If y = (sin x)^{tan x},then frac{dy}{dx} is e | vedclass

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(A) Given $y = (\sin x)^{	an x}$.Taking logarithm on both sides,we get $\log y = 	an x \cdot \log(\sin x)$.Differentiating both sides with respect t

Vedclass · Accepted Answer

$(\sin x)^{	an x} (1 + \sec^2 x \cdot \log \sin x)$

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(A) Given $y = (\sin x)^{	an x}$.Taking logarithm on both sides,we get $\log y = 	an x \cdot \log(\sin x)$.Differentiating both sides with respect to $x$ using the product rule:$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(	an x) \cdot \log(\sin x) + 	an x \cdot \frac{d}{dx}(\log(\sin x))$.$\frac{1}{y} \frac{dy}{dx} = \sec^2 x \cdot \log(\sin x) + 	an x \cdot \frac{1}{\sin x} \cdot \cos x$.Since $\frac{\cos x}{\sin x} = \cot x$,we have $	an x \cdot \cot x = 1$.Therefore,$\frac{1}{y} \frac{dy}{dx} = \sec^2 x \cdot \log(\sin x) + 1$.$\frac{dy}{dx} = y [1 + \sec^2 x \cdot \log(\sin x)]$.Substituting $y = (\sin x)^{	an x}$,we get $\frac{dy}{dx} = (\sin x)^{	an x} [1 + \sec^2 x \cdot \log(\sin x)]$.

$X = x$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X = x)$	$0.15$	$0.23$	$K$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$

If $y = (\sin x)^{\tan x}$,then $\frac{dy}{dx}$ is equal to

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