The derivative of $x^{\sin x}$ is, x>0
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The derivative of $x^{\sin x}$ is, $x>0$

(A) $x^{\sin x-1} \sin x-x^{\sin x} \cos x \log x$

(B) $x^{\sin x-1} \sin x+x^{\sin x} \cos x \log x$

(C) $x^{\sin x-1} \sin x$

(D) None of the above

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Best answer

SOLUTION —

Let $y=x^{\sin x}$

$\Rightarrow  \quad \log y  =\sin x \log x \\$

$\Rightarrow   \frac{1}{y} \frac{d y}{d x}  =\frac{\sin x}{x}+\log x \cos x \\$

$\Rightarrow  \frac{d y}{d x}  =x^{\sin x}\left[\frac{\sin x}{x}+\cos x \log x\right] \\$

$ =x^{\sin x-1} \sin x+x^{\sin x} \cos x \log x$

So, The correct option will be (B).

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