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Question:

Find $\frac{\mathrm{dy}}{\mathrm{dx}}$, when

$y=x^{x}+(\sin x)^{x}$

Solution:

let $y=x^{x}+(\sin x)^{x}$

$\Rightarrow y=a+b$

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

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

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

$a=x^{x}$

Taking log both the sides:

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

$\Rightarrow \log a=x \log x$

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

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

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

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

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

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

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

$\Rightarrow \frac{1}{\mathrm{a}} \frac{\mathrm{da}}{\mathrm{dx}}=1+\log \mathrm{x}$

$\Rightarrow \frac{\mathrm{da}}{\mathrm{dx}}=\mathrm{a}\{1+\log \mathrm{x}\}$

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

$\Rightarrow \frac{d a}{d x}=x^{x}\{1+\log x\}$

$b=(\sin x)^{x}$

Taking log both the sides:

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

$\Rightarrow \log b=x \log (\sin x)$

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

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

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

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

$\left\{\right.$ Using product rule, $\left.\frac{d(u v)}{d x}=u \frac{d v}{d x}+v \frac{d u}{d x}\right\}$

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

$\left\{\frac{d(\log u)}{d x}=\frac{1}{u} \frac{d u}{d x} ; \frac{d(\sin x)}{d x}=\cos x\right\}$

$\Rightarrow \frac{1}{b} \frac{d b}{d x}=\frac{x}{\sin x}(\cos x)+\log (\sin x)$

$\left\{\cot x=\frac{\cos x}{\sin x}\right\}$

$\Rightarrow \frac{\mathrm{d} \mathrm{b}}{\mathrm{dx}}=\mathrm{b}\{\mathrm{x} \cot \mathrm{x}+\log (\sin \mathrm{x})\}$

$\Rightarrow \frac{\mathrm{db}}{\mathrm{dx}}=\mathrm{b}\{\mathrm{x} \cot \mathrm{x}+\log (\sin \mathrm{x})\}$

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

$\Rightarrow \frac{\mathrm{db}}{\mathrm{dx}}=(\sin \mathrm{x})^{\mathrm{x}}\{x \cot \mathrm{x}+\log (\sin \mathrm{x})\}$

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

$\Rightarrow \frac{d y}{d x}=x^{x}\{1+\log x\}+(\sin x)^{x}\{x \cot x+\log (\sin x)\}$

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