If x^y=y^{sin x}(tan x)^{cos x},then left(log | vedclass

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(A) Given the equation: $x^y = y^{\sin x} (	an x)^{\cos x}$ Taking the natural logarithm on both sides: $y \log x = \sin x \log y + \cos x \log (	an

Vedclass · Accepted Answer

$\cos x \log y-\sin x \log (	an x)+\operatorname{cosec} x-\frac{y}{x}$

Vedclass · Answer

(A) Given the equation: $x^y = y^{\sin x} (	an x)^{\cos x}$ Taking the natural logarithm on both sides: $y \log x = \sin x \log y + \cos x \log (	an x)$ Differentiating both sides with respect to $x$: $\frac{d}{dx}(y \log x) = \frac{d}{dx}(\sin x \log y) + \frac{d}{dx}(\cos x \log 	an x)$ $\frac{dy}{dx} \log x + y \cdot \frac{1}{x} = (\cos x \log y + \sin x \cdot \frac{1}{y} \frac{dy}{dx}) + (-\sin x \log 	an x + \cos x \cdot \frac{1}{	an x} \cdot \sec^2 x)$ Simplifying the derivative term: $\cos x \cdot \frac{1}{	an x} \cdot \sec^2 x = \cos x \cdot \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\sin x} = \operatorname{cosec} x$ Rearranging the terms to isolate $\left(\log x - \frac{\sin x}{y}ight) \frac{dy}{dx}$: $\left(\log x - \frac{\sin x}{y}ight) \frac{dy}{dx} = \cos x \log y - \sin x \log (	an x) + \operatorname{cosec} x - \frac{y}{x}$

If $x^y=y^{\sin x}(\tan x)^{\cos x}$,then $\left(\log x-\frac{\sin x}{y}\right) \frac{d y}{d x}=$

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