If y = x^ x , then dy dx = - Vidyadip

MCQ

If $y = {x^{\sin x}},$ then ${{dy} \over {dx}} = $

✓
${{x\cos x.\log x + \sin x} \over x}.{x^{\sin x}}$
B
${{y[x\cos x.\log x + \cos x]} \over x}$
C
$y[x\sin x.\log x + \cos x]$
D
None of these

✓

Answer

Correct option: A.

${{x\cos x.\log x + \sin x} \over x}.{x^{\sin x}}$

a
(a) $y = {x^{\sin x}} \Rightarrow {\log _e}y = \sin x{\log _e}x$

$\therefore$ ${{dy} \over {dx}} = {x^{\sin x}}\left[ {\frac{{\sin x + x\cos x{{\log }_e}x}}{x}} \right]$.

$={{x}^{\sin x}}\left[ \frac{\sin x+x\cos x{{\log }_{e}}x}{x} \right]$

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