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Mathematics Question on Logarithmic Differentiation

The derivative of y=xsin⁡ xy = x^{\sin\,x} is

A

cos⁡ x xsin⁡ x−1\cos\,x\,x^{\sin\,x-1}

B

sin⁡ 2x2xsin⁡ x−1\frac{\sin\,2x}{2} x^{\sin\,x-1}

C

xsin⁡ x(cos⁡ x log⁡ x+sin⁡ xx)x^{\sin\,x}\left(\cos\,x\, \log\,x+\frac{\sin\,x}{x}\right)

D

cos⁡ x log⁡ x+sin⁡ xx\cos\,x\, \log\,x+\frac{\sin\,x}{x}

Answer

xsin⁡ x(cos⁡ x log⁡ x+sin⁡ xx)x^{\sin\,x}\left(\cos\,x\, \log\,x+\frac{\sin\,x}{x}\right)

Explanation

Solution

The correct option is(C): xsin⁡x(sin⁡xx+log⁡xcos⁡x)x^{\sin x}\left(\frac{\sin x}{x}+\log x \cos x\right).

Given, y=xsin⁡xy=x^{\sin x}
On taking log both sides, we get
log⁡y=sin⁡xlog⁡x\log y=\sin x \log x
On differentiating both sides w.r.t. xx, we get
1y⋅dydx=sin⁡xddxlog⁡x+log⁡xddxsin⁡x\frac{1}{y} \cdot \frac{d y}{d x}= \sin x \frac{d}{d x} \log x+\log x \frac{d}{d x} \sin x
⇒1y⋅dydx=sin⁡x⋅1x+log⁡x⋅cos⁡x\Rightarrow \frac{1}{y} \cdot \frac{d y}{d x}=\sin x \cdot \frac{1}{x}+\log x \cdot \cos x
⇒dydx=y(sin⁡xx+log⁡x⋅cos⁡x)\Rightarrow \frac{d y}{d x}=y\left(\frac{\sin x}{x}+\log x \cdot \cos x\right)
⇒dydx=xsin⁡x(sin⁡xx+log⁡xcos⁡x)\Rightarrow \frac{d y}{d x}=x^{\sin x}\left(\frac{\sin x}{x}+\log x \cos x\right)