\sqrt{\frac{a-x}{a+x}}, \] so that \[ y=\tan^{-1}(u) \] and by the chain rule, \[ \frac{dy}{dx}=\frac{1}{1+u^2}\,\frac{du}{dx}. \] Step 1. Compute \(u^2\) and \(1+u^2\): We have \[ u^2=\frac{a-x}{a+x}. \] Therefore, \[ 1+u^2=1+\frac{a-x}{a+x}=\frac{(a+x)+(a-x)}{a+x}=\frac{2a}{a+x}. \] Step 2. Differentiate \(u\) with respect to \(x\): Write \[ u=\left(\frac{a-x}{a+x}\right)^{\frac{1}{2}}. \] Then by the chain rule, \[ \frac{du}{dx}=\frac{1}{2}\left(\frac{a-x}{a+x}\right)^{-\frac{1}{2}}\cdot \frac{d}{dx}\left(\frac{a-x}{a+x}\right). \] To differentiate \(\frac{a-x}{a+x}\), let \[ F(x)=a-x\quad \text{and}\quad G(x)=a+x, \] so that \(F'(x)=-1\) and \(G'(x)=1\). Then, by the quotient rule, \[ \frac{d}{dx}\left(\frac{a-x}{a+x}\right)=\frac{G(x)F'(x)-F(x)G'(x)}{(a+x)^2}=\frac{(a+x)(-1)-(a-x)(1)}{(a+x)^2}=\frac{-2a}{(a+x)^2}. \] Hence, \[ \frac{du}{dx}=\frac{1}{2}\left(\frac{a-x}{a+x}\right)^{-\frac{1}{2}}\cdot \frac{-2a}{(a+x)^2} =-\frac{a}{(a+x)^2}\left(\frac{a-x}{a+x}\right)^{-\frac{1}{2}}. \] Recognize that \[ \left(\frac{a-x}{a+x}\right)^{-\frac{1}{2}}=\sqrt{\frac{a+x}{a-x}}, \] so that \[ \frac{du}{dx}=-\frac{a}{(a+x)^2}\sqrt{\frac{a+x}{a-x}}. \] Step 3. Combine the results to find \(\frac{dy}{dx}\): Substitute the expressions for \(1+u^2\) and \(\frac{du}{dx}\) into the chain rule: \[ \frac{dy}{dx}=\frac{1}{\frac{2a}{a+x}}\left(-\frac{a}{(a+x)^2}\sqrt{\frac{a+x}{a-x}}\right) =\frac{a+x}{2a}\left(-\frac{a}{(a+x)^2}\sqrt{\frac{a+x}{a-x}}\right). \] Cancel the factor \(a\) and simplify: \[ \frac{dy}{dx}=-\frac{1}{2}\,\frac{1}{a+x}\sqrt{\frac{a+x}{a-x}}. \] Notice that \[ \sqrt{\frac{a+x}{a-x}}=\frac{\sqrt{a+x}}{\sqrt{a-x}}, \] so that \[ \frac{dy}{dx}=-\frac{1}{2}\,\frac{\sqrt{a+x}}{(a+x)\sqrt{a-x}} =-\frac{1}{2}\,\frac{1}{\sqrt{(a+x)(a-x)}}. \] Since \[ (a+x)(a-x)=a^2-x^2, \] it follows that \[ \frac{dy}{dx}=-\frac{1}{2\sqrt{a^2-x^2}}. \]

Solution: We start with \[ y = 2\tan^{-1}\sqrt{\frac{x-a}{b-x}}. \] Let \[ w = \sqrt{\frac{x-a}{b-x}}, \] so that \[ y = 2\tan^{-1}(w). \] Differentiating \(y\) with respect to \(x\) by the chain rule gives \[ \frac{dy}{dx} = 2\cdot\frac{1}{1+w^2}\cdot\frac{dw}{dx}. \] Step 1. Compute \(1+w^2\): Since \[ w^2 = \frac{x-a}{b-x}, \] it follows that \[ 1+w^2 = 1+\frac{x-a}{b-x} = \frac{(b-x)+(x-a)}{b-x} = \frac{b-a}{b-x}. \] Thus, \[ \frac{1}{1+w^2} = \frac{b-x}{b-a}. \] Step 2. Differentiate \(w\) with respect to \(x\): Write \[ w = \left(\frac{x-a}{b-x}\right)^{\frac{1}{2}}. \] Taking natural logarithms, we have \[ \ln w = \frac{1}{2}\Bigl[\ln(x-a)-\ln(b-x)\Bigr]. \] Differentiating with respect to \(x\), \[ \frac{w’}{w} = \frac{1}{2}\left(\frac{1}{x-a}+\frac{1}{b-x}\right) = \frac{1}{2}\cdot\frac{(b-x)+(x-a)}{(x-a)(b-x)} = \frac{b-a}{2(x-a)(b-x)}. \] Hence, \[ \frac{dw}{dx} = \frac{w(b-a)}{2(x-a)(b-x)}. \] Step 3. Substitute into the derivative formula: Combining the results, we obtain \[ \frac{dy}{dx} = 2\cdot\frac{b-x}{b-a}\cdot\frac{w(b-a)}{2(x-a)(b-x)} = \frac{w}{x-a}. \] Since \[ w = \sqrt{\frac{x-a}{b-x}}, \] we have \[ \frac{dy}{dx} = \frac{1}{x-a}\sqrt{\frac{x-a}{b-x}} = \frac{1}{\sqrt{(x-a)(b-x)}}. \] Note that for \(a < x < b\), we have \(x-b < 0\) and therefore \[ (x-a)(x-b) = -\,(x-a)(b-x). \] Squaring the derivative, we get \[ \left(\frac{dy}{dx}\right)^2 = \frac{1}{(x-a)(b-x)} = -\frac{1}{(x-a)(x-b)}. \] Rearranging yields the required result: \[ \left(\frac{dy}{dx}\right)^2 + \frac{1}{(x-a)(x-b)} = 0. \]

Solution: Given \[ \tan y = \frac{2t}{1-t^2}, \] recall the trigonometric identity \[ \tan\bigl(2\tan^{-1}t\bigr)=\frac{2t}{1-t^2}, \] which implies that, for appropriate domain, \[ y=2\tan^{-1}t. \] Similarly, we are given \[ \sin x = \frac{2t}{1+t^2}. \] It is known that \[ \sin\bigl(2\tan^{-1}t\bigr)=\frac{2t}{1+t^2}, \] so that \[ x=2\tan^{-1}t. \] Therefore, we have: \[ y=2\tan^{-1}t \quad \text{and} \quad x=2\tan^{-1}t, \] which directly shows that \[ y=x. \] Differentiating with respect to \(x\), we obtain \[ \frac{dy}{dx}=1. \]

Solution: We are given \[ y = a^{t+\frac{1}{t}}, \] and we differentiate \(y\) with respect to \(t\). Using the formula \[ \frac{d}{dt}\Bigl(a^{f(t)}\Bigr) = a^{f(t)}\,\text{log}\,a\, f'(t), \] where \[ f(t)=t+\frac{1}{t}, \quad \text{so} \quad f'(t)=1-\frac{1}{t^2}, \] we obtain \[ \frac{dy}{dt}=a^{t+\frac{1}{t}}\,\text{log}\,a\,\left(1-\frac{1}{t^2}\right). \] Next, we differentiate \[ x = \Bigl(t+\frac{1}{t}\Bigr)^a. \] Let \[ h(t)=t+\frac{1}{t}, \] so that \(x=[h(t)]^a\). Then, by the chain rule, \[ \frac{dx}{dt}=a\,[h(t)]^{a-1}\, h'(t), \] and since \[ h'(t)=1-\frac{1}{t^2}, \] it follows that \[ \frac{dx}{dt}=a\left(t+\frac{1}{t}\right)^{a-1}\left(1-\frac{1}{t^2}\right). \] Now, using the chain rule to obtain \(\frac{dy}{dx}\), \[ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}} =\frac{a^{t+\frac{1}{t}}\,\text{log}\,a\,\left(1-\frac{1}{t^2}\right)} {a\left(t+\frac{1}{t}\right)^{a-1}\left(1-\frac{1}{t^2}\right)}. \] Cancel the common factor \(\left(1-\frac{1}{t^2}\right)\) (assuming \(t\neq 0\)): \[ \frac{dy}{dx}=\frac{a^{t+\frac{1}{t}}\,\text{log}\,a}{a\left(t+\frac{1}{t}\right)^{a-1}}. \]

Solution: We write \[ y = A + B, \] where \[ A=\sin^{-1}\!\left(\frac{1}{\sqrt{1+x^2}}\right) \quad\text{and}\quad B=\tan^{-1}\!\left(\frac{\sqrt{1+x^2}-1}{x}\right). \] Step 1. Differentiate \(A\): Let \[ u(x)=\frac{1}{\sqrt{1+x^2}}=(1+x^2)^{-1/2}. \] Then, \[ A=\sin^{-1}\bigl(u(x)\bigr) \quad \Rightarrow \quad A’=\frac{u'(x)}{\sqrt{1-u(x)^2}}. \] Differentiate \(u(x)\): \[ u'(x)=-\frac{1}{2}(1+x^2)^{-3/2}\cdot 2x=-\frac{x}{(1+x^2)^{3/2}}. \] Note that \[ u(x)^2=\frac{1}{1+x^2}\quad\text{so that}\quad 1-u(x)^2=\frac{x^2}{1+x^2}, \] and hence \[ \sqrt{1-u(x)^2}=\frac{|x|}{\sqrt{1+x^2}}. \] Assuming \(x>0\) (a similar argument holds for \(x<0\) with appropriate absolute values), we have \[ \sqrt{1-u(x)^2}=\frac{x}{\sqrt{1+x^2}}. \] Therefore, \[ A'=\frac{-\frac{x}{(1+x^2)^{3/2}}}{\frac{x}{\sqrt{1+x^2}}} = -\frac{1}{1+x^2}. \] Step 2. Differentiate \(B\): Let \[ v(x)=\frac{\sqrt{1+x^2}-1}{x} \quad \text{so that} \quad B=\tan^{-1}\bigl(v(x)\bigr). \] Then, \[ B’=\frac{v'(x)}{1+v(x)^2}. \] Instead of performing tedious algebra, we note the following trigonometric fact: It is known that \[ \tan^{-1}\!\left(\frac{\sqrt{1+x^2}-1}{x}\right)+\cot^{-1}\!\left(\frac{\sqrt{1+x^2}-1}{x}\right) =\frac{\pi}{2}, \] and in a previous result, it was shown that \[ \frac{d}{dx}\left[\cot^{-1}\!\left(\frac{\sqrt{1+x^2}-1}{x}\right)\right] = -\frac{1}{2(1+x^2)}. \] Therefore, differentiating the identity gives \[ B’ = -\left[-\frac{1}{2(1+x^2)}\right] = \frac{1}{2(1+x^2)}. \] Step 3. Combine the derivatives: Now, \[ \frac{dy}{dx}=A’+B’=-\frac{1}{1+x^2}+\frac{1}{2(1+x^2)} =-\frac{2}{2(1+x^2)}+\frac{1}{2(1+x^2)} = -\frac{1}{2(1+x^2)}. \]

Solution: Let \[ x=\sin\theta \quad \text{and} \quad y=\sin\phi. \] Then \[ \sqrt{1-x^2}=\cos\theta \quad \text{and} \quad \sqrt{1-y^2}=\cos\phi. \] The given equation \[ y\sqrt{1-x^2}+x\sqrt{1-y^2}=c \] becomes \[ \sin\phi\,\cos\theta+\sin\theta\,\cos\phi = \sin(\theta+\phi)=c. \] Thus, \(\theta+\phi\) is a constant: \[ \theta+\phi = \sin^{-1}c. \] Differentiating both sides with respect to \(x\) and noting that \(\sin^{-1}c\) is constant, we have \[ \frac{d\theta}{dx}+\frac{d\phi}{dx}=0 \quad\Rightarrow\quad \frac{d\phi}{dx}=-\frac{d\theta}{dx}. \] Since \[ x=\sin\theta \quad \Rightarrow \quad \frac{dx}{dx}=\cos\theta\,\frac{d\theta}{dx}=1, \] it follows that \[ \frac{d\theta}{dx}=\frac{1}{\cos\theta}. \] Now, differentiating \(y=\sin\phi\) with respect to \(x\) gives \[ \frac{dy}{dx}=\cos\phi\,\frac{d\phi}{dx} = \cos\phi\left(-\frac{1}{\cos\theta}\right) = -\frac{\cos\phi}{\cos\theta}. \] Noting that \(\cos\phi=\sqrt{1-\sin^2\phi}=\sqrt{1-y^2}\) and \(\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-x^2}\), we deduce \[ \frac{dy}{dx}=-\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}} = -\sqrt{\frac{1-y^2}{1-x^2}}. \]

Solution: We start with the function \[ f(x)=\tan\left(x^x\right). \] Using the chain rule, \[ f'(x)=\sec^2\left(x^x\right) \cdot \frac{d}{dx}\left(x^x\right). \] To differentiate \(x^x\), express it in exponential form: \[ x^x=e^{x\,\text{log}\, x}. \] Differentiating, we have \[ \frac{d}{dx}\left(e^{x\,\text{log}\, x}\right)=e^{x\,\text{log}\, x}\cdot\frac{d}{dx}\left(x\,\text{log}\, x\right). \] Now, differentiate \(x\,\text{log}\, x\): \[ \frac{d}{dx}\left(x\,\text{log}\, x\right)=\text{log}\, x+1. \] Thus, \[ \frac{d}{dx}\left(x^x\right)=x^x\left(\text{log}\, x+1\right). \] Therefore, the derivative of \(f(x)\) is \[ f'(x)=\sec^2\left(x^x\right)x^x\left(\text{log}\, x+1\right). \]

Solution: Let \[ y=(5x)^{3\cos 2x}. \] Taking full \(\text{log}\) on both sides, we have \[ \text{log}\,y=3\cos 2x\,\text{log}(5x). \] Differentiating both sides with respect to \(x\) using the chain rule, we obtain \[ \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}\Bigl[3\cos 2x\,\text{log}(5x)\Bigr]. \] Using the product rule on the right-hand side, we differentiate: \[ \frac{d}{dx}\Bigl[3\cos 2x\,\text{log}(5x)\Bigr]=3\frac{d}{dx}\bigl[\cos 2x\bigr]\,\text{log}(5x)+3\cos 2x\,\frac{d}{dx}\bigl[\text{log}(5x)\bigr]. \] Now, compute each derivative: \[ \frac{d}{dx}\bigl[\cos 2x\bigr]=-2\sin 2x, \] and \[ \frac{d}{dx}\bigl[\text{log}(5x)\bigr]=\frac{1}{x}, \] since \(\text{log}(5x)=\text{log}5+\text{log}x\) and the derivative of \(\text{log}5\) is zero. Hence, \[ \frac{d}{dx}\Bigl[3\cos 2x\,\text{log}(5x)\Bigr]=3(-2\sin 2x)\,\text{log}(5x)+\frac{3\cos 2x}{x}. \] This simplifies to: \[ \frac{d}{dx}\Bigl[3\cos 2x\,\text{log}(5x)\Bigr]=-\;6\sin 2x\,\text{log}(5x)+\frac{3\cos 2x}{x}. \] Thus, \[ \frac{1}{y}\frac{dy}{dx}=-6\sin 2x\,\text{log}(5x)+\frac{3\cos 2x}{x}. \] Multiplying both sides by \(y=(5x)^{3\cos 2x}\), we obtain: \[ \frac{dy}{dx}=(5x)^{3\cos 2x}\left[\frac{3\cos 2x}{x}-6\sin 2x\,\text{log}(5x)\right]. \]

Solution: Let \[ y=(\sin x-\cos x)^{\sin x-\cos x}. \] Taking the full \(\text{log}\) on both sides, we obtain: \[ \text{log}\,y=(\sin x-\cos x)\,\text{log}(\sin x-\cos x). \] Differentiating both sides with respect to \(x\) using the chain and product rules: \[ \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}\Bigl[(\sin x-\cos x)\,\text{log}(\sin x-\cos x)\Bigr]. \] Let \[ u=\sin x-\cos x. \] Then, \[ \frac{du}{dx}=\cos x+\sin x. \] Notice that: \[ \frac{d}{dx}\left[u\,\text{log}\,u\right]=\frac{du}{dx}\,\text{log}\,u+u\cdot\frac{1}{u}\frac{du}{dx}=\frac{du}{dx}\left(\text{log}\,u+1\right). \] Hence, \[ \frac{1}{y}\frac{dy}{dx}=(\cos x+\sin x)\left(\text{log}(\sin x-\cos x)+1\right). \] Multiplying both sides by \(y\), we have: \[ \frac{dy}{dx}=(\sin x-\cos x)^{\sin x-\cos x}(\cos x+\sin x)\left(\text{log}(\sin x-\cos x)+1\right). \]

Solution: Let \[ f(x)=\left(x+\frac{1}{x}\right)^{x}+x^{\left(1+\frac{1}{x}\right)}. \] We differentiate \(f(x)\) by differentiating each term separately.

First Term:
Let \[ y_1=\left(x+\frac{1}{x}\right)^{x}. \] Taking the full \(\text{log}\) on both sides, we have \[ \text{log}\,y_1=x\,\text{log}\left(x+\frac{1}{x}\right). \] Differentiating both sides with respect to \(x\): \[ \frac{1}{y_1}\frac{dy_1}{dx}=\text{log}\left(x+\frac{1}{x}\right)+x\cdot\frac{1}{\left(x+\frac{1}{x}\right)}\frac{d}{dx}\left(x+\frac{1}{x}\right). \] Now, \[ \frac{d}{dx}\left(x+\frac{1}{x}\right)=1-\frac{1}{x^2}. \] Thus, \[ \frac{1}{y_1}\frac{dy_1}{dx}=\text{log}\left(x+\frac{1}{x}\right)+\frac{x\left(1-\frac{1}{x^2}\right)}{\left(x+\frac{1}{x}\right)}. \] Simplify the fraction: \[ x\left(1-\frac{1}{x^2}\right)=x-\frac{1}{x}, \quad \text{and} \quad x+\frac{1}{x}=\frac{x^2+1}{x}. \] Therefore, \[ \frac{x-\frac{1}{x}}{\frac{x^2+1}{x}}=\frac{x^2-1}{x^2+1}. \] Hence, \[ \frac{dy_1}{dx}=y_1\left[\text{log}\left(x+\frac{1}{x}\right)+\frac{x^2-1}{x^2+1}\right], \] or, \[ \frac{dy_1}{dx}=\left(x+\frac{1}{x}\right)^{x}\left[\text{log}\left(x+\frac{1}{x}\right)+\frac{x^2-1}{x^2+1}\right]. \]

Second Term:
Let \[ y_2=x^{\left(1+\frac{1}{x}\right)}. \] Taking the full \(\text{log}\) on both sides: \[ \text{log}\,y_2=\left(1+\frac{1}{x}\right)\text{log}\,x. \] Differentiating with respect to \(x\): \[ \frac{1}{y_2}\frac{dy_2}{dx}=\frac{d}{dx}\left[\text{log}\,x+\frac{\text{log}\,x}{x}\right]. \] Here, \[ \frac{d}{dx}\left(\text{log}\,x\right)=\frac{1}{x}, \] and \[ \frac{d}{dx}\left(\frac{\text{log}\,x}{x}\right)=\frac{1}{x^2}-\frac{\text{log}\,x}{x^2}. \] Therefore, \[ \frac{1}{y_2}\frac{dy_2}{dx}=\frac{1}{x}+\frac{1}{x^2}-\frac{\text{log}\,x}{x^2}, \] and hence, \[ \frac{dy_2}{dx}=x^{\left(1+\frac{1}{x}\right)}\left[\frac{1}{x}+\frac{1}{x^2}-\frac{\text{log}\,x}{x^2}\right]. \]

Finally, combining both derivatives, we obtain \[ f'(x)=\left(x+\frac{1}{x}\right)^{x}\left[\text{log}\left(x+\frac{1}{x}\right)+\frac{x^2-1}{x^2+1}\right]+x^{\left(1+\frac{1}{x}\right)}\left[\frac{1}{x}+\frac{1}{x^2}-\frac{\text{log}\,x}{x^2}\right]. \]

Solution: Since the infinite exponentiation defines a self-referential expression, we have \[ y=(\cos x)^y. \] Taking the full \(\text{log}\) on both sides: \[ \text{log}\,y = y\,\text{log}(\cos x). \] Differentiate both sides with respect to \(x\). On the left-hand side, using the chain rule: \[ \frac{1}{y}\frac{dy}{dx}. \] On the right-hand side, applying the product rule: \[ \frac{d}{dx}\Bigl[y\,\text{log}(\cos x)\Bigr]=\frac{dy}{dx}\,\text{log}(\cos x) + y\,\frac{d}{dx}\Bigl[\text{log}(\cos x)\Bigr]. \] Note that \[ \frac{d}{dx}\Bigl[\text{log}(\cos x)\Bigr]=\frac{-\sin x}{\cos x}=-\tan x. \] Therefore, the differentiated equation becomes: \[ \frac{1}{y}\frac{dy}{dx}=\frac{dy}{dx}\,\text{log}(\cos x) – y\,\tan x. \] Rearranging the terms to isolate \(\frac{dy}{dx}\): \[ \frac{dy}{dx}\left(\frac{1}{y}-\text{log}(\cos x)\right)=-y\,\tan x. \] Hence, \[ \frac{dy}{dx}=-y\,\tan x \cdot \frac{1}{\frac{1}{y}-\text{log}(\cos x)} = -y\,\tan x \cdot \frac{y}{1-y\,\text{log}(\cos x)}. \] Simplifying, we obtain: \[ \frac{dy}{dx}=-\frac{y^{2}\tan x}{1-y\,\text{log}(\cos x)}. \]

Solution: We start with \[ \sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=6. \] Multiply both sides by \(\sqrt{x}\sqrt{y}\) to eliminate the fractional radicals: \[ \sqrt{x}\sqrt{y}\left(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}\right)=6\sqrt{x}\sqrt{y}. \] Notice that \[ \sqrt{x}\sqrt{y}\sqrt{\frac{x}{y}} = x \quad \text{and} \quad \sqrt{x}\sqrt{y}\sqrt{\frac{y}{x}} = y. \] Thus, the equation becomes \[ x+y=6\sqrt{xy}. \] Taking natural logarithms on both sides gives \[ \ln(x+y)=\ln6+\frac{1}{2}\ln(xy). \] Differentiate both sides with respect to \(x\). The left-hand side, by the chain rule, is \[ \frac{d}{dx}\ln(x+y)=\frac{1}{x+y}\left(1+\frac{dy}{dx}\right). \] The right-hand side differentiates as \[ \frac{d}{dx}\left[\ln6+\frac{1}{2}\ln x+\frac{1}{2}\ln y\right] = \frac{1}{2x}+\frac{1}{2y}\frac{dy}{dx}, \] since \(\ln6\) is constant. Equate the two derivatives: \[ \frac{1}{x+y}\left(1+\frac{dy}{dx}\right)=\frac{1}{2x}+\frac{1}{2y}\frac{dy}{dx}. \] Multiply both sides by \(2(x+y)\) to clear the fractions: \[ 2\left(1+\frac{dy}{dx}\right)=\frac{x+y}{x}+\frac{x+y}{y}\frac{dy}{dx}. \] Rearranging, we have \[ 2+2\frac{dy}{dx}=\frac{x+y}{x}+\frac{x+y}{y}\frac{dy}{dx}. \] Bring the terms involving \(\frac{dy}{dx}\) together: \[ 2\frac{dy}{dx}-\frac{x+y}{y}\frac{dy}{dx} = \frac{x+y}{x}-2. \] Factor out \(\frac{dy}{dx}\) on the left: \[ \frac{dy}{dx}\left(2-\frac{x+y}{y}\right)=\frac{x+y}{x}-2. \] Now, simplify the factors: \[ 2-\frac{x+y}{y}=\frac{2y-(x+y)}{y}=\frac{y-x}{y}, \] and \[ \frac{x+y}{x}-2=\frac{x+y-2x}{x}=\frac{y-x}{x}. \] Therefore, the equation becomes \[ \frac{dy}{dx}\left(\frac{y-x}{y}\right)=\frac{y-x}{x}. \] Cancel the common factor \((y-x)\) (assuming \(y\ne x\)): \[ \frac{dy}{dx}\cdot\frac{1}{y}=\frac{1}{x}. \] Multiplying both sides by \(y\) yields \[ \frac{dy}{dx}=\frac{y}{x}. \]

Solution: We start with \[ y=\tan^{2}(3x-2). \] Differentiate using the chain rule. First, compute the first derivative: \[ \frac{dy}{dx}=2\tan(3x-2)\cdot\frac{d}{dx}\left[\tan(3x-2)\right]. \] Since \[ \frac{d}{dx}\left[\tan(3x-2)\right]=3\sec^{2}(3x-2), \] it follows that \[ \frac{dy}{dx}=2\tan(3x-2)\cdot 3\sec^{2}(3x-2)=6\tan(3x-2)\sec^{2}(3x-2). \] Next, differentiate \( y’ = 6\tan(3x-2)\sec^{2}(3x-2) \) with respect to \(x\): \[ \frac{d^{2}y}{dx^{2}}=6\frac{d}{dx}\left[\tan(3x-2)\sec^{2}(3x-2)\right]. \] Applying the product rule: \[ \frac{d}{dx}\left[\tan(3x-2)\sec^{2}(3x-2)\right] = \frac{d}{dx}\left[\tan(3x-2)\right]\cdot\sec^{2}(3x-2) + \tan(3x-2)\cdot\frac{d}{dx}\left[\sec^{2}(3x-2)\right]. \] We already have: \[ \frac{d}{dx}\left[\tan(3x-2)\right]=3\sec^{2}(3x-2). \] Thus, the first term is: \[ 3\sec^{2}(3x-2)\cdot\sec^{2}(3x-2)=3\sec^{4}(3x-2). \] For the second term, recall that \[ \frac{d}{dx}\left[\sec^{2}(3x-2)\right] =2\sec^{2}(3x-2)\tan(3x-2)\cdot\frac{d}{dx}(3x-2) =2\sec^{2}(3x-2)\tan(3x-2)\cdot3 =6\sec^{2}(3x-2)\tan(3x-2). \] Hence, the second term becomes: \[ \tan(3x-2)\cdot6\sec^{2}(3x-2)\tan(3x-2) =6\tan^{2}(3x-2)\sec^{2}(3x-2). \] Combining these, we have: \[ \frac{d}{dx}\left[\tan(3x-2)\sec^{2}(3x-2)\right] =3\sec^{4}(3x-2)+6\tan^{2}(3x-2)\sec^{2}(3x-2). \] Thus, the second derivative is: \[ \frac{d^{2}y}{dx^{2}} =6\left[3\sec^{4}(3x-2)+6\tan^{2}(3x-2)\sec^{2}(3x-2)\right]. \] Simplifying further: \[ \frac{d^{2}y}{dx^{2}} =18\sec^{4}(3x-2)+36\tan^{2}(3x-2)\sec^{2}(3x-2). \]

Solution: First, note that the denominator factors as \[ x^{2}-3x+2 = (x-1)(x-2). \] It is given (or found by partial fractions) that \[ \frac{x^{2}+2 x-1}{x^{2}-3 x+2} = 1-\frac{2}{x-1}+\frac{7}{x-2}. \] Denote the function by: \[ f(x)=1-\frac{2}{x-1}+\frac{7}{x-2}. \] **Step 1.** Find the first derivative \( f'(x) \). The derivative of the constant 1 is 0. For the term \( -\frac{2}{x-1} \): \[ \frac{d}{dx}\left(-\frac{2}{x-1}\right) = -2 \cdot \frac{d}{dx}\left((x-1)^{-1}\right) = -2 \cdot \left(-\frac{1}{(x-1)^{2}}\right) = \frac{2}{(x-1)^{2}}. \] For the term \( \frac{7}{x-2} \): \[ \frac{d}{dx}\left(\frac{7}{x-2}\right) = 7 \cdot \frac{d}{dx}\left((x-2)^{-1}\right) = 7 \cdot \left(-\frac{1}{(x-2)^{2}}\right) = -\frac{7}{(x-2)^{2}}. \] Thus, the first derivative is: \[ f'(x)=\frac{2}{(x-1)^{2}}-\frac{7}{(x-2)^{2}}. \] **Step 2.** Find the second derivative \( f”(x) \). Differentiate each term of \( f'(x) \) with respect to \(x\). For the term \( \frac{2}{(x-1)^{2}} \): \[ \frac{d}{dx}\left(\frac{2}{(x-1)^{2}}\right) = 2 \cdot \frac{d}{dx}\left((x-1)^{-2}\right) = 2 \cdot \left(-2(x-1)^{-3}\right) = -\frac{4}{(x-1)^{3}}. \] For the term \( -\frac{7}{(x-2)^{2}} \): \[ \frac{d}{dx}\left(-\frac{7}{(x-2)^{2}}\right) = -7 \cdot \frac{d}{dx}\left((x-2)^{-2}\right) = -7 \cdot \left(-2(x-2)^{-3}\right) = \frac{14}{(x-2)^{3}}. \] Therefore, the second derivative is: \[ f”(x)=-\frac{4}{(x-1)^{3}}+\frac{14}{(x-2)^{3}}. \]

Solution: We begin with the function \[ y=\cos\left(3\cos^{-1} x\right). \] Let \(\theta=\cos^{-1}x\) so that \(x=\cos\theta\). Then, using the multiple-angle formula for cosine, \[ \cos 3\theta=4\cos^3\theta-3\cos\theta, \] it follows that \[ y=4\cos^3\theta-3\cos\theta=4x^3-3x. \] Now, differentiate \( y=4x^3-3x \) with respect to \(x\) to obtain the first derivative: \[ \frac{dy}{dx}=\frac{d}{dx}\left(4x^3-3x\right)=12x^2-3. \] Next, differentiate the first derivative to compute the second derivative: \[ \frac{d^2y}{dx^2}=\frac{d}{dx}\left(12x^2-3\right)=24x. \] Thus, we have shown that the second derivative of the function is \[ \frac{d^2y}{dx^2}=24x. \]

Solution: We start with the equation: \[ \sin(x+y)=ky. \] Step 1. Differentiate both sides with respect to \(x\). Differentiating the left-hand side using the chain rule, we have: \[ \frac{d}{dx}\left[\sin(x+y)\right] = \cos(x+y)\cdot\frac{d}{dx}(x+y) = \cos(x+y)\Bigl(1+\frac{dy}{dx}\Bigr) = \cos(x+y)\Bigl(1+y_{1}\Bigr). \] Differentiating the right-hand side: \[ \frac{d}{dx}(ky)= k\,\frac{dy}{dx} = k\,y_{1}. \] Equate the two derivatives: \[ \cos(x+y)(1+y_{1})= k\,y_{1}. \tag{1} \] Step 2. Differentiate equation \((1)\) with respect to \(x\) to obtain the second derivative. Differentiating the left-hand side using the product rule: \[ \frac{d}{dx}\Bigl[\cos(x+y)(1+y_{1})\Bigr] = \frac{d}{dx}\bigl[\cos(x+y)\bigr](1+y_{1}) + \cos(x+y)\frac{d}{dx}(1+y_{1}). \] Compute each derivative separately: – For \(\cos(x+y)\): \[ \frac{d}{dx}\bigl[\cos(x+y)\bigr] = -\sin(x+y)\cdot\frac{d}{dx}(x+y) = -\sin(x+y)(1+y_{1}). \] – For \(1+y_{1}\): \[ \frac{d}{dx}(1+y_{1})=y_{2}. \] Thus, the derivative of the left-hand side becomes: \[ -\sin(x+y)(1+y_{1})^{2}+\cos(x+y)y_{2}. \tag{2} \] The derivative of the right-hand side of \((1)\) is: \[ \frac{d}{dx}\bigl[k\,y_{1}\bigr]=k\,y_{2}. \tag{3} \] Equate equations \((2)\) and \((3)\): \[ -\sin(x+y)(1+y_{1})^{2}+\cos(x+y)y_{2}= k\,y_{2}. \tag{4} \] Rearranging \((4)\), we obtain: \[ \cos(x+y)y_{2}-k\,y_{2}= \sin(x+y)(1+y_{1})^{2}, \] or equivalently, \[ y_{2}\Bigl[\cos(x+y)-k\Bigr]= \sin(x+y)(1+y_{1})^{2}. \tag{5} \] Step 3. Eliminate the trigonometric functions using the original relation. From the original equation, we have: \[ \sin(x+y)= ky. \tag{6} \] Also, from equation \((1)\), \[ \cos(x+y)(1+y_{1})= k\,y_{1} \quad\Longrightarrow\quad \cos(x+y)=\frac{k\,y_{1}}{1+y_{1}}. \tag{7} \] Substitute \((7)\) into \((5)\): \[ y_{2}\left[\frac{k\,y_{1}}{1+y_{1}}-k\right] = \sin(x+y)(1+y_{1})^{2}. \] Simplify the bracket on the left: \[ \frac{k\,y_{1}}{1+y_{1}}-k = k\left(\frac{y_{1}- (1+y_{1})}{1+y_{1}}\right) = -\frac{k}{1+y_{1}}. \] Hence, equation \((5)\) becomes: \[ y_{2}\left(-\frac{k}{1+y_{1}}\right) = \sin(x+y)(1+y_{1})^{2}. \] Now substitute \((6)\) for \(\sin(x+y)\): \[ -\frac{k\,y_{2}}{1+y_{1}}= k\,y(1+y_{1})^{2}. \] Provided \(k\neq0\), we can cancel \(k\) from both sides: \[ -\frac{y_{2}}{1+y_{1}}= y(1+y_{1})^{2}. \] Multiply both sides by \(1+y_{1}\) to obtain: \[ -y_{2}= y(1+y_{1})^{3}. \] Rearranging gives: \[ y_{2}+ y(1+y_{1})^{3}=0. \]

Solution: We are given: \[ xy=\sin x. \] Differentiate both sides with respect to \(x\): \[ \frac{d}{dx}(xy)=\frac{d}{dx}(\sin x). \] Applying the product rule on the left-hand side: \[ y + x\,\frac{dy}{dx} = \cos x. \] This can be written as: \[ y + xy_{1}=\cos x,\quad \text{where } y_{1}=\frac{dy}{dx}. \] Differentiate the resulting equation with respect to \(x\) again: \[ \frac{d}{dx}\left(y+xy_{1}\right)=\frac{d}{dx}(\cos x). \] The derivative of the left-hand side (applying the product rule to \(xy_{1}\)): \[ \frac{dy}{dx}+ \left( y_{1}+ x\,\frac{d y_{1}}{dx}\right) = y_{1}+ y_{1}+ x\,y_{2} =2y_{1}+xy_{2}, \] where \(y_{2}=\frac{d^{2}y}{dx^{2}}\). The derivative of the right-hand side is: \[ -\sin x. \] Therefore, we have: \[ 2y_{1}+xy_{2}=-\sin x. \tag{1} \] Recall that the original equation is: \[ xy=\sin x \quad \Longrightarrow \quad \sin x=xy. \tag{2} \] Substituting \((2)\) into \((1)\) gives: \[ 2y_{1}+xy_{2}=-xy. \] Dividing both sides by \(x\) (with \(x\neq 0\)): \[ y_{2}+\frac{2}{x}y_{1}=-y. \] Rearranging, we obtain: \[ y_{2}+\frac{2}{x}y_{1}+y=0. \] This is the required result.

Solution: We are given: \[ x = (a + bt)e^{-nt}. \] Step 1: First derivative Using the product rule: \[ \frac{dx}{dt} = \frac{d}{dt}[(a + bt)e^{-nt}] = (a + bt)\frac{d}{dt}(e^{-nt}) + e^{-nt}\frac{d}{dt}(a + bt). \] Now compute derivatives: \[ \frac{d}{dt}(a + bt) = b, \quad \frac{d}{dt}(e^{-nt}) = -n e^{-nt}. \] So, \[ \frac{dx}{dt} = (a + bt)(-n e^{-nt}) + b e^{-nt} = e^{-nt} (b – n(a + bt)). \] Step 2: Second derivative Differentiate again: \[ \frac{d^{2}x}{dt^{2}} = \frac{d}{dt} \left[ e^{-nt}(b – n(a + bt)) \right]. \] Apply the product rule: \[ \frac{d^{2}x}{dt^{2}} = (-n e^{-nt})(b – n(a + bt)) + e^{-nt}(-n b). \] Simplify: \[ \frac{d^{2}x}{dt^{2}} = -n e^{-nt} (b – n(a + bt)) – n b e^{-nt} = -n e^{-nt} (2b – n(a + bt)). \] Step 3: Evaluate the given expression We are to prove: \[ \frac{d^{2}x}{dt^{2}} + 2n \frac{dx}{dt} + n^{2} x = 0. \] Substitute all: \[ \frac{d^{2}x}{dt^{2}} = -n e^{-nt} (2b – n(a + bt)), \] \[ \frac{dx}{dt} = e^{-nt} (b – n(a + bt)), \] \[ x = (a + bt) e^{-nt}. \] Plug into the expression: \[ \frac{d^{2}x}{dt^{2}} + 2n \frac{dx}{dt} + n^{2} x = -n e^{-nt} (2b – n(a + bt)) + 2n e^{-nt} (b – n(a + bt)) + n^{2}(a + bt) e^{-nt}. \] Factor out \( e^{-nt} \): \[ = e^{-nt} \left[ -n(2b – n(a + bt)) + 2n(b – n(a + bt)) + n^{2}(a + bt) \right]. \] Now simplify the expression in brackets: First expand: \[ -n(2b – n(a + bt)) = -2nb + n^{2}(a + bt), \] \[ 2n(b – n(a + bt)) = 2nb – 2n^{2}(a + bt), \] \[ n^{2}(a + bt) = n^{2}(a + bt). \] Now add them: \[ (-2nb + n^{2}(a + bt)) + (2nb – 2n^{2}(a + bt)) + n^{2}(a + bt) = 0. \] Therefore: \[ \frac{d^{2}x}{dt^{2}} + 2n \frac{dx}{dt} + n^{2} x = e^{-nt} \cdot 0 = 0. \]

Solution: We start with \[ x=(a+b\,t)e^{-n\,t}. \] Differentiate using the product rule: \[ \frac{dx}{dt}=\frac{d}{dt}\left[(a+b\,t)e^{-n\,t}\right] = (a+b\,t)’\,e^{-n\,t}+ (a+b\,t)\,\frac{d}{dt}(e^{-n\,t}). \] Since \[ (a+b\,t)’=b \quad \text{and} \quad \frac{d}{dt}(e^{-n\,t})=-n\,e^{-n\,t}, \] we have \[ \frac{dx}{dt}=b\,e^{-n\,t}-n(a+b\,t)e^{-n\,t} =e^{-n\,t}\Bigl(b-n(a+b\,t)\Bigr). \] Next, differentiate \(\frac{dx}{dt}\): \[ \frac{d^{2}x}{dt^{2}}=\frac{d}{dt}\left[e^{-n\,t}\Bigl(b-n(a+b\,t)\Bigr)\right]. \] Let \[ u(t)=e^{-n\,t}\quad \text{and} \quad v(t)=b-n(a+b\,t). \] Then, \[ u'(t)=-n\,e^{-n\,t}\quad \text{and} \quad v'(t)=-n\,b. \] Applying the product rule yields: \[ \frac{d^{2}x}{dt^{2}}=u'(t)v(t)+u(t)v'(t) =-n\,e^{-n\,t}\Bigl(b-n(a+b\,t)\Bigr)-n\,b\,e^{-n\,t}. \] Factor \(e^{-n\,t}\): \[ \frac{d^{2}x}{dt^{2}} =e^{-n\,t}\Bigl[-n\Bigl(b-n(a+b\,t)\Bigr)-n\,b\Bigr]. \] Expand the expression inside the bracket: \[ -n\Bigl(b-n(a+b\,t)\Bigr)-n\,b =-n\,b+n^{2}(a+b\,t)-n\,b =n^{2}a+n^{2}b\,t-2n\,b. \] Thus, \[ \frac{d^{2}x}{dt^{2}}=e^{-n\,t}\Bigl(n^{2}a+n^{2}b\,t-2n\,b\Bigr). \] Now, consider \[ \frac{d^{2}x}{dt^{2}}+2n\,\frac{dx}{dt}+n^{2}x. \] Substitute the expressions found: \[ \frac{d^{2}x}{dt^{2}}=e^{-n\,t}\Bigl(n^{2}a+n^{2}b\,t-2n\,b\Bigr), \] \[ \frac{dx}{dt}=e^{-n\,t}\Bigl(b-n(a+b\,t)\Bigr), \] \[ x=(a+b\,t)e^{-n\,t}. \] Therefore, \[ \begin{aligned} \frac{d^{2}x}{dt^{2}}+2n\,\frac{dx}{dt}+n^{2}x &= e^{-n\,t}\Bigl(n^{2}a+n^{2}b\,t-2n\,b\Bigr) +2n\,e^{-n\,t}\Bigl(b-n(a+b\,t)\Bigr)\\[1mm] &\quad + n^{2}(a+b\,t)e^{-n\,t}. \end{aligned} \] Factor \(e^{-n\,t}\): \[ e^{-n\,t}\Bigl[\underbrace{n^{2}a+n^{2}b\,t-2n\,b}_{\text{from } \frac{d^{2}x}{dt^{2}}} +\underbrace{2n\,(b-n\,a- n\,b\,t)}_{\text{from } 2n\,\frac{dx}{dt}} +\underbrace{n^{2}(a+b\,t)}_{\text{from } n^{2}x}\Bigr]. \] Combine like terms: \[ \begin{aligned} \text{Coefficient of } a: & \quad n^{2}a-2n^{2}a+n^{2}a=0,\\[1mm] \text{Coefficient of } t: & \quad n^{2}b\,t-2n^{2}b\,t+n^{2}b\,t=0,\\[1mm] \text{Constants:} & \quad -2n\,b+2n\,b=0. \end{aligned} \] Hence, \[ \frac{d^{2}x}{dt^{2}}+2n\,\frac{dx}{dt}+n^{2}x = e^{-n\,t}\cdot 0 = 0. \]

Solution: We start by differentiating \(y\) and \(x\) with respect to the parameter \(t\). First, for \[ y=\frac{3at}{1+t}, \] applying the quotient rule, \[ \frac{dy}{dt}=\frac{(3a)(1+t)-3at\cdot(1)}{(1+t)^{2}}=\frac{3a(1+t-t)}{(1+t)^{2}}=\frac{3a}{(1+t)^{2}}. \] Next, for \[ x=\frac{2at^{2}}{1+t}, \] again using the quotient rule, \[ \frac{dx}{dt}=\frac{(1+t)\cdot(4at)-2at^{2}\cdot(1)}{(1+t)^{2}} =\frac{4at(1+t)-2at^{2}}{(1+t)^{2}}. \] Factor \(2at\) from the numerator: \[ \frac{dx}{dt}=\frac{2at\Bigl[2(1+t)-t\Bigr]}{(1+t)^{2}} =\frac{2at(2+t)}{(1+t)^{2}}. \] Thus, the first derivative \(\frac{dy}{dx}\) is given by \[ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}} =\frac{\displaystyle \frac{3a}{(1+t)^{2}}}{\displaystyle \frac{2at(2+t)}{(1+t)^{2}}} =\frac{3}{2t(2+t)}. \] To find the second derivative \(\frac{d^{2}y}{dx^{2}}\), we differentiate \(\frac{dy}{dx}\) with respect to \(t\) and then divide by \(\frac{dx}{dt}\). Write \[ \frac{dy}{dx}=\frac{3}{2t(2+t)}. \] Let \[ u(t)=\frac{3}{2t(2+t)}. \] Differentiating \(u(t)\) with respect to \(t\): \[ u'(t)=-\frac{3}{\Bigl[2t(2+t)\Bigr]^{2}}\cdot\frac{d}{dt}[2t(2+t)]. \] First, compute: \[ 2t(2+t)=4t+2t^{2}. \] Its derivative is: \[ \frac{d}{dt}[4t+2t^{2}]=4+4t=4(t+1). \] Thus, \[ u'(t)=-\frac{3\cdot 4(t+1)}{\Bigl[2t(2+t)\Bigr]^{2}} =-\frac{12(t+1)}{4t^{2}(2+t)^{2}} =-\frac{3(t+1)}{t^{2}(2+t)^{2}}. \] Now, using the relation: \[ \frac{d^{2}y}{dx^{2}}=\frac{u'(t)}{\frac{dx}{dt}}, \] and substituting \(\frac{dx}{dt}=\frac{2at(2+t)}{(1+t)^{2}}\), we have: \[ \frac{d^{2}y}{dx^{2}}=\frac{-\frac{3(t+1)}{t^{2}(2+t)^{2}}}{\frac{2at(2+t)}{(1+t)^{2}}} =-\frac{3(t+1)}{t^{2}(2+t)^{2}}\cdot\frac{(1+t)^{2}}{2at(2+t)}. \] Simplify by combining like factors: \[ \frac{d^{2}y}{dx^{2}}=-\frac{3(t+1)^{3}}{2at^{3}(2+t)^{3}}. \]

Solution: We first rewrite the function by using a trigonometric identity. Note that \[ f(x)=2\cos\left(2x-\frac{\pi}{2}\right)=2\sin2x. \] Since \( \sin2x \) is continuous and differentiable for all \( x \), the function is continuous on \([0,\pi]\) and differentiable on \((0,\pi)\). Next, we check the values at the endpoints: \[ f(0)=2\sin(0)=0,\quad f(\pi)=2\sin(2\pi)=0. \] Since \( f(0)=f(\pi) \), the hypothesis of Rolle’s theorem is satisfied. Now, we find \( c \) in \( (0,\pi) \) such that \( f'(c)=0 \). Differentiate: \[ f'(x)=\frac{d}{dx}[2\sin2x]=4\cos2x. \] Setting \( f'(x)=0 \): \[ 4\cos2x=0 \quad \Longrightarrow \quad \cos2x=0. \] The general solution for \( \cos2x=0 \) is \[ 2x=\frac{\pi}{2}+k\pi,\quad k\in \mathbb{Z}, \] leading to \[ x=\frac{\pi}{4}+\frac{k\pi}{2}. \] For \( x \) in \( (0,\pi) \), the valid solutions are: \[ x=\frac{\pi}{4} \quad (k=0) \quad \text{and} \quad x=\frac{3\pi}{4} \quad (k=1). \]

Solution: The Lagrange’s Mean Value Theorem states that if \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then there exists at least one number \(c\) in \((a,b)\) such that \[ f'(c)=\frac{f(b)-f(a)}{b-a}. \] For \(f(x)=\text{log}\, x\) on the interval \([1,2e]\): \[ f(1)=\text{log}\,1=0 \quad \text{and} \quad f(2e)=\text{log}\,(2e)=\text{log}\,2+\text{log}\,e=\text{log}\,2+1. \] Thus, \[ \frac{f(2e)-f(1)}{2e-1}=\frac{(\text{log}\,2+1)-0}{2e-1}=\frac{\text{log}\,2+1}{2e-1}. \] Since \[ f'(x)=\frac{d}{dx}\left(\text{log}\, x\right)=\frac{1}{x}, \] we set \[ f'(c)=\frac{1}{c}=\frac{\text{log}\,2+1}{2e-1}. \] Solving for \(c\): \[ c=\frac{2e-1}{\text{log}\,2+1}. \]