Question #cab31

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Question #cab31

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Answer 1 · 2018-01-26T16:42:34

$- cot x . ln x - \frac{1}{x}$ $-cotx. lnx -1/x$

Explanation:

$y = ln (x^{sin x})$ $y = ln (x^sinx)$
$y = sin x ln x$ $y= sinx lnx$

Differentiating wrt to $cos x$ $cosx$ :

Using product rule:

$\frac{d y}{d (cos x)} = \frac{d sin x}{d (cos x)} . ln x + \frac{d ln x}{d (cos x)} . sin x$ $dy/(d(cosx)) = (dsinx) /(d (cosx)). lnx + (dlnx)/(d(cosx)). sinx$

Differentiating $sin x$ $sinx$ wrt to $cos x$ $cosx$ is equivalent to differenting both wrt to $x$ $x$
i.e $\frac{\frac{d sin x}{d x}}{\frac{d cos x}{d x}}$ $((dsinx)/dx )/((dcosx)/dx)$

$\frac{d y}{d (cos x)} = \frac{cos x}{- sin x} . ln x + \frac{\frac{1}{x}}{- sin x} . sin x$ $dy/(d(cosx)) = cosx /-sinx. lnx + (1/x)/-sinx. sinx$

$\frac{d y}{d (cos x)} = - cot x . ln x - \frac{1}{x}$ $dy/(d(cosx)) = -cotx. lnx - 1/x$

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Question #cab31

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