# Differentiate the following functions with respect to x :

Question:

Differentiate the following functions with respect to $x$ :

$(x \cos x)^{x}+(x \sin x)^{1 / x}$

Solution:

Let $y=(x \cos x)^{x}+(x \sin x)^{\frac{1}{x}}$

$\Rightarrow y=a+b$

where $a=(x \cos x)^{x} ; b=(x \sin x)^{\frac{1}{x}}$

$\frac{d y}{d x}=\frac{d a}{d x}+\frac{d b}{d x}$

$\left\{\right.$ Using chain rule, $\frac{d(u+a)}{d x}=\frac{d u}{d x}+\frac{d a}{d x}$ where a and $u$ are any variables $\}$

$a=(x \cos x)^{x}$

Taking log both the sides:

$\Rightarrow \log a=\log (x \cos x)^{x}$

$\Rightarrow \log a=x \log (x \cos x)$

$\left\{\log x^{a}=a \log x\right\}$

Differentiating with respect to $x$ :

$\Rightarrow \frac{\mathrm{d}(\log \mathrm{a})}{\mathrm{dx}}=\frac{\mathrm{d}(\mathrm{x} \log (\mathrm{x} \cos \mathrm{x}))}{\mathrm{dx}}$

$\Rightarrow \frac{\mathrm{d}(\log \mathrm{a})}{\mathrm{dx}}=\mathrm{x} \times \frac{\mathrm{d}(\log (\mathrm{x} \cos \mathrm{x}))}{\mathrm{dx}}+\log (\mathrm{x} \cos \mathrm{x}) \times \frac{\mathrm{dx}}{\mathrm{dx}}$

$\left\{\right.$ Using product rule,$\left.\frac{\mathrm{d}(\mathrm{uv})}{\mathrm{dx}}=\mathrm{u} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

$\Rightarrow \frac{1}{a} \frac{d a}{d x}=x \times \frac{1}{x \cos x} \frac{d(x \cos x)}{d x}+\log (x \cos x)$

$\left\{\frac{\mathrm{d}(\log \mathrm{u})}{\mathrm{dx}}=\frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

$\Rightarrow \frac{1}{a} \frac{d a}{d x}=\frac{x}{x \cos x}\left\{x \frac{d(\cos x)}{d x}+\cos x\right\}+\log (x \cos x)$

$\left\{\right.$ Again using product rule, $\left.\frac{\mathrm{d}(\mathrm{uv})}{\mathrm{dx}}=\mathrm{u} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

$\left.\Rightarrow \frac{1}{a} \frac{d a}{d x}=\frac{1}{\cos x}\{x(-\sin x)+\cos x\}\right\}+\log (x \cos x)$

$\left\{\frac{d(\cos x)}{d x}=-\sin x\right\}$

$\Rightarrow \frac{\mathrm{da}}{\mathrm{dx}}=\mathrm{a}\left\{\frac{\cos \mathrm{x}-\mathrm{x} \sin \mathrm{x}}{\cos \mathrm{x}}+\log (\mathrm{x} \cos \mathrm{x})\right\}$

Put the value of $\mathrm{a}=(\mathrm{x} \cos \mathrm{x})^{\mathrm{x}}$ :

$\Rightarrow \frac{\mathrm{da}}{\mathrm{dx}}=(\mathrm{x} \cos \mathrm{x})^{\mathrm{x}}\left\{\frac{\cos \mathrm{x}-\mathrm{x} \sin \mathrm{x}}{\cos \mathrm{x}}+\log (\mathrm{x} \cos \mathrm{x})\right\}$

$\Rightarrow \frac{\mathrm{da}}{\mathrm{dx}}=(\mathrm{x} \cos \mathrm{x})^{\mathrm{x}}\{1-\mathrm{x} \tan \mathrm{x}+\log (\mathrm{x} \cos \mathrm{x})\}$

$\mathrm{b}=(x \sin \mathrm{x})^{\frac{1}{\mathrm{x}}}$

Taking log both the sides:

$\Rightarrow \log \mathrm{b}=\log (\mathrm{x} \sin \mathrm{x})^{\frac{1}{\mathrm{x}}}$

$\Rightarrow \log \mathrm{b}=\frac{1}{\mathrm{x}} \log (\mathrm{x} \sin \mathrm{x})\left\{\log \mathrm{x}^{\mathrm{a}}=\operatorname{alog} \mathrm{x}\right\}$

Differentiating with respect to $\mathrm{x}$ :

$\Rightarrow \frac{\mathrm{d}(\log \mathrm{b})}{\mathrm{dx}}=\frac{\mathrm{d}\left(\frac{1}{\mathrm{x}} \log (\mathrm{x} \sin \mathrm{x})\right)}{\mathrm{dx}}$

$\Rightarrow \frac{d(\log b)}{d x}=\frac{1}{x} \times \frac{d(\log (x \sin x))}{d x}+\log (x \sin x) \times \frac{d\left(x^{-1}\right)}{d x}$

$\left\{\right.$ Using product rule, $\left.\frac{\mathrm{d}(\mathrm{uv})}{\mathrm{dx}}=\mathrm{u} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

$\Rightarrow \frac{1}{b} \frac{d b}{d x}=\frac{1}{x} \times \frac{1}{x \sin x} \frac{d(x \sin x)}{d x}+\log (x \sin x)\left(-x^{-2}\right)$

$\left\{\right.$ Again using product rule, $\left.\frac{\mathrm{d}(\mathrm{uv})}{\mathrm{dx}}=\mathrm{u} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v} \frac{\mathrm{du}}{\mathrm{dx}} ; \frac{\mathrm{d}\left(\mathrm{u}^{\mathrm{n}}\right)}{\mathrm{dx}}=\mathrm{nu}^{\mathrm{n}-1} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

$\Rightarrow \frac{1}{b} \frac{d b}{d x}=\frac{1}{x^{2} \sin x}\left(x \frac{d(\sin x)}{d x}+\sin x \frac{d x}{d x}\right)-\frac{\log (x \sin x)}{x^{2}}$

$\Rightarrow \frac{\mathrm{db}}{\mathrm{dx}}=\mathrm{b}\left\{\frac{\mathrm{x} \cos \mathrm{x}+\sin \mathrm{x}}{\mathrm{x}^{2} \sin \mathrm{x}}-\frac{\log (\mathrm{x} \sin \mathrm{x})}{\mathrm{x}^{2}}\right\}$

$\left\{\frac{d(\sin x)}{d x}=\cos x\right\}$

Put the value of $b=(x \sin x)^{\frac{1}{x}}$ :

$\Rightarrow \frac{\mathrm{db}}{\mathrm{dx}}=(\mathrm{x} \sin \mathrm{x})^{\frac{1}{\mathrm{x}}}\left\{\frac{\mathrm{x} \cos \mathrm{x}+\sin \mathrm{x}}{\mathrm{x}^{2} \sin \mathrm{x}}-\frac{\log (\mathrm{x} \sin \mathrm{x})}{\mathrm{x}^{2}}\right\}$

$\Rightarrow \frac{\mathrm{db}}{\mathrm{dx}}=(\mathrm{x} \sin \mathrm{x})^{\frac{1}{\mathrm{x}}}\left\{\frac{\mathrm{x} \cot \mathrm{x}+1}{\mathrm{x}^{2}}-\frac{\log (\mathrm{x} \sin \mathrm{x})}{\mathrm{x}^{2}}\right\}$

$\Rightarrow \frac{\mathrm{db}}{\mathrm{dx}}=(\mathrm{x} \sin \mathrm{x})^{\frac{1}{x}}\left\{\frac{\mathrm{x} \cot \mathrm{x}+1-\log (\mathrm{x} \sin \mathrm{x})}{\mathrm{x}^{2}}\right\}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{da}}{\mathrm{dx}}+\frac{\mathrm{db}}{\mathrm{dx}}$

$\Rightarrow \frac{d y}{d x}=(x \cos x)^{x}\{1-x \tan x+\log (x \cos x)\}$

$+(x \sin x)^{\frac{1}{x}}\left\{\frac{x \cot x+1-\log (x \sin x)}{x^{2}}\right\}$