TS EAMCET · Maths · Differentiation
Derivative of \((\sin x) x\) with respect to \(x^{(\sin x)}\) is
- 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 \(\frac{x^{\sin x}[x \cos x(\log x)+\sin x]}{(\sin x)^x[(\sin x) \log (\sin x)+x \cos x]}\)
Answer & Solution
Correct Answer
(A) \(\frac{(\sin x)^{x-1}[(\sin x) \log (\sin x)+x \cos x]}{x^{(\sin x-1)}[x \cos x(\log x)+\sin x]}\)
Step-by-step Solution
Detailed explanation
\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…
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