The rate of change of $x^{\sin x}$ with respect to $(\sin x)^{\mathrm{x}}$ is

The rate of change of $x^{\sin x}$ with respect to $(\sin x)^{\mathrm{x}}$ is
  1. $\frac{x^{\sin x}\left(\frac{\sin x}{x}+\cos x \cdot \log x\right)}{(\sin x)^x(x \cdot \cot x+\log \sin x)}$
  2. $\frac{x^{\sin x}(x \cot x+\log \sin x)}{x^{\sin x}\left(\frac{\sin x}{x}+\cos x \cdot \log x\right)}$
  3. $y\left(\frac{\sin x}{x}+\cos x \cdot \log x\right)$
  4. $(\sin x)^{\mathrm{x}}(x \cot x+\log \sin x)$

Solution

Let $\mathrm{u}=x^{\sin x} \Rightarrow \log \mathrm{u}=\sin x \log x^{\prime}$
Differentiating w.r.t. $x$ $\begin{aligned} & \frac{1}{u} \frac{d u}{d x}=\cos x \log x+\frac{\sin x}{x} \\ & \Rightarrow \frac{d u}{d x}=x^{\sin x}\left[\cos x \log x+\frac{\sin x}{x}\right] \end{aligned}$
Let $\mathrm{v}=(\sin x)^x \Rightarrow \log \mathrm{v}=x \log \sin x$ Differentiating w.r.t. $x$ $\begin{aligned} & \frac{1}{v} \frac{d v}{d x}=\log \sin x+x \cot x \\ & \frac{d v}{d x}=(\sin x)^x[x \cot x+\log \sin x] \end{aligned}$
Now, $\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{x^{\sin x}\left[\cos x \log x+\frac{\sin x}{x}\right]}{(\sin x)^x[x \cot x+\log \sin x]}$

Asked in: AP EAMCET 2024 (21 May Shift 1)

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