Ex 5.11 – Continuity and Differentiability | ML Aggarwal Class 12 Solutions

Solution: Let \[ y = (x+3)^{2}(x+4)^{3}(x+5)^{4}. \] Taking the natural logarithm of both sides, we have: \[ \log y = 2\log(x+3) + 3\log(x+4) + 4\log(x+5). \] Differentiating both sides with respect to \( x \): \[ \frac{1}{y}\frac{dy}{dx} = \frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5}. \] Multiplying through by \( y \) gives: \[ \frac{dy}{dx} = (x+3)^{2}(x+4)^{3}(x+5)^{4}\left(\frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5}\right). \]

Solution: Let \[ y = \cos x \cos 2x \cos 3x. \] Taking the natural logarithm on both sides, we write: \[ \text{log}\, y = \text{log} (\cos x) + \text{log} (\cos 2x) + \text{log} (\cos 3x). \] Differentiating with respect to \( x \): \[ \frac{1}{y}\frac{dy}{dx} = -\tan x – 2\tan 2x – 3\tan 3x. \] Multiplying both sides by \( y \) gives: \[ \frac{dy}{dx} = -\cos x \cos 2x \cos 3x \left(\tan x + 2\tan 2x + 3\tan 3x\right). \]

Solution: Let \[ y = x\,(x^2+1)^{1/2}\,(x+1)^{-2/3}. \] Taking the natural logarithm on both sides, we have: \[ \text{log}\, y = \text{log}\, x + \frac{1}{2}\,\text{log}(x^2+1) – \frac{2}{3}\,\text{log}(x+1). \] Differentiating with respect to \( x \): \[ \frac{1}{y}\frac{dy}{dx} = \frac{1}{x} + \frac{1}{2}\cdot\frac{2x}{x^2+1} – \frac{2}{3}\cdot\frac{1}{x+1}. \] Simplifying: \[ \frac{1}{y}\frac{dy}{dx} = \frac{1}{x} + \frac{x}{x^2+1} – \frac{2}{3(x+1)}. \] Multiplying by \( y \): \[ \frac{dy}{dx} = \frac{x\sqrt{x^2+1}}{(x+1)^{2/3}}\left(\frac{1}{x} + \frac{x}{x^2+1} – \frac{2}{3(x+1)}\right). \]

Solution: Let \[ y=\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}. \] We rewrite \( y \) as \[ y=\left[\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}\right]^{\frac{1}{2}}. \] Taking the natural logarithm on both sides: \[ \text{log}\, y=\frac{1}{2}\left\{\text{log}\,\big[(x-1)(x-2)\big]-\text{log}\,\big[(x-3)(x-4)(x-5)\big]\right\}. \] Differentiating with respect to \( x \): \[ \frac{1}{y}\frac{dy}{dx}=\frac{1}{2}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right]. \] Multiplying both sides by \( y \) gives: \[ \frac{dy}{dx}=\frac{1}{2}\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right]. \]

Solution: We begin with \[ f(x)=e^{x}\cos^{3}x\,\sin^{2}x. \] Using the product rule, \[ f'(x)=e^{x}\cos^{3}x\,\sin^{2}x+e^{x}\frac{d}{dx}\Bigl(\cos^{3}x\,\sin^{2}x\Bigr). \] Let \[ u(x)=\cos^{3}x\,\sin^{2}x. \] Then, \[ u'(x)=\frac{d}{dx}\Bigl(\cos^{3}x\Bigr)\sin^{2}x+\cos^{3}x\,\frac{d}{dx}\Bigl(\sin^{2}x\Bigr). \] Since \[ \frac{d}{dx}\Bigl(\cos^{3}x\Bigr)=3\cos^{2}x\cdot\Bigl(-\sin x\Bigr)=-3\cos^{2}x\,\sin x, \] and \[ \frac{d}{dx}\Bigl(\sin^{2}x\Bigr)=2\sin x\,\cos x, \] we have \[ u'(x)=-3\cos^{2}x\,\sin x\cdot \sin^{2}x+2\cos^{3}x\,\sin x\,\cos x. \] This simplifies to: \[ u'(x)=-3\cos^{2}x\,\sin^{3}x+2\cos^{4}x\,\sin x. \] Substituting back, \[ f'(x)=e^{x}\cos^{3}x\,\sin^{2}x+e^{x}\Bigl[-3\cos^{2}x\,\sin^{3}x+2\cos^{4}x\,\sin x\Bigr]. \] Factoring out \( e^{x}\cos^{2}x\,\sin x \): \[ f'(x)=e^{x}\cos^{2}x\,\sin x\Bigl[\cos x\,\sin x-3\sin^{2}x+2\cos^{2}x\Bigr]. \] Recognizing that \[ \tan x=\frac{\sin x}{\cos x} \quad \text{and} \quad \cot x=\frac{\cos x}{\sin x}, \] we can write: \[ f'(x)=e^{x}\cos^{3}x\,\sin^{2}x\Bigl[1-3\tan x+2\cot x\Bigr]. \]

Solution: We express \[ f(x)=\left[\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}\right]^{\frac{1}{2}}. \] Taking natural logarithms on both sides, we obtain: \[ \log f(x)=\frac{1}{2}\Bigl[\log\bigl((x-1)(x-2)\bigr)-\log\bigl((x-3)(x-4)(x-5)\bigr)\Bigr]. \] Differentiating both sides with respect to \( x \) using the chain rule: \[ \frac{f'(x)}{f(x)}=\frac{1}{2}\Biggl[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\Biggr]. \] Multiplying by \( f(x) \) yields: \[ f'(x)=\frac{1}{2}f(x)\Biggl[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\Biggr]. \] Substituting back \( f(x)=\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} \), we have: \[ f'(x)=\frac{1}{2}\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\Biggl[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\Biggr]. \]

Solution: Let \[ f(x) = (\sin x)^{\cos x}. \] Taking natural logarithm on both sides: \[ \log f(x) = \cos x \cdot \log(\sin x). \] Differentiating both sides using the chain rule and product rule: \[ \frac{f'(x)}{f(x)} = \frac{d}{dx} \left[ \cos x \cdot \log(\sin x) \right] = -\sin x \cdot \log(\sin x) + \cos x \cdot \frac{\cos x}{\sin x}. \] Simplifying: \[ \frac{f'(x)}{f(x)} = -\sin x \cdot \log(\sin x) + \frac{\cos^2 x}{\sin x}. \] Multiply both sides by \( f(x) = (\sin x)^{\cos x} \): \[ f'(x) = (\sin x)^{\cos x} \left( -\sin x \cdot \log(\sin x) + \frac{\cos^2 x}{\sin x} \right). \] We can also factor as: \[ f'(x) = (\sin x)^{\cos x} \left( \cos x \cdot \cot x – \sin x \cdot \log(\sin x) \right). \]

Solution: Let \[ f(x) = (\sin x)^{\sin x}. \] Taking logarithm on both sides: \[ \log f(x) = \sin x \cdot \log(\sin x). \] Differentiating both sides with respect to \( x \): \[ \frac{f'(x)}{f(x)} = \cos x \cdot \log(\sin x) + \sin x \cdot \frac{\cos x}{\sin x}. \] Simplify: \[ \frac{f'(x)}{f(x)} = \cos x \cdot \log(\sin x) + \cos x = \cos x \left(1 + \log(\sin x)\right). \] Multiplying both sides by \( f(x) = (\sin x)^{\sin x} \), we get: \[ f'(x) = (\sin x)^{\sin x} \cdot \cos x \left(1 + \log(\sin x)\right). \]

Solution: We rewrite the function as: \[ f(x)=\exp\Bigl(\sin x\cdot \log x\Bigr), \] where \(\log x\) denotes the full logarithm. Differentiating using the chain rule: \[ f'(x)=\exp\Bigl(\sin x\cdot \log x\Bigr) \cdot \frac{d}{dx}\Bigl(\sin x\cdot \log x\Bigr). \] Applying the product rule to \(\sin x\cdot \log x\): \[ \frac{d}{dx}\Bigl(\sin x\cdot \log x\Bigr)=\cos x\cdot \log x+\sin x\cdot \frac{1}{x}. \] Therefore, the derivative is: \[ f'(x)=x^{\sin x}\left(\cos x\cdot \log x+\frac{\sin x}{x}\right). \]

Solution: We write \[ f(x) = (2x+3)^{x-5}. \] Taking logarithm on both sides, we have: \[ \log f(x) = (x-5) \cdot \log(2x+3). \] Differentiating both sides with respect to \( x \) using the product rule: \[ \frac{f'(x)}{f(x)} = \frac{d}{dx}\Bigl[(x-5) \cdot \log(2x+3)\Bigr] = \log(2x+3) + (x-5)\cdot \frac{2}{2x+3}. \] Multiplying both sides by \( f(x) \), we obtain: \[ f'(x) = (2x+3)^{x-5}\left[\log(2x+3) + \frac{2(x-5)}{2x+3}\right]. \]

Solution: Let \[ f(x) = (\log x)^{\cos x}. \] Taking logarithm on both sides: \[ \log f(x) = \cos x \cdot \log(\log x). \] Differentiating both sides: \[ \frac{f'(x)}{f(x)} = -\sin x \cdot \log(\log x) + \cos x \cdot \frac{1}{\log x} \cdot \frac{1}{x}. \] Simplifying: \[ \frac{f'(x)}{f(x)} = \frac{\cos x}{x \log x} – \sin x \cdot \log(\log x). \] Multiply both sides by \( f(x) = (\log x)^{\cos x} \): \[ f'(x) = (\log x)^{\cos x} \left[\frac{\cos x}{x \log x} – \sin x \cdot \log(\log x)\right]. \]

Solution: We begin with \[ f(x) = (\log x)^{\log x}. \] Taking logarithm on both sides: \[ \log f(x) = \log x \cdot \log(\log x). \] Differentiating both sides with respect to \( x \) using the product rule, we obtain: \[ \frac{f'(x)}{f(x)} = \frac{d}{dx}\Bigl[\log x \cdot \log(\log x)\Bigr] = \frac{1}{x} \cdot \log(\log x) + \log x \cdot \frac{1}{\log x}\cdot\frac{1}{x}. \] Simplify the second term: \[ \log x \cdot \frac{1}{\log x}\cdot\frac{1}{x} = \frac{1}{x}. \] Thus, \[ \frac{f'(x)}{f(x)} = \frac{1}{x}\Bigl[\log(\log x)+1\Bigr]. \] Multiplying both sides by \( f(x) \) gives: \[ f'(x) = (\log x)^{\log x}\cdot\frac{1}{x}\Bigl[\log(\log x)+1\Bigr]. \]

Solution: Let \[ y = x^{\sin x+\cos x}. \] Taking the natural logarithm on both sides, we have \[ \text{log}\,y = (\sin x+\cos x)\,\text{log}\,x. \] Differentiating both sides with respect to \( x \) yields \[ \frac{1}{y}\frac{dy}{dx} = \frac{d}{dx}\left[(\sin x+\cos x)\,\text{log}\,x\right]. \] Applying the product rule on the right side, we obtain \[ \frac{1}{y}\frac{dy}{dx} = (\cos x-\sin x)\,\text{log}\,x + (\sin x+\cos x)\frac{1}{x}. \] Multiplying both sides by \( y \) gives \[ \frac{dy}{dx} = y\left[(\cos x-\sin x)\,\text{log}\,x + \frac{\sin x+\cos x}{x}\right]. \] Substituting back \( y = x^{\sin x+\cos x} \), we finally have \[ \frac{dy}{dx} = x^{\sin x+\cos x}\left[\frac{\sin x+\cos x}{x}+(\cos x-\sin x)\,\text{log}\,x\right]. \]

Solution: Let \[ y=\left(x^{2}\sin x\right)^{\frac{1}{x}}. \] Taking the natural logarithm on both sides, we obtain \[ \text{log}\,y=\frac{1}{x}\,\text{log}\,(x^{2}\sin x). \] Differentiate both sides with respect to \( x \): \[ \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}\left(\frac{1}{x}\,\text{log}\,(x^{2}\sin x)\right). \] Write the right-hand side as \[ \frac{1}{x}\,\text{log}\,(x^{2}\sin x)=\frac{2\,\text{log}\,x+\text{log}\,(\sin x)}{x}. \] Differentiate using the quotient rule. Let \[ f(x)=2\,\text{log}\,x+\text{log}\,(\sin x) \quad \text{and} \quad g(x)=x. \] Then, \[ f'(x)=\frac{2}{x}+\cot x \quad \text{and} \quad g'(x)=1. \] The quotient rule gives \[ \frac{d}{dx}\left(\frac{f(x)}{x}\right)=\frac{x\left(\frac{2}{x}+\cot x\right)-\left(2\,\text{log}\,x+\text{log}\,(\sin x)\right)}{x^{2}}. \] Simplify the numerator: \[ x\left(\frac{2}{x}+\cot x\right)=2+x\cot x. \] Hence, we obtain \[ \frac{1}{y}\frac{dy}{dx}=\frac{2+x\cot x-\left(2\,\text{log}\,x+\text{log}\,(\sin x)\right)}{x^{2}}. \] Recognize that \[ 2\,\text{log}\,x+\text{log}\,(\sin x)=\text{log}\,(x^{2}\sin x). \] Therefore, \[ \frac{1}{y}\frac{dy}{dx}=\frac{2+x\cot x-\text{log}(x^{2}\sin x)}{x^{2}}. \] Multiplying both sides by \( y \), we get \[ \frac{dy}{dx}=\frac{1}{x^{2}}\left(x^{2}\sin x\right)^{\frac{1}{x}}\left(2+x\cot x-\text{log}(x^{2}\sin x)\right). \]

Solution: We are given \[ y=x^y. \] Take the natural logarithm on both sides: \[ \text{log}\,y = y\,\text{log}\,x. \] Differentiate both sides with respect to \( x \): \[ \frac{1}{y}\frac{dy}{dx} = \frac{dy}{dx}\,\text{log}\,x + \frac{y}{x}. \] Rearranging, we have: \[ \frac{1}{y}\frac{dy}{dx} – \frac{dy}{dx}\,\text{log}\,x = \frac{y}{x}. \] Factor out \(\frac{dy}{dx}\) on the left-hand side: \[ \frac{dy}{dx}\left(\frac{1}{y}-\text{log}\,x\right)=\frac{y}{x}. \] Solve for \(\frac{dy}{dx}\): \[ \frac{dy}{dx}=\frac{y}{x\left(\frac{1}{y}-\text{log}\,x\right)}. \] Multiply the numerator and denominator by \( y \) to simplify: \[ \frac{dy}{dx}=\frac{y^2}{x\left(1-y\,\text{log}\,x\right)}. \] Finally, multiplying both sides by \( x \) gives: \[ x\frac{dy}{dx}=\frac{y^2}{1-y\,\text{log}\,x}. \]

Solution: We are given \[ x=e^{\frac{x}{y}}. \] Taking the natural logarithm on both sides, we get \[ \text{log}\,x=\frac{x}{y}. \] Differentiate both sides with respect to \( x \): \[ \frac{d}{dx}\left(\text{log}\,x\right)=\frac{d}{dx}\left(\frac{x}{y}\right). \] The derivative of the left-hand side is \[ \frac{1}{x}. \] For the right-hand side, apply the quotient rule. Let \[ u=x \quad \text{and} \quad v=y, \] so that \[ u’=1 \quad \text{and} \quad v’=\frac{dy}{dx}. \] Then, \[ \frac{d}{dx}\left(\frac{x}{y}\right)=\frac{y\cdot 1 – x\frac{dy}{dx}}{y^{2}}. \] Equating the derivatives, we have \[ \frac{1}{x}=\frac{y-x\frac{dy}{dx}}{y^{2}}. \] Multiply both sides by \( y^{2} \): \[ \frac{y^{2}}{x}=y-x\frac{dy}{dx}. \] Rearranging the terms, we obtain \[ x\frac{dy}{dx}=y-\frac{y^{2}}{x}. \] Factor \( y \) in the right-hand side: \[ x\frac{dy}{dx}=\frac{y(x-y)}{x}. \] Dividing both sides by \( x \) gives \[ \frac{dy}{dx}=\frac{y(x-y)}{x^{2}}. \] From the original equation \( \text{log}\,x=\frac{x}{y} \), we can express \( y \) as \[ y=\frac{x}{\text{log}\,x}. \] Substituting this into our expression for \( \frac{dy}{dx} \) yields \[ \frac{dy}{dx}=\frac{\frac{x}{\text{log}\,x}(x-y)}{x^{2}}=\frac{x-y}{x\,\text{log}\,x}. \]

Solution: We are given \[ x^{y}=e^{x-y}. \] Taking the natural logarithm on both sides, we have \[ \text{log}\,(x^{y})=\text{log}\,(e^{x-y}). \] Using the logarithmic identity \(\text{log}\,(a^{b})=b\,\text{log}\,a\) and \(\text{log}\,(e^{u})=u\), this becomes \[ y\,\text{log}\,x=x-y. \] Rearranging, we get \[ y\,\text{log}\,x + y = x \quad \Longrightarrow \quad y\left(\text{log}\,x+1\right)=x. \] Thus, \[ y=\frac{x}{\text{log}\,x+1}. \] Note that \(\text{log}\,x+1\) can be written as \(\text{log}\,(x\,e)\) because \[ \text{log}\,(x\,e)=\text{log}\,x+\text{log}\,e=\text{log}\,x+1. \] Therefore, we can also express \(y\) as \[ y=\frac{x}{\text{log}\,(x\,e)}. \] Differentiating \( y=\frac{x}{\text{log}\,(x\,e)} \) with respect to \( x \) using the quotient rule, let \[ u=x \quad \text{and} \quad v=\text{log}\,(x\,e). \] We have \[ u’=1 \quad \text{and} \quad v’=\frac{d}{dx}\left(\text{log}\,(x\,e)\right)=\frac{1}{x\,e}\cdot e=\frac{1}{x}, \] since \( \text{log}\,(x\,e)=\text{log}\,x+1 \) and the derivative of \(\text{log}\,x\) is \(\frac{1}{x}\). The quotient rule gives: \[ \frac{dy}{dx}=\frac{v\cdot u’-u\cdot v’}{v^{2}}=\frac{\text{log}\,(x\,e)\cdot 1-x\cdot\frac{1}{x}}{\left[\text{log}\,(x\,e)\right]^{2}}. \] Simplifying the numerator: \[ \text{log}\,(x\,e)-1=\text{log}\,x+1-1=\text{log}\,x. \] Therefore, we have \[ \frac{dy}{dx}=\frac{\text{log}\,x}{\left[\text{log}\,(x\,e)\right]^{2}}, \] which is the required result.

Solution: We start with the given equation: \[ x^{16}y^{9}=(x^{2}+y)^{17}. \] Take the natural logarithm on both sides: \[ \text{log}\,(x^{16}y^{9})=\text{log}\,(x^{2}+y)^{17}. \] Using the logarithm properties, this becomes: \[ 16\,\text{log}\,x+9\,\text{log}\,y=17\,\text{log}\,(x^{2}+y). \] Differentiate both sides with respect to \(x\). The derivative of the left-hand side is: \[ \frac{16}{x}+\frac{9}{y}\frac{dy}{dx}. \] For the right-hand side, apply the chain rule: \[ 17\cdot\frac{1}{x^{2}+y}\left(2x+\frac{dy}{dx}\right). \] Thus, we have the equation: \[ \frac{16}{x}+\frac{9}{y}\frac{dy}{dx}=\frac{17\left(2x+\frac{dy}{dx}\right)}{x^{2}+y}. \] Multiply both sides by \(x^{2}+y\) to clear the denominator: \[ \frac{16(x^{2}+y)}{x}+\frac{9(x^{2}+y)}{y}\frac{dy}{dx}=17\left(2x+\frac{dy}{dx}\right). \] Rearrange to collect the terms containing \(\frac{dy}{dx}\): \[ \frac{9(x^{2}+y)}{y}\frac{dy}{dx}-17\frac{dy}{dx}=17\cdot 2x-\frac{16(x^{2}+y)}{x}. \] Factor \(\frac{dy}{dx}\) on the left: \[ \frac{dy}{dx}\left[\frac{9(x^{2}+y)}{y}-17\right]=34x-\frac{16(x^{2}+y)}{x}. \] Write both sides with common denominators: \[ \frac{dy}{dx}\left[\frac{9(x^{2}+y)-17y}{y}\right]=\frac{34x^{2}-16(x^{2}+y)}{x}. \] Simplify the numerator on the left: \[ 9(x^{2}+y)-17y=9x^{2}+9y-17y=9x^{2}-8y, \] and on the right: \[ 34x^{2}-16(x^{2}+y)=34x^{2}-16x^{2}-16y=18x^{2}-16y. \] Thus, we have: \[ \frac{dy}{dx}\cdot\frac{9x^{2}-8y}{y}=\frac{18x^{2}-16y}{x}. \] Solving for \(\frac{dy}{dx}\), we get: \[ \frac{dy}{dx}=\frac{18x^{2}-16y}{x}\cdot\frac{y}{9x^{2}-8y}. \] Notice that the numerator \(18x^{2}-16y\) can be factored as: \[ 18x^{2}-16y=2(9x^{2}-8y). \] Substituting this back, we have: \[ \frac{dy}{dx}=\frac{2(9x^{2}-8y)y}{x(9x^{2}-8y)}=\frac{2y}{x}. \]

Solution: We differentiate \( f(x) \) term by term. 1. For \( x^{x} \): Write \[ x^{x}=e^{x\,\text{log}\,x}. \] Differentiating using the chain rule: \[ \frac{d}{dx}\left(e^{x\,\text{log}\,x}\right)=e^{x\,\text{log}\,x}\cdot \frac{d}{dx}\left(x\,\text{log}\,x\right). \] Now, \[ \frac{d}{dx}\left(x\,\text{log}\,x\right)=\text{log}\,x+1. \] Therefore, \[ \frac{d}{dx}\left(x^{x}\right)=x^{x}\left(\text{log}\,x+1\right). \] 2. For \( a^{x} \): Since \( a \) is a constant, \[ \frac{d}{dx}\left(a^{x}\right)=a^{x}\,\text{log}\,a. \] 3. For \( x^{a} \): Here \( a \) is constant so the derivative is \[ \frac{d}{dx}\left(x^{a}\right)=a\,x^{a-1}. \] 4. For \( a^{a} \): This is a constant and its derivative is zero. Combining all these results, the derivative of \( f(x) \) is \[ f'(x)=x^{x}\left(\text{log}\,x+1\right)+a^{x}\,\text{log}\,a+a\,x^{a-1}. \]

Solution: We start by differentiating each term separately. First, consider \[ x^{\text{log } x} = e^{\text{log } x \cdot \text{log } x} = e^{(\text{log } x)^2}. \] Differentiating, we have \[ \frac{d}{dx}\left(e^{(\text{log } x)^2}\right) = e^{(\text{log } x)^2} \cdot \frac{d}{dx}\left((\text{log } x)^2\right) = e^{(\text{log } x)^2} \cdot 2\,\text{log } x \cdot \frac{1}{x}. \] Therefore, \[ \frac{d}{dx}\left(x^{\text{log } x}\right) = \frac{2\,\text{log } x}{x}\,x^{\text{log } x}. \] Next, consider the second term: \[ (\text{log } x)^{x} = e^{x\,\text{log}(\text{log } x)}. \] Differentiating the exponent, we obtain \[ \frac{d}{dx}\left(x\,\text{log}(\text{log } x)\right) = \text{log}(\text{log } x) + x \cdot \frac{1}{\text{log } x} \cdot \frac{1}{x} = \text{log}(\text{log } x) + \frac{1}{\text{log } x}. \] Hence, \[ \frac{d}{dx}\left((\text{log } x)^{x}\right) = (\text{log } x)^{x} \left(\text{log}(\text{log } x) + \frac{1}{\text{log } x}\right). \] Combining both derivatives, we obtain \[ \frac{d}{dx}\left(x^{\text{log } x} + (\text{log } x)^{x}\right) = \frac{2\,\text{log } x}{x}\,x^{\text{log } x} + (\text{log } x)^{x} \left(\text{log}(\text{log } x) + \frac{1}{\text{log } x}\right). \]

Solution: We differentiate each term separately. \[ (\sin x)^{\cos x} = e^{\cos x\,\log (\sin x)}. \] Differentiating the exponent: \[ \frac{d}{dx}\left[\cos x\,\log (\sin x)\right] = -\sin x\,\log (\sin x) + \cos x\,\frac{\cos x}{\sin x} = -\sin x\,\log (\sin x) + \frac{\cos^{2} x}{\sin x}. \] Thus, the derivative is: \[ \frac{d}{dx}\left[(\sin x)^{\cos x}\right] = (\sin x)^{\cos x}\left(-\sin x\,\log (\sin x) + \frac{\cos^{2} x}{\sin x}\right). \] Note that \(\frac{\cos^{2} x}{\sin x} = \cos x\,\cot x\). Next, for the term: \[ x^{\sin x} = e^{\sin x\,\log x}. \] Differentiating the exponent: \[ \frac{d}{dx}\left[\sin x\,\log x\right] = \cos x\,\log x + \sin x\,\frac{1}{x}. \] Therefore, the derivative is: \[ \frac{d}{dx}\left[x^{\sin x}\right] = x^{\sin x}\left(\cos x\,\log x + \frac{\sin x}{x}\right). \] Combining the two results, we obtain: \[ \frac{d}{dx}\left[(\sin x)^{\cos x} + x^{\sin x}\right] = (\sin x)^{\cos x}\left(\cos x\,\cot x – \sin x\,\log (\sin x)\right) + x^{\sin x}\left(\cos x\,\log x + \frac{\sin x}{x}\right). \]

Solution: We differentiate each term separately. For the first term: \[ x^{\cos x} = e^{\cos x\,\log x}. \] Differentiating the exponent: \[ \frac{d}{dx}\left[\cos x\,\log x\right] = -\sin x\,\log x + \frac{\cos x}{x}. \] Hence, \[ \frac{d}{dx}\left[x^{\cos x}\right] = x^{\cos x}\left(\frac{\cos x}{x} – \sin x\,\log x\right). \] For the second term: \[ (\cos x)^{x} = e^{x\,\log (\cos x)}. \] Differentiating the exponent: \[ \frac{d}{dx}\left[x\,\log (\cos x)\right] = \log (\cos x) + x\left(-\frac{\sin x}{\cos x}\right) = \log (\cos x) – x\,\tan x. \] Thus, \[ \frac{d}{dx}\left[(\cos x)^{x}\right] = (\cos x)^{x}\left(\log (\cos x) – x\,\tan x\right). \] Combining both derivatives, we obtain: \[ \frac{d}{dx}\left[x^{\cos x}+(\cos x)^{x}\right] = x^{\cos x}\left(\frac{\cos x}{x} – \sin x\,\log x\right) + (\cos x)^{x}\left(\log (\cos x) – x\,\tan x\right). \]

Solution: We differentiate each term one by one. For the first term: \[ x^{\cos x} = e^{\cos x \log x}. \] Differentiating the exponent: \[ \frac{d}{dx}[\cos x \log x] = -\sin x \log x + \frac{\cos x}{x}. \] Therefore, \[ \frac{d}{dx}\left[x^{\cos x}\right] = x^{\cos x}\left(\frac{\cos x}{x} – \sin x \log x\right). \] For the second term: \[ (\sin x)^{\tan x} = e^{\tan x \log (\sin x)}. \] Differentiating the exponent: \[ \frac{d}{dx}[\tan x \log (\sin x)] = \sec^2 x \log (\sin x) + \tan x \cdot \frac{\cos x}{\sin x} = \sec^2 x \log (\sin x) + \tan x \cot x. \] Hence, \[ \frac{d}{dx}\left[(\sin x)^{\tan x}\right] = (\sin x)^{\tan x} \left( \sec^2 x \log (\sin x) + 1\right). \] So, the derivative of the full expression is: \[ \frac{d}{dx}\left[ x^{\cos x} + (\sin x)^{\tan x} \right] = x^{\cos x} \left( \frac{\cos x}{x} – \sin x \log x \right) + (\sin x)^{\tan x} \left( \sec^2 x \log (\sin x) + 1\right). \]

Solution: First, express the term \( (\sin x)^{x} \) in exponential form: \[ (\sin x)^{x} = e^{x\,\log (\sin x)}. \] Differentiating the exponent: \[ \frac{d}{dx}\Bigl[x\,\log (\sin x)\Bigr] = \log (\sin x) + x\,\frac{1}{\sin x}\cos x = \log (\sin x) + x\,\cot x. \] Hence, the derivative of the first term is: \[ \frac{d}{dx}\left[(\sin x)^{x}\right] = (\sin x)^{x}\left(\log (\sin x) + x\,\cot x\right). \] Next, consider the second term: \[ \sin^{-1}\sqrt{x} = \arcsin(\sqrt{x}). \] Let \( u = \sqrt{x} \), so that \( u’ = \frac{1}{2\sqrt{x}} \). The derivative of \( \arcsin u \) is: \[ \frac{d}{dx}\left[\arcsin u\right] = \frac{u’}{\sqrt{1-u^{2}}} = \frac{\frac{1}{2\sqrt{x}}}{\sqrt{1-x}} = \frac{1}{2\sqrt{x(1-x)}}. \] Combining both parts, the derivative is: \[ \frac{d}{dx}\left[(\sin x)^{x} + \sin^{-1}\sqrt{x}\right] = (\sin x)^{x}\left(\log (\sin x) + x\,\cot x\right) + \frac{1}{2\sqrt{x(1-x)}}. \]

Solution: We write the function as the sum of two terms and differentiate each separately. \[ \frac{d}{dx}\left(e^{\sin x}\right)=e^{\sin x}\cos x. \] For the second term, express it in exponential form: \[ (\tan x)^{x}=e^{x\,\log (\tan x)}. \] Differentiating the exponent, we have: \[ \frac{d}{dx}\Bigl[x\,\log (\tan x)\Bigr]=\log (\tan x)+x\,\frac{d}{dx}\left[\log (\tan x)\right]. \] Now, since \[ \frac{d}{dx}\left[\log (\tan x)\right]=\frac{1}{\tan x}\cdot\sec^{2}x, \] it follows that: \[ \frac{d}{dx}\Bigl[x\,\log (\tan x)\Bigr]=\log (\tan x)+\frac{x\,\sec^{2}x}{\tan x}. \] Notice that \[ \frac{x\,\sec^{2}x}{\tan x}=\frac{x}{\sin x\cos x}, \] and using the double-angle identity \[ \sin 2x=2\sin x\cos x, \] we can rewrite: \[ \frac{x}{\sin x\cos x}=\frac{2x}{2\sin x\cos x}=2x\,\mathrm{cosec}\,2x. \] Therefore, the derivative of the second term becomes: \[ \frac{d}{dx}\left[(\tan x)^{x}\right]= (\tan x)^{x}\left(\log (\tan x)+2x\,\mathrm{cosec}\,2x\right). \] Combining both derivatives, the overall derivative is: \[ \frac{d}{dx}\left[e^{\sin x}+(\tan x)^{x}\right]=e^{\sin x}\cos x+ (\tan x)^{x}\left(\log (\tan x)+2x\,\mathrm{cosec}\,2x\right). \]

Solution: To differentiate \( x^{x} \), we first express it as \[ x^{x} = e^{x\,\text{log}\,x}. \] Differentiating using the chain rule, we have \[ \frac{d}{dx}\left(e^{x\,\text{log}\,x}\right) = e^{x\,\text{log}\,x} \cdot \frac{d}{dx}(x\,\text{log}\,x). \] Now, by the product rule, \[ \frac{d}{dx}(x\,\text{log}\,x) = 1\cdot\text{log}\,x + x\cdot\frac{1}{x} = \text{log}\,x + 1. \] Thus, \[ \frac{d}{dx}\left(x^{x}\right) = x^{x}\left(1 + \text{log}\,x\right). \] Next, for \( 2^{\sin x} \), rewrite it as \[ 2^{\sin x} = e^{\sin x\,\text{log}\,2}. \] Differentiating, we obtain \[ \frac{d}{dx}\left(2^{\sin x}\right) = e^{\sin x\,\text{log}\,2} \cdot \text{log}\,2 \cdot \cos x = 2^{\sin x}\,\text{log}\,2\,\cos x. \] Combining the two results and applying the subtraction, the derivative of \( x^{x} – 2^{\sin x} \) is \[ \frac{d}{dx}\left(x^{x} – 2^{\sin x}\right) = x^{x}\left(1 + \text{log}\,x\right) – 2^{\sin x}\,\text{log}\,2\,\cos x. \]

Solution: We first rewrite the function in an exponential form for the term \( x^{x\cos x} \): \[ x^{x\cos x} = e^{x\cos x\,\text{log}\,x}. \] Let \[ u(x) = x\cos x\,\text{log}\,x. \] Differentiating \( u(x) \) using the product rule: \[ u'(x) = \frac{d}{dx}(x\cos x)\cdot \text{log}\,x + x\cos x\cdot \frac{1}{x}. \] Note that \[ \frac{d}{dx}(x\cos x) = \cos x – x\sin x. \] Hence, \[ u'(x) = (\cos x – x\sin x)\,\text{log}\,x + \cos x. \] Therefore, the derivative of \( x^{x\cos x} \) is \[ \frac{d}{dx}\left(x^{x\cos x}\right) = x^{x\cos x}\left[(\cos x – x\sin x)\,\text{log}\,x + \cos x\right]. \] Next, for the term \[ \frac{x^{2}+1}{x^{2}-1}, \] applying the quotient rule with \[ u(x) = x^{2}+1,\quad v(x) = x^{2}-1, \] we have \[ u'(x)=2x,\quad v'(x)=2x. \] Then, \[ \frac{d}{dx}\left(\frac{x^{2}+1}{x^{2}-1}\right) = \frac{2x(x^{2}-1)-2x(x^{2}+1)}{(x^{2}-1)^{2}}. \] Simplifying the numerator: \[ 2x\left[(x^{2}-1)-(x^{2}+1)\right] = 2x(-2) = -4x, \] so that \[ \frac{d}{dx}\left(\frac{x^{2}+1}{x^{2}-1}\right) = -\frac{4x}{(x^{2}-1)^{2}}. \] Combining both derivatives, the overall derivative is: \[ \frac{d}{dx}\left(x^{x\cos x} + \frac{x^{2}+1}{x^{2}-1}\right) = x^{x\cos x}\left[(\cos x – x\sin x)\,\text{log}\,x + \cos x\right] – \frac{4x}{(x^{2}-1)^{2}}. \] For a compact form, we can rewrite the expression in the bracket as: \[ (\cos x – x\sin x)\,\text{log}\,x + \cos x = (1+\text{log}\,x)\cos x – x\sin x\,\text{log}\,x. \] Hence, the derivative becomes: \[ \frac{d}{dx}\left(x^{x\cos x} + \frac{x^{2}+1}{x^{2}-1}\right) = x^{x\cos x}\left[(1+\text{log}\,x)\cos x – x\sin x\,\text{log}\,x\right] – \frac{4x}{(x^{2}-1)^{2}}. \]

Solution: First, express the first term in exponential form: \[ (\log x)^{\cos x} = e^{\cos x\,\text{log}(\log x)}. \] Let \[ u(x)=\cos x\,\text{log}(\log x). \] Differentiating \( u(x) \) using the product rule: \[ u'(x)=\frac{d}{dx}(\cos x)\cdot\text{log}(\log x)+\cos x\cdot\frac{d}{dx}\left[\text{log}(\log x)\right]. \] We have \[ \frac{d}{dx}(\cos x)=-\sin x, \] and for the second factor, \[ \frac{d}{dx}\left[\text{log}(\log x)\right] = \frac{1}{\log x}\cdot\frac{1}{x}=\frac{1}{x\,\log x}. \] Thus, \[ u'(x)=-\sin x\,\text{log}(\log x)+\frac{\cos x}{x\,\text{log} x}. \] Consequently, the derivative of the first term is: \[ \frac{d}{dx}\left[(\log x)^{\cos x}\right] = (\log x)^{\cos x}\left[-\sin x\,\text{log}(\log x)+\frac{\cos x}{x\,\text{log} x}\right]. \] Next, differentiate the second term \(\frac{x^{2}+1}{x^{2}-1}\) using the quotient rule. Let \[ u(x)=x^{2}+1,\quad v(x)=x^{2}-1. \] Then, \[ u'(x)=2x,\quad v'(x)=2x. \] By the quotient rule, \[ \frac{d}{dx}\left(\frac{x^{2}+1}{x^{2}-1}\right)=\frac{2x(x^{2}-1)-2x(x^{2}+1)}{(x^{2}-1)^{2}}. \] Simplify the numerator: \[ 2x\left[(x^{2}-1)-(x^{2}+1)\right]=2x(-2)=-4x, \] so that \[ \frac{d}{dx}\left(\frac{x^{2}+1}{x^{2}-1}\right)=-\frac{4x}{(x^{2}-1)^{2}}. \] Combining both results, we obtain: \[ \frac{dy}{dx}=(\log x)^{\cos x}\left[-\sin x\,\text{log}(\log x)+\frac{\cos x}{x\,\text{log} x}\right]-\frac{4x}{(x^{2}-1)^{2}}. \]

Solution: First, take the natural logarithm of both sides: \[ \ln\left(x^{y}y^{x}\right)=\ln\left(a^{b}\right). \] Using logarithm properties, we get: \[ y\,\text{log}\,x + x\,\text{log}\,y = b\,\text{log}\,a. \] Since \(b\,\text{log}\,a\) is constant, differentiating both sides with respect to \(x\) yields: \[ \frac{d}{dx}\left(y\,\text{log}\,x + x\,\text{log}\,y\right)=0. \] Differentiating term by term: \[ \frac{d}{dx}\left(y\,\text{log}\,x\right)= y’\,\text{log}\,x + \frac{y}{x}, \] and \[ \frac{d}{dx}\left(x\,\text{log}\,y\right)= \text{log}\,y + x\cdot\frac{y’}{y}. \] Therefore, we have: \[ y’\,\text{log}\,x + \frac{y}{x} + \text{log}\,y + \frac{x\,y’}{y}=0. \] Collecting the terms involving \(y’\): \[ y’\left(\text{log}\,x+\frac{x}{y}\right)=-\left(\frac{y}{x}+\text{log}\,y\right). \] Hence, the derivative is: \[ \frac{dy}{dx}=-\frac{\frac{y}{x}+\text{log}\,y}{\text{log}\,x+\frac{x}{y}}. \] Multiplying numerator and denominator by \(xy\) to simplify, we obtain: \[ \frac{dy}{dx}=-\frac{y\,(y+x\,\text{log}\,y)}{x\,(x+y\,\text{log}\,x)}. \]

Solution: Since \(\text{log}\,a^{b}\) is a constant, differentiate both sides with respect to \(x\): \[ \frac{d}{dx}\left(x^{y}+y^{x}\right)=0. \] Write \(x^{y}\) and \(y^{x}\) in exponential form: \[ x^{y}=e^{y\,\text{log}\,x},\quad y^{x}=e^{x\,\text{log}\,y}. \] Differentiate \(x^{y}\): \[ \frac{d}{dx}\left(x^{y}\right)=x^{y}\left(y’\,\text{log}\,x+\frac{y}{x}\right). \] Differentiate \(y^{x}\): \[ \frac{d}{dx}\left(y^{x}\right)=y^{x}\left(\text{log}\,y+\frac{x\,y’}{y}\right). \] Therefore, we have: \[ x^{y}\left(y’\,\text{log}\,x+\frac{y}{x}\right)+y^{x}\left(\text{log}\,y+\frac{x\,y’}{y}\right)=0. \] Collect the terms containing \(y’\): \[ y’\left(x^{y}\,\text{log}\,x+\frac{x\,y^{x}}{y}\right)=-\left(\frac{y\,x^{y}}{x}+y^{x}\,\text{log}\,y\right). \] Notice that \(\frac{y\,x^{y}}{x}=y\,x^{y-1}\) and \(\frac{x\,y^{x}}{y}=x\,y^{x-1}\). Thus, \[ y’\left(x^{y}\,\text{log}\,x+x\,y^{x-1}\right)=-\left(y\,x^{y-1}+y^{x}\,\text{log}\,y\right). \] Dividing both sides by \(x^{y}\,\text{log}\,x+x\,y^{x-1}\), we get: \[ y’=-\frac{y\,x^{y-1}+y^{x}\,\text{log}\,y}{x^{y}\,\text{log}\,x+x\,y^{x-1}}. \] To express the answer in a compact form, factor \(y\) in the numerator and \(x\) in the denominator: \[ y’=-\frac{y\left(x^{y-1}+y^{x-1}\,\text{log}\,y\right)}{x\left(x^{y-1}\,\text{log}\,x+y^{x-1}\right)}. \]

Solution: Differentiate both sides with respect to \(x\): \[ \frac{d}{dx}(x^x + y^x) = \frac{d}{dx}(1) \] First, differentiate \(x^x\): \[ \frac{d}{dx}(x^x) = x^x(1 + \log x) \] Now differentiate \(y^x\). Since \(y\) is a function of \(x\), use the chain rule: \[ \frac{d}{dx}(y^x) = \frac{d}{dx}(e^{x \log y}) = e^{x \log y} \left( \log y + x \cdot \frac{1}{y} \cdot \frac{dy}{dx} \right) \] \[ \Rightarrow \frac{d}{dx}(y^x) = y^x \left( \log y + \frac{x}{y} \cdot \frac{dy}{dx} \right) \] Substitute into the main equation: \[ x^x(1 + \log x) + y^x \left( \log y + \frac{x}{y} \cdot \frac{dy}{dx} \right) = 0 \] Rearranging: \[ y^x \cdot \frac{x}{y} \cdot \frac{dy}{dx} = -x^x(1 + \log x) – y^x \log y \] Solve for \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = -\frac{y \left[x^x(1 + \log x) + y^x \log y\right]}{x y^x} \] Now simplify using \( \frac{y}{y^x} = y^{1 – x} \), and write it with \( y^{x-1} \) in denominator: \[ \frac{dy}{dx} = -\frac{x^x(1 + \log x) + y^x \log y}{x y^{x – 1}} \]

Solution: Take logarithm on both sides: \[ \log\left((\sin x)^y\right) = \log(x + y). \] Using log laws: \[ y \cdot \log(\sin x) = \log(x + y). \] Now differentiate both sides with respect to \(x\). Use product rule on the left and chain rule on the right: **LHS:** \[ \frac{d}{dx} \left[ y \cdot \log(\sin x) \right] = y \cdot \frac{\cos x}{\sin x} + \log(\sin x) \cdot \frac{dy}{dx} = y \cdot \cot x + \log(\sin x) \cdot \frac{dy}{dx}. \] **RHS:** \[ \frac{d}{dx} \left[\log(x + y)\right] = \frac{1}{x + y} \cdot \left(1 + \frac{dy}{dx}\right). \] Equating both derivatives: \[ y \cot x + \log(\sin x) \cdot \frac{dy}{dx} = \frac{1 + \frac{dy}{dx}}{x + y}. \] Multiply both sides by \(x + y\): \[ (x + y) y \cot x + (x + y) \log(\sin x) \cdot \frac{dy}{dx} = 1 + \frac{dy}{dx}. \] Bring all terms to one side: \[ (x + y) \log(\sin x) \cdot \frac{dy}{dx} – \frac{dy}{dx} = 1 – (x + y) y \cot x. \] Factor \(\frac{dy}{dx}\) on the left: \[ \frac{dy}{dx} \left[(x + y) \log(\sin x) – 1 \right] = 1 – (x + y) y \cot x. \] Therefore: \[ \frac{dy}{dx} = \frac{1 – (x + y) y \cot x}{(x + y) \log(\sin x) – 1}. \]

Solution: Differentiate both sides with respect to \(x\). For the left-hand side, using the product rule: \[ \frac{d}{dx}(xy)= y+x\,\frac{dy}{dx}. \] For the right-hand side, differentiate using the chain rule: \[ \frac{d}{dx}\left(e^{x-y}\right)= e^{x-y}\cdot\frac{d}{dx}(x-y)= e^{x-y}\left(1-\frac{dy}{dx}\right). \] Equate the derivatives: \[ y+x\,\frac{dy}{dx}= e^{x-y}\left(1-\frac{dy}{dx}\right). \] Rearranging, bring the \( \frac{dy}{dx} \) terms together: \[ x\,\frac{dy}{dx}+e^{x-y}\frac{dy}{dx}= e^{x-y}-y. \] Factor out \(\frac{dy}{dx}\): \[ \frac{dy}{dx}\left(x+e^{x-y}\right)= e^{x-y}-y. \] Therefore, \[ \frac{dy}{dx}=\frac{e^{x-y}-y}{x+e^{x-y}}. \] Now, note from the original equation \(xy = e^{x-y}\), we have: \[ e^{x-y}=xy. \] Substituting this into our expression: \[ \frac{dy}{dx}=\frac{xy-y}{x+xy}=\frac{y(x-1)}{x(1+y)}. \]

Solution: First, take the natural logarithm of both sides: \[ \text{log}((\cos x)^{y}) = \text{log}((\cos y)^{x}). \] Using logarithmic properties, we have: \[ y\,\text{log}(\cos x) = x\,\text{log}(\cos y). \] Differentiate both sides with respect to \(x\). For the left-hand side, applying the product rule: \[ \frac{d}{dx}\left[y\,\text{log}(\cos x)\right] = y’\,\text{log}(\cos x) + y\cdot\frac{d}{dx}\left[\text{log}(\cos x)\right]. \] Since \[ \frac{d}{dx}\left[\text{log}(\cos x)\right] = -\tan x, \] it follows that: \[ \frac{d}{dx}\left[y\,\text{log}(\cos x)\right] = y’\,\text{log}(\cos x) – y\,\tan x. \] Similarly, for the right-hand side: \[ \frac{d}{dx}\left[x\,\text{log}(\cos y)\right] = \text{log}(\cos y) + x\cdot\frac{d}{dx}\left[\text{log}(\cos y)\right]. \] Here, \[ \frac{d}{dx}\left[\text{log}(\cos y)\right] = -\tan y\,y’, \] so that: \[ \frac{d}{dx}\left[x\,\text{log}(\cos y)\right] = \text{log}(\cos y) – x\,\tan y\,y’. \] Equating the derivatives: \[ y’\,\text{log}(\cos x) – y\,\tan x = \text{log}(\cos y) – x\,\tan y\,y’. \] Collecting the \(y’\) terms: \[ y’\,\text{log}(\cos x) + x\,\tan y\,y’ = \text{log}(\cos y) + y\,\tan x. \] Factor out \(y’\): \[ y’\left[\text{log}(\cos x) + x\,\tan y\right] = \text{log}(\cos y) + y\,\tan x. \] Finally, solving for \(y’\) yields: \[ \frac{dy}{dx} = \frac{y\,\tan x + \text{log}(\cos y)}{x\,\tan y + \text{log}(\cos x)}. \]