Ex 5.10 – Continuity and Differentiability | ML Aggarwal Class 12 Solutions

Solution: We differentiate each term individually: \[ \frac{d}{dx}\left(e^{x}\right) = e^{x}, \] \[ \frac{d}{dx}\left(3\sin x\right) = 3\cos x. \] Hence, by the sum rule: \[ \frac{d}{dx}\left(e^{x}+3\sin x\right) = e^{x}+3\cos x. \]

Solution: We differentiate term-by-term: \[ \frac{d}{dx}\left(10^{x}\right)=10^{x}\text{ log }10, \] \[ \frac{d}{dx}\left(\frac{1}{3}e^{x}\right)=\frac{1}{3}e^{x}, \] \[ \frac{d}{dx}\left(-2\log x\right)=-2\cdot\frac{1}{x}=-\frac{2}{x}. \] Therefore, by the sum rule, \[ \frac{d}{dx}\left(10^{x}+\frac{1}{3}e^{x}-2\log x\right)=10^{x}\text{ log }10+\frac{1}{3}e^{x}-\frac{2}{x}. \]

Solution: We apply the chain rule. For the exponential function, \[ \frac{d}{dx}\left(e^{-x}\right) = e^{-x}\cdot\frac{d}{dx}(-x). \] Since, \[ \frac{d}{dx}(-x) = -1, \] it follows that \[ \frac{d}{dx}\left(e^{-x}\right) = -e^{-x}. \]

Solution: Let \[ u=\log x. \] Then the given function becomes \[ \sin u. \] Differentiating using the chain rule: \[ \frac{d}{dx}\left(\sin u\right)=\cos u\cdot\frac{du}{dx}. \] Since \[ \frac{du}{dx}=\frac{1}{x}, \] it follows that \[ \frac{d}{dx}\left(\sin\bigl(\log x\bigr)\right)=\cos\bigl(\log x\bigr)\cdot\frac{1}{x}=\frac{\cos\bigl(\log x\bigr)}{x}. \]

Solution: Using the chain rule, we have: \[ \frac{d}{dx}\left(e^{\cos x}\right)=e^{\cos x}\cdot \frac{d}{dx}(\cos x). \] Since, \[ \frac{d}{dx}(\cos x)=-\sin x, \] it follows that: \[ \frac{d}{dx}\left(e^{\cos x}\right)=-e^{\cos x}\sin x. \]

Solution:

Let \[ y = e^{\sin^{-1} x}. \]

To differentiate \( y \) with respect to \( x \), we apply the chain rule. The outer function is \( e^u \) with \( u = \sin^{-1} x \), and its derivative is \( e^u \). The inner function is \( u = \sin^{-1} x \), whose derivative is \[ \frac{d}{dx} (\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}}. \]

Thus, by the chain rule, the derivative is: \[ \frac{dy}{dx} = e^{\sin^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}. \]

Solution: Let \[ u=x^3, \] so that \[ \frac{du}{dx}=3x^2. \] Applying the chain rule: \[ \frac{d}{dx}\left(e^{x^3}\right)=e^{x^3}\cdot\frac{du}{dx}=3x^2e^{x^3}. \]

Solution: We first rewrite the function in an exponential form: \[ \sqrt{e^{\sqrt{x}}} = e^{\frac{1}{2}\sqrt{x}}. \] Let \[ u=\sqrt{x}, \] then \[ \frac{du}{dx}=\frac{1}{2\sqrt{x}}. \] The function becomes: \[ e^{\frac{1}{2}u}. \] Differentiating with respect to \( u \): \[ \frac{d}{du}\left(e^{\frac{1}{2}u}\right)=\frac{1}{2}e^{\frac{1}{2}u}. \] Now, applying the chain rule: \[ \frac{d}{dx}\left(e^{\frac{1}{2}\sqrt{x}}\right)=\frac{1}{2}e^{\frac{1}{2}\sqrt{x}}\cdot\frac{1}{2\sqrt{x}}=\frac{e^{\frac{1}{2}\sqrt{x}}}{4\sqrt{x}}. \] This expression can be rewritten as: \[ \frac{e^{\sqrt{x}}}{4\sqrt{x}\sqrt{e^{\sqrt{x}}}}, \] since \[ e^{\frac{1}{2}\sqrt{x}}=\sqrt{e^{\sqrt{x}}}. \]

Solution: Let \[ u=\log x. \] Then the function becomes \[ \log u. \] Differentiating with respect to \( u \): \[ \frac{d}{du}\bigl(\log u\bigr)=\frac{1}{u}. \] Also, \[ \frac{du}{dx}=\frac{1}{x}. \] By the chain rule: \[ \frac{d}{dx}\bigl(\log(\log x)\bigr)=\frac{1}{u}\cdot\frac{1}{x}=\frac{1}{x\log x}. \]

Solution: First, express the logarithm in base \( 7 \) in terms of the natural logarithm: \[ \log_{7}(\log x)=\frac{\log(\log x)}{\log 7}. \] Since \(\frac{1}{\log 7}\) is a constant, we differentiate: \[ \frac{d}{dx}\left(\log_{7}(\log x)\right)=\frac{1}{\log 7}\cdot\frac{d}{dx}\bigl(\log(\log x)\bigr). \] Let \[ u=\log x,\quad\text{so that}\quad \frac{du}{dx}=\frac{1}{x}. \] Then, \[ \frac{d}{dx}\bigl(\log u\bigr)=\frac{1}{u}\cdot\frac{du}{dx}=\frac{1}{\log x}\cdot\frac{1}{x}=\frac{1}{x\log x}. \] Therefore, \[ \frac{d}{dx}\left(\log_{7}(\log x)\right)=\frac{1}{\log 7}\cdot\frac{1}{x\log x}=\frac{1}{x\log 7\log x}. \]

Solution: Let \[ u=\cos\bigl(e^{x}\bigr). \] Then \[ \frac{d}{dx}\bigl(\log u\bigr)=\frac{1}{u}\cdot\frac{du}{dx}. \] Now, differentiating \( u \) using the chain rule: \[ \frac{du}{dx}=-\sin\bigl(e^{x}\bigr)\cdot\frac{d}{dx}\bigl(e^{x}\bigr)=-\sin\bigl(e^{x}\bigr)e^{x}. \] Therefore, \[ \frac{d}{dx}\left(\log\Bigl(\cos\bigl(e^{x}\bigr)\Bigr)\right) =\frac{-\sin\bigl(e^{x}\bigr)e^{x}}{\cos\bigl(e^{x}\bigr)} =-e^{x}\tan\bigl(e^{x}\bigr). \]

Solution: Let \( y = \tan^{-1}\left(e^{\sin x}\right) \). Using the chain rule, we have: \[ \frac{dy}{dx} = \frac{1}{1+\left(e^{\sin x}\right)^2} \cdot \frac{d}{dx}\left(e^{\sin x}\right). \] Since \[ \frac{d}{dx}\left(e^{\sin x}\right) = e^{\sin x} \cdot \cos x, \] it follows that: \[ \frac{dy}{dx} = \frac{e^{\sin x} \cdot \cos x}{1+e^{2\sin x}}. \]

Solution: The domain of \( f(x)=\log\bigl(2x-1\bigr) \) requires: \[ 2x-1>0\quad\Longrightarrow\quad x>\frac{1}{2}. \] Since \( x=0 \) does not satisfy \( x>\frac{1}{2} \), the function is not even defined at \( x=0 \). Hence, \( f(x) \) cannot be differentiable at \( x=0 \).

Solution: Given \( f(x)=\log(\log x) \), we differentiate using the chain rule. Let \( u=\log x \), then \( f(x)=\log u \) and: \[ \frac{df}{dx}=\frac{1}{u}\cdot \frac{du}{dx}=\frac{1}{\log x}\cdot \frac{1}{x}=\frac{1}{x\log x}. \] Evaluating at \( x=e \), note that \( \log e = 1 \): \[ f'(e)=\frac{1}{e\cdot 1}=\frac{1}{e}. \]

Solution: Differentiate \( y = \log (\tan x) \) using the chain rule: \[ \frac{dy}{dx} = \frac{1}{\tan x} \cdot \frac{d}{dx} (\tan x). \] Since \[ \frac{d}{dx} (\tan x) = \sec^2 x, \] it follows that \[ \frac{dy}{dx} = \frac{\sec^2 x}{\tan x}. \] Write \(\tan x\) and \(\sec^2 x\) in terms of sine and cosine: \[ \tan x = \frac{\sin x}{\cos x} \quad \text{and} \quad \sec^2 x = \frac{1}{\cos^2 x}. \] Thus, \[ \frac{dy}{dx} = \frac{1/\cos^2 x}{\sin x/\cos x} = \frac{1}{\sin x\,\cos x}. \] Using the double-angle identity \[ \sin 2x = 2 \sin x\,\cos x, \] we get \[ \frac{dy}{dx} = \frac{2}{\sin 2x} = 2\, \mathrm{cosec}\,2x. \]

Solution: Let \[ y=\log f(e^x). \] Differentiate \( y \) with respect to \( x \) using the chain rule: \[ \frac{dy}{dx}=\frac{1}{f(e^x)}\cdot f'(e^x)\cdot \frac{d}{dx}(e^x). \] Since \[ \frac{d}{dx}(e^x)=e^x, \] we obtain: \[ \frac{dy}{dx}=\frac{f'(e^x)e^x}{f(e^x)}. \] Now, evaluate at \( x=0 \): \[ e^0=1. \] Therefore: \[ \frac{dy}{dx}\Big|_{x=0}=\frac{f'(1)\cdot 1}{f(1)}=\frac{2}{4}=\frac{1}{2}. \]

Solution: We have: \[ f(x)=e^x\,g(x). \] Differentiate \( f(x) \) using the product rule: \[ f'(x)=\frac{d}{dx}\bigl(e^x\bigr)g(x)+e^x\,\frac{d}{dx}\bigl(g(x)\bigr)=e^x\,g(x)+e^x\,g'(x). \] This simplifies to: \[ f'(x)=e^x\left(g(x)+g'(x)\right). \] Evaluating at \( x=0 \): \[ f'(0)=e^0\left(g(0)+g'(0)\right)=1\cdot (2+1)=3. \]

Solution: Let \[ y=\frac{e^x}{\sin x}. \] Using the quotient rule: \[ \frac{dy}{dx}=\frac{\sin x\,\frac{d}{dx}(e^x)-e^x\,\frac{d}{dx}(\sin x)}{(\sin x)^2}. \] Since \[ \frac{d}{dx}(e^x)=e^x \quad \text{and} \quad \frac{d}{dx}(\sin x)=\cos x, \] we have: \[ \frac{dy}{dx}=\frac{\sin x\,e^x-e^x\,\cos x}{\sin^2 x}=\frac{e^x(\sin x-\cos x)}{\sin^2 x}. \]

Solution: Let \[ y=\frac{e^x}{1+\sin x}. \] Using the quotient rule: \[ \frac{dy}{dx}=\frac{(1+\sin x)\,\frac{d}{dx}(e^x)-e^x\,\frac{d}{dx}(1+\sin x)}{(1+\sin x)^2}. \] Since \[ \frac{d}{dx}(e^x)=e^x \quad \text{and} \quad \frac{d}{dx}(1+\sin x)=\cos x, \] we have: \[ \frac{dy}{dx}=\frac{(1+\sin x)e^x-e^x\cos x}{(1+\sin x)^2}=\frac{e^x(1+\sin x-\cos x)}{(1+\sin x)^2}. \]

Solution: Let \[ y = 2x \tan^{-1}x – \log\left(1+x^2\right). \] Differentiate term by term: For the first term, \( 2x \tan^{-1}x \), use the product rule: \[ \frac{d}{dx}\left(2x \tan^{-1}x\right) = 2 \tan^{-1}x + 2x \cdot \frac{1}{1+x^2}. \] For the second term, \( \log\left(1+x^2\right) \), use the chain rule: \[ \frac{d}{dx}\left[\log\left(1+x^2\right)\right] = \frac{1}{1+x^2} \cdot 2x = \frac{2x}{1+x^2}. \] Thus, combining the derivatives: \[ \frac{dy}{dx} = \left[2 \tan^{-1}x + \frac{2x}{1+x^2}\right] – \frac{2x}{1+x^2}. \] Notice that the terms \(\frac{2x}{1+x^2}\) cancel out, yielding: \[ \frac{dy}{dx} = 2 \tan^{-1}x. \]

Solution: Let \[ y=\frac{\cos x}{\log x}. \] Using the quotient rule: \[ \frac{dy}{dx}=\frac{(\log x)\cdot\frac{d}{dx}(\cos x)-\cos x\cdot\frac{d}{dx}(\log x)}{(\log x)^2}. \] We compute the derivatives: \[ \frac{d}{dx}(\cos x)=-\sin x \quad \text{and} \quad \frac{d}{dx}(\log x)=\frac{1}{x}. \] Thus, \[ \frac{dy}{dx}=\frac{(\log x)(-\sin x)-\cos x\left(\frac{1}{x}\right)}{(\log x)^2}=\frac{-\sin x\log x-\frac{\cos x}{x}}{(\log x)^2}. \]

Solution: Let \[ y = e^{\sec^2 x} + 3 \cos^{-1}x. \] Differentiate the first term using the chain rule: \[ \frac{d}{dx}\left(e^{\sec^2 x}\right) = e^{\sec^2 x} \cdot \frac{d}{dx}\left(\sec^2 x\right). \] Since \[ \frac{d}{dx}\left(\sec^2 x\right) = 2\sec x \cdot \frac{d}{dx}\left(\sec x\right) = 2\sec x \cdot \sec x \tan x = 2\sec^2 x \tan x, \] we have \[ \frac{d}{dx}\left(e^{\sec^2 x}\right) = 2\sec^2 x \tan x\, e^{\sec^2 x}. \] Next, differentiate the second term: \[ \frac{d}{dx}\left(3\cos^{-1}x\right) = 3\cdot\frac{d}{dx}\left(\cos^{-1}x\right). \] Recall that \[ \frac{d}{dx}\left(\cos^{-1}x\right) = -\frac{1}{\sqrt{1-x^2}}, \] hence, \[ \frac{d}{dx}\left(3\cos^{-1}x\right) = -\frac{3}{\sqrt{1-x^2}}. \] Combining the two results: \[ \frac{dy}{dx} = 2\sec^2 x \tan x\, e^{\sec^2 x} – \frac{3}{\sqrt{1-x^2}}. \]

Solution: We differentiate the function term-by-term. \[ \frac{d}{dx}\left(e^x\right)=e^x. \] For the general term \( e^{x^n} \) where \( n \) is a positive integer, using the chain rule: \[ \frac{d}{dx}\left(e^{x^n}\right)=e^{x^n}\cdot\frac{d}{dx}\left(x^n\right)=e^{x^n}\cdot n\,x^{n-1}. \] Applying this to each term: \[ \begin{aligned} \frac{d}{dx}\left(e^x\right)&=e^x,\\[1mm] \frac{d}{dx}\left(e^{x^2}\right)&=2x\,e^{x^2},\\[1mm] \frac{d}{dx}\left(e^{x^3}\right)&=3x^2\,e^{x^3},\\[1mm] \frac{d}{dx}\left(e^{x^4}\right)&=4x^3\,e^{x^4},\\[1mm] \frac{d}{dx}\left(e^{x^5}\right)&=5x^4\,e^{x^5}. \end{aligned} \] Therefore, the derivative of the sum is: \[ \frac{dy}{dx}=e^x+2x\,e^{x^2}+3x^2\,e^{x^3}+4x^3\,e^{x^4}+5x^4\,e^{x^5}. \]

Solution: Write \[ y = \frac{u}{v},\quad \text{where} \quad u=5^{x}\,\log x,\quad v=x^{2}+1. \] Differentiate \( u \) using the product and chain rules: \[ u’ = \frac{d}{dx}\left(5^{x}\right)\,\log x + 5^{x}\,\frac{d}{dx}\left(\log x\right) = 5^{x}\,\log 5\,\log x + \frac{5^{x}}{x}. \] Also, differentiate \( v \): \[ v’ = 2x. \] Now, applying the quotient rule: \[ \frac{dy}{dx} = \frac{u’v – uv’}{v^{2}} = \frac{5^{x}\left(\log 5\,\log x + \frac{1}{x}\right)(x^{2}+1) – 5^{x}\,\log x\,(2x)}{(x^{2}+1)^{2}}. \] Multiply the term \( \frac{1}{x}(x^{2}+1) \) to get: \[ \frac{dy}{dx} = \frac{5^{x}\left[(x^{2}+1)\,\log 5\,\log x + \frac{x^{2}+1}{x} – 2x\,\log x\right]}{(x^{2}+1)^{2}}. \] Express the terms over a common denominator \( x \): \[ \frac{dy}{dx} = \frac{5^{x}}{x\,(x^{2}+1)^{2}} \left[(x^{2}+1)(x\,\log 5\,\log x + 1) – 2x^{2}\,\log x\right]. \]

Solution:

We start with \[ y = \cos\left(\log x + e^x\right). \] Let \[ u = \log x + e^x. \] Then, \( y = \cos u \). Applying the chain rule, we have \[ \frac{dy}{dx} = -\sin u \cdot \frac{du}{dx}. \]

Next, differentiate \( u \) with respect to \( x \): \[ \frac{du}{dx} = \frac{d}{dx} (\log x) + \frac{d}{dx} (e^x) = \frac{1}{x} + e^x. \]

Substituting back into the derivative of \( y \): \[ \frac{dy}{dx} = -\sin\left(\log x + e^x\right) \left(\frac{1}{x} + e^x\right). \]

This can be rewritten as: \[ \frac{dy}{dx} = -\frac{1 + x e^x}{x} \sin\left(\log x + e^x\right). \]

Solution:

We start by rewriting the expression using the logarithm property: \[ y = \log\left(x+\sqrt{x^2-a^2}\right) – \log\left(x-\sqrt{x^2-a^2}\right). \]

Differentiate the first term: \[ \frac{d}{dx}\left[\log\left(x+\sqrt{x^2-a^2}\right)\right] = \frac{1}{x+\sqrt{x^2-a^2}} \cdot \left(1+\frac{x}{\sqrt{x^2-a^2}}\right). \] Notice that \[ 1+\frac{x}{\sqrt{x^2-a^2}} = \frac{\sqrt{x^2-a^2}+x}{\sqrt{x^2-a^2}}. \] Thus, the derivative becomes \[ \frac{\sqrt{x^2-a^2}+x}{\left(x+\sqrt{x^2-a^2}\right)\sqrt{x^2-a^2}} = \frac{1}{\sqrt{x^2-a^2}}. \]

Similarly, differentiate the second term: \[ \frac{d}{dx}\left[\log\left(x-\sqrt{x^2-a^2}\right)\right] = \frac{1}{x-\sqrt{x^2-a^2}} \cdot \left(1-\frac{x}{\sqrt{x^2-a^2}}\right). \] Since \[ 1-\frac{x}{\sqrt{x^2-a^2}} = \frac{\sqrt{x^2-a^2}-x}{\sqrt{x^2-a^2}}, \] the derivative becomes \[ \frac{\sqrt{x^2-a^2}-x}{\left(x-\sqrt{x^2-a^2}\right)\sqrt{x^2-a^2}} = -\frac{1}{\sqrt{x^2-a^2}}. \]

Now, subtract the second derivative from the first: \[ \frac{dy}{dx} = \frac{1}{\sqrt{x^2-a^2}} – \left(-\frac{1}{\sqrt{x^2-a^2}}\right) = \frac{1}{\sqrt{x^2-a^2}} + \frac{1}{\sqrt{x^2-a^2}} = \frac{2}{\sqrt{x^2-a^2}}. \]

Solution:

First, rewrite the given function as: \[ y = \log\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right) = \frac{1}{2}\log\left(\frac{1-\cos x}{1+\cos x}\right). \]

Notice that the expression inside the logarithm can be simplified using the trigonometric identity: \[ \frac{1-\cos x}{1+\cos x} = \tan^2\frac{x}{2}. \] Thus, we have: \[ y = \frac{1}{2}\log\left(\tan^2\frac{x}{2}\right) = \log\left(\tan\frac{x}{2}\right). \]

Now, differentiate: \[ y = \log\left(\tan\frac{x}{2}\right). \] Using the chain rule, \[ \frac{dy}{dx} = \frac{1}{\tan\frac{x}{2}} \cdot \frac{d}{dx}\left(\tan\frac{x}{2}\right). \]

Differentiate \(\tan\frac{x}{2}\) with respect to \(x\): \[ \frac{d}{dx}\left(\tan\frac{x}{2}\right) = \sec^2\frac{x}{2} \cdot \frac{1}{2}. \] Therefore, \[ \frac{dy}{dx} = \frac{1}{\tan\frac{x}{2}} \cdot \frac{1}{2}\sec^2\frac{x}{2}. \]

Simplify the expression: \[ \frac{\sec^2\frac{x}{2}}{\tan\frac{x}{2}} = \frac{1}{\cos^2\frac{x}{2}} \cdot \frac{\cos\frac{x}{2}}{\sin\frac{x}{2}} = \frac{1}{\cos\frac{x}{2}\sin\frac{x}{2}} = \frac{2}{\sin x}, \] since \(\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}\).

Finally, the derivative is: \[ \frac{dy}{dx} = \frac{1}{2}\cdot\frac{2}{\sin x} = \frac{1}{\sin x} = \mathrm{cosec}\, x. \]

Solution:

Let \[ y = e^{\cot^{-1}(x^2)}. \] Using the chain rule, we have: \[ \frac{dy}{dx} = e^{\cot^{-1}(x^2)} \cdot \frac{d}{dx}\left[\cot^{-1}(x^2)\right]. \]

Recall that the derivative of \( \cot^{-1} u \) with respect to \( u \) is: \[ \frac{d}{du}\cot^{-1}u = -\frac{1}{1+u^2}. \] Here, \( u = x^2 \) so that: \[ \frac{du}{dx} = 2x. \] Hence, by the chain rule: \[ \frac{d}{dx}\left[\cot^{-1}(x^2)\right] = -\frac{1}{1+(x^2)^2}\cdot 2x = -\frac{2x}{1+x^4}. \]

Substitute this result into the derivative for \( y \): \[ \frac{dy}{dx} = e^{\cot^{-1}(x^2)} \cdot \left(-\frac{2x}{1+x^4}\right) = -\frac{2x}{1+x^4}\, e^{\cot^{-1}(x^2)}. \]

Solution:

We are given \[ y = x\sqrt{1+x^2} + \log\left(x+\sqrt{x^2+1}\right). \] We differentiate each term separately.

First term: \( f(x)=x\sqrt{1+x^2} \).
Using the product rule: \[ f'(x)=\sqrt{1+x^2}+x\cdot\frac{d}{dx}\left(\sqrt{1+x^2}\right). \] Since \[ \frac{d}{dx}\left(\sqrt{1+x^2}\right)=\frac{1}{2\sqrt{1+x^2}}\cdot 2x=\frac{x}{\sqrt{1+x^2}}, \] we have: \[ f'(x)=\sqrt{1+x^2}+x\cdot\frac{x}{\sqrt{1+x^2}} =\sqrt{1+x^2}+\frac{x^2}{\sqrt{1+x^2}} =\frac{1+x^2+x^2}{\sqrt{1+x^2}} =\frac{1+2x^2}{\sqrt{1+x^2}}. \]

Second term: \( g(x)=\log\left(x+\sqrt{x^2+1}\right) \).
Differentiating using the chain rule: \[ g'(x)=\frac{1}{x+\sqrt{x^2+1}}\cdot\left(1+\frac{x}{\sqrt{x^2+1}}\right). \] Notice that: \[ 1+\frac{x}{\sqrt{x^2+1}}=\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}}, \] hence: \[ g'(x)=\frac{\sqrt{x^2+1}+x}{\left(x+\sqrt{x^2+1}\right)\sqrt{x^2+1}}. \] Since the numerator and the factor \( x+\sqrt{x^2+1} \) in the denominator are identical, this simplifies to: \[ g'(x)=\frac{1}{\sqrt{x^2+1}}. \]

Adding the derivatives of the two terms: \[ \frac{dy}{dx}=f'(x)+g'(x) = \frac{1+2x^2}{\sqrt{1+x^2}}+\frac{1}{\sqrt{x^2+1}}. \] Noting that \(\sqrt{1+x^2}=\sqrt{x^2+1}\), we combine: \[ \frac{dy}{dx}=\frac{1+2x^2+1}{\sqrt{1+x^2}} = \frac{2(1+x^2)}{\sqrt{1+x^2}} = 2\sqrt{1+x^2}. \]

Solution:

Let \[ y=\frac{u}{v},\quad \text{with } u=\log\left(x+\sqrt{x^2+1}\right) \quad \text{and} \quad v=\sqrt{x^2+1}. \]

Step 1: Differentiate \( u \).
We have: \[ u=\log\left(x+\sqrt{x^2+1}\right). \] Differentiating with respect to \( x \): \[ u’=\frac{1}{x+\sqrt{x^2+1}}\left(1+\frac{x}{\sqrt{x^2+1}}\right). \] Notice that \[ 1+\frac{x}{\sqrt{x^2+1}}=\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}}, \] and since \( x+\sqrt{x^2+1}=\sqrt{x^2+1}+x \), it follows that: \[ u’=\frac{1}{\sqrt{x^2+1}}. \]

Step 2: Differentiate \( v \).
Here, \[ v=\sqrt{x^2+1}, \] so \[ v’=\frac{1}{2\sqrt{x^2+1}}(2x)=\frac{x}{\sqrt{x^2+1}}. \]

Step 3: Apply the Quotient Rule.
The quotient rule states: \[ y’=\frac{u’v-uv’}{v^2}. \] Substituting the computed derivatives: \[ y’=\frac{\displaystyle \frac{1}{\sqrt{x^2+1}}\cdot\sqrt{x^2+1}-\log\left(x+\sqrt{x^2+1}\right)\cdot\frac{x}{\sqrt{x^2+1}}}{x^2+1}. \] Simplify the numerator: \[ \frac{1}{\sqrt{x^2+1}}\cdot\sqrt{x^2+1}=1, \] hence: \[ y’=\frac{1-\frac{x\log\left(x+\sqrt{x^2+1}\right)}{\sqrt{x^2+1}}}{x^2+1}. \]

Step 4: Form the Expression \(\left(x^2+1\right)y’+xy\).
Multiply \( y’ \) by \( (x^2+1) \): \[ \left(x^2+1\right)y’ = 1-\frac{x\log\left(x+\sqrt{x^2+1}\right)}{\sqrt{x^2+1}}. \] Also, \[ y=\frac{\log\left(x+\sqrt{x^2+1}\right)}{\sqrt{x^2+1}}, \] so \[ xy=\frac{x\log\left(x+\sqrt{x^2+1}\right)}{\sqrt{x^2+1}}. \] Adding these results: \[ \left(x^2+1\right)y’+xy = \left[1-\frac{x\log\left(x+\sqrt{x^2+1}\right)}{\sqrt{x^2+1}}\right]+\frac{x\log\left(x+\sqrt{x^2+1}\right)}{\sqrt{x^2+1}} = 1. \]

Thus, we have proved that: \[ \left(x^2+1\right)\frac{dy}{dx}+xy=1. \]

Solution:

We start with \[ y = e^{2\log x + 3x}. \] Notice that using the property \(e^{2\log x}=x^2\), we can rewrite: \[ y = x^2 e^{3x}. \]

Now, differentiate \(y=x^2e^{3x}\) using the product rule: \[ \frac{dy}{dx} = \frac{d}{dx}\left(x^2\right) e^{3x} + x^2 \frac{d}{dx}\left(e^{3x}\right). \]

Compute the derivatives: \[ \frac{d}{dx}\left(x^2\right)=2x, \quad \frac{d}{dx}\left(e^{3x}\right)=3e^{3x}. \]

Substitute these into the product rule: \[ \frac{dy}{dx} = 2x\, e^{3x} + x^2\cdot 3e^{3x} = e^{3x}\left(2x + 3x^2\right). \]

Factor out \(x\) from the expression inside the parentheses: \[ \frac{dy}{dx} = x\left(2+3x\right)e^{3x}. \]

Thus, we have proved that: \[ \frac{dy}{dx} = x(2+3x)e^{3x}. \]

Solution:

Let \[ u=\frac{2^{x+1}}{1-4^x}. \] Then, \[ y=\tan^{-1}(u) \] and by the chain rule, \[ \frac{dy}{dx}=\frac{u’}{1+u^2}. \]

Step 1: Simplify \( u \).
Note that \[ 2^{x+1}=2\cdot2^x \quad \text{and} \quad 4^x=(2^2)^x=2^{2x}. \] Thus, \[ u=\frac{2\cdot2^x}{1-2^{2x}}. \]

Step 2: Differentiate \( u \) using the quotient rule.
Let \[ N(x)=2\cdot2^x \quad \text{and} \quad D(x)=1-2^{2x}. \] Then, \[ N'(x)=2\cdot2^x\log2, \] and \[ D'(x)=-\frac{d}{dx}(2^{2x})=-2^{2x}\cdot(2\log2)=-2^{2x+1}\log2. \] By the quotient rule: \[ u’=\frac{N'(x)D(x)-N(x)D'(x)}{[D(x)]^2}. \] Substituting the expressions: \[ u’=\frac{2\cdot2^x\log2\,(1-2^{2x})-2\cdot2^x\left(-2^{2x+1}\log2\right)}{(1-2^{2x})^2}. \]

Simplify the numerator: \[ 2\cdot2^x\log2\,(1-2^{2x})+2\cdot2^x\cdot2^{2x+1}\log2 =2\cdot2^x\log2\left[(1-2^{2x})+2^{2x+1}\right]. \] Notice that: \[ 2^{2x+1}=2\cdot2^{2x}, \] so: \[ (1-2^{2x})+2\cdot2^{2x}=1+2^{2x}. \] Thus, \[ u’=\frac{2\cdot2^x\log2\,(1+2^{2x})}{(1-2^{2x})^2}. \]

Step 3: Compute \( 1+u^2 \).
We have: \[ u=\frac{2\cdot2^x}{1-2^{2x}}, \] so \[ u^2=\frac{4\cdot2^{2x}}{(1-2^{2x})^2}. \] Therefore, \[ 1+u^2=\frac{(1-2^{2x})^2+4\cdot2^{2x}}{(1-2^{2x})^2}. \] Expand and simplify the numerator: \[ (1-2^{2x})^2+4\cdot2^{2x}=1-2\cdot2^{2x}+2^{4x}+4\cdot2^{2x} =1+2^{4x}+2\cdot2^{2x}. \] Recognize that: \[ 1+2\cdot2^{2x}+2^{4x}=\left(1+2^{2x}\right)^2. \] Hence, \[ 1+u^2=\frac{\left(1+2^{2x}\right)^2}{(1-2^{2x})^2}. \]

Step 4: Write the derivative \( \frac{dy}{dx} \).
Substitute \( u’ \) and \( 1+u^2 \) into the chain rule expression: \[ \frac{dy}{dx}=\frac{u’}{1+u^2} =\frac{\displaystyle \frac{2\cdot2^x\log2\,(1+2^{2x})}{(1-2^{2x})^2}}{\displaystyle \frac{\left(1+2^{2x}\right)^2}{(1-2^{2x})^2}} =\frac{2\cdot2^x\log2\,(1+2^{2x})}{\left(1+2^{2x}\right)^2}. \] Cancel the common factor \(1+2^{2x}\): \[ \frac{dy}{dx}=\frac{2\cdot2^x\log2}{1+2^{2x}}. \] Writing \(2\cdot2^x\) as \(2^{x+1}\), we have: \[ \frac{dy}{dx}=\frac{2^{x+1}\log2}{1+2^{2x}}. \]

Solution:

We differentiate the equation \[ xy + xe^{-y} + ye^{x} = x^2 \] implicitly with respect to \(x\).

Step 1: Differentiate each term.
Term 1: \(xy\) \[ \frac{d}{dx}(xy) = y + x\,y’. \] Term 2: \(xe^{-y}\) (using the product rule) \[ \frac{d}{dx}\bigl(xe^{-y}\bigr) = \frac{d}{dx}(x)\cdot e^{-y} + x\cdot\frac{d}{dx}\bigl(e^{-y}\bigr) = e^{-y} + x\Bigl(-e^{-y}y’\Bigr) = e^{-y} – x\,e^{-y}\,y’. \] Term 3: \(ye^{x}\) (again using the product rule) \[ \frac{d}{dx}\bigl(ye^{x}\bigr) = y’\,e^{x} + y\,e^{x}. \] Right-hand side: \(x^2\) \[ \frac{d}{dx}\left(x^2\right) = 2x. \]

Step 2: Combine the differentiated terms.
The derivative of the left-hand side becomes: \[ \left(y + xy’\right) + \left(e^{-y} – xe^{-y}\,y’\right) + \left(e^{x}y’ + ye^{x}\right) = 2x. \]

Step 3: Group the terms containing \(y’\).
Collect the terms with \(y’\): \[ xy’ – xe^{-y}\,y’ + e^{x}y’ + \left(y + e^{-y} + ye^{x}\right) = 2x. \] Factor out \(y’\): \[ y’\Bigl(x – xe^{-y} + e^{x}\Bigr) + \Bigl(y + e^{-y} + ye^{x}\Bigr) = 2x. \]

Step 4: Solve for \(y’\).
Isolate the \(y’\) term: \[ y’\Bigl(x – xe^{-y} + e^{x}\Bigr) = 2x – \Bigl(y + e^{-y} + ye^{x}\Bigr). \] Thus, \[ y’ = \frac{2x – \left(y + e^{-y} + ye^{x}\right)}{x – xe^{-y} + e^{x}}. \]

We can also rewrite the denominator as: \[ x – xe^{-y} + e^{x} = x + e^{x} – xe^{-y}, \] which matches the provided answer.

Solution:

We start with the equation \[ e^{x-y} = \log\left(\frac{x}{y}\right). \] We will differentiate both sides with respect to \(x\) using implicit differentiation.

Step 1: Differentiate the left-hand side.
The left-hand side is \[ e^{x-y}. \] Differentiating, we use the chain rule: \[ \frac{d}{dx}\left(e^{x-y}\right) = e^{x-y}\cdot\frac{d}{dx}(x-y) = e^{x-y}\left(1 – \frac{dy}{dx}\right). \]

Step 2: Differentiate the right-hand side.
The right-hand side is \[ \log\left(\frac{x}{y}\right). \] Using the property of logarithms, we write: \[ \log\left(\frac{x}{y}\right) = \log x – \log y. \] Differentiating term by term: \[ \frac{d}{dx}(\log x) = \frac{1}{x},\quad \frac{d}{dx}(\log y) = \frac{1}{y}\frac{dy}{dx}. \] Hence, the derivative is: \[ \frac{d}{dx}\left(\log\left(\frac{x}{y}\right)\right) = \frac{1}{x} – \frac{1}{y}\frac{dy}{dx}. \]

Step 3: Equate the derivatives.
Equating the derivatives from both sides, we have: \[ e^{x-y}\left(1 – \frac{dy}{dx}\right) = \frac{1}{x} – \frac{1}{y}\frac{dy}{dx}. \]

Step 4: Solve for \(\frac{dy}{dx}\).
Expand the left-hand side: \[ e^{x-y} – e^{x-y}\frac{dy}{dx} = \frac{1}{x} – \frac{1}{y}\frac{dy}{dx}. \] Bring the terms involving \(\frac{dy}{dx}\) to one side: \[ e^{x-y} – \frac{1}{x} = e^{x-y}\frac{dy}{dx} – \frac{1}{y}\frac{dy}{dx}. \] Factor out \(\frac{dy}{dx}\) on the right: \[ e^{x-y} – \frac{1}{x} = \frac{dy}{dx}\left(e^{x-y} – \frac{1}{y}\right). \] Therefore, solving for \(\frac{dy}{dx}\) yields: \[ \frac{dy}{dx} = \frac{e^{x-y} – \frac{1}{x}}{e^{x-y} – \frac{1}{y}}. \]

Solution:

We start with the given equation: \[ y \log x = x – y. \]

Step 1: Bring the terms involving \( y \) together: \[ y \log x + y = x. \] Factor out \( y \): \[ y (1+\log x) = x. \]

Step 2: Solve for \( y \): \[ y = \frac{x}{1+\log x}. \]

Step 3: Differentiate \( y \) with respect to \( x \) using the quotient rule. Recall the quotient rule: \[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} – u \frac{dv}{dx}}{v^2}, \] where \( u = x \) and \( v = 1+\log x \).

Compute the derivatives: \[ \frac{du}{dx} = 1 \quad \text{and} \quad \frac{dv}{dx} = \frac{1}{x}. \]

Applying the quotient rule: \[ \frac{dy}{dx} = \frac{(1+\log x)(1) – x\left(\frac{1}{x}\right)}{(1+\log x)^2}. \]

Simplify the numerator: \[ (1+\log x) – 1 = \log x. \]

Thus, the derivative is: \[ \frac{dy}{dx} = \frac{\log x}{(1+\log x)^2}. \]

Solution:

First, express the logarithm with base \( x \) in terms of natural logarithms: \[ \log_{x}(2x-3)=\frac{\log(2x-3)}{\log x}. \]

Let \[ u=\log(2x-3) \quad \text{and} \quad v=\log x. \] Then \[ y=\frac{u}{v}. \]

Compute the derivatives \( u’ \) and \( v’ \):

Differentiating \( u \): \[ u’=\frac{d}{dx}[\log(2x-3)]=\frac{1}{2x-3}\cdot2=\frac{2}{2x-3}. \]

Differentiating \( v \): \[ v’=\frac{d}{dx}[\log x]=\frac{1}{x}. \]

Now, apply the quotient rule: \[ \frac{dy}{dx}=\frac{v\,u’-u\,v’}{v^2}. \]

Substitute the derivatives: \[ \frac{dy}{dx}=\frac{(\log x)\left(\frac{2}{2x-3}\right)-\log(2x-3)\left(\frac{1}{x}\right)}{(\log x)^2}. \]

Simplify the numerator: \[ \frac{dy}{dx}=\frac{\frac{2\log x}{2x-3}-\frac{\log(2x-3)}{x}}{(\log x)^2}. \]

To combine the two fractions in the numerator, find a common denominator, which is \( x(2x-3) \): \[ \frac{dy}{dx}=\frac{\frac{2x\log x-(2x-3)\log(2x-3)}{x(2x-3)}}{(\log x)^2}. \]

Thus, the derivative is: \[ \frac{dy}{dx}=\frac{2x\log x-(2x-3)\log(2x-3)}{x(2x-3)(\log x)^2}. \]

Solution:

First, rewrite the logarithm with base \(\cos x\) in terms of natural logarithms: \[ \log_{\cos x} \sin x = \frac{\log(\sin x)}{\log(\cos x)}. \]

Let \[ u = \log(\sin x) \quad \text{and} \quad v = \log(\cos x). \] Then the function becomes: \[ y = \frac{u}{v}. \]

Differentiate \( u \) with respect to \( x \): \[ u’ = \frac{d}{dx} \log(\sin x) = \frac{1}{\sin x} \cdot \cos x = \cot x. \]

Differentiate \( v \) with respect to \( x \): \[ v’ = \frac{d}{dx} \log(\cos x) = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x. \]

Apply the quotient rule: \[ \frac{dy}{dx} = \frac{v\,u’ – u\,v’}{v^2}. \]

Substitute the computed derivatives: \[ \frac{dy}{dx} = \frac{\log(\cos x)\,\cot x – \log(\sin x)\,(-\tan x)}{[\log(\cos x)]^2}. \]

Simplify the numerator: \[ \frac{dy}{dx} = \frac{\log(\cos x)\,\cot x + \log(\sin x)\,\tan x}{[\log(\cos x)]^2}. \]