Q106 / 160 · TS EAMCET 2024 (11 May Shift 1) · Mathematics

Derivative of $(\sin x) x$ with respect to $x^{(\sin x)}$ is

  • Correct answer
    A
    $\frac{(\sin x)^{x-1}[(\sin x) \log (\sin x)+x \cos x]}{x^{(\sin x-1)}[x \cos x(\log x)+\sin x]}$
  • B
    $\frac{(\sin x)^x[(\sin x) \log (\sin x)+x \cos x]}{x^{(\sin x)}[x \cos x(\log x)+\sin x]}$
  • C
    $\frac{x^{\sin x-1}[x \cos x(\log x)+\sin x]}{(\sin x)^{x-1}[(\sin x) \log (\sin x)+x \cos x]}$
  • D
    $\frac{x^{\sin x}[x \cos x(\log x)+\sin x]}{(\sin x)^x[(\sin x) \log (\sin x)+x \cos x]}$

Solution

$\begin{aligned} & u=(\sin x) x \\ & \log u=x \log (\sin x) \\ & \frac{1}{u} \frac{d u}{d x}=\log (\sin x)+\frac{x \cos x}{\sin x} \\ & \Rightarrow \frac{d u}{d x}=(\sin x)^{x-1}[\sin x \log (\sin x)+x \cos x] \\ & v=x^{\sin x} \Rightarrow \log v=\sin x \log x \\ & \Rightarrow \frac{1}{v} \frac{d v}{d x}=\frac{\sin x}{x}+\cos x \log x \\ & \Rightarrow \frac{d v}{d x}=x^{\sin x-1}(\sin x+x \cos x \log x) \\ & \therefore \frac{d u}{d v}=\frac{(\sin x)^{x-1}[\sin x \log (\sin x)+x \cos x]}{x^{\sin x-1}(x \cos x \log x+\sin x)}\end{aligned}$

Hence, the correct answer is option: 'A'.

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