If y = (sin x)^{(sin x)^{(sin x)^{...infty}}} | vedclass

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Question

(A) Given $y = (\sin x)^{(\sin x)^{(\sin x)^{...\infty}}}$.Since the exponent is an infinite series,we can write $y = (\sin x)^y$.Taking the natural l

Vedclass · Accepted Answer

$\frac{y^2 \cot x}{1 - y \log \sin x}$

Vedclass · Answer

(A) Given $y = (\sin x)^{(\sin x)^{(\sin x)^{...\infty}}}$.Since the exponent is an infinite series,we can write $y = (\sin x)^y$.Taking the natural logarithm on both sides,we get $\ln y = y \ln(\sin x)$.Differentiating both sides with respect to $x$ using the product rule:$\frac{1}{y} \frac{dy}{dx} = \frac{dy}{dx} \ln(\sin x) + y \cdot \frac{1}{\sin x} \cdot \cos x$.$\frac{1}{y} \frac{dy}{dx} - \frac{dy}{dx} \ln(\sin x) = y \cot x$.$\frac{dy}{dx} \left( \frac{1}{y} - \ln(\sin x) \right) = y \cot x$.$\frac{dy}{dx} \left( \frac{1 - y \ln(\sin x)}{y} \right) = y \cot x$.Therefore,$\frac{dy}{dx} = \frac{y^2 \cot x}{1 - y \ln(\sin x)}$.

If $y = (\sin x)^{(\sin x)^{(\sin x)^{...\infty}}}$,then $\frac{dy}{dx} = $

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