If y^{cos x}=x^{sin y}, then frac{d y}{d x}= | vedclass

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(A) Given equation: $y^{\cos x}=x^{\sin y}$ Taking logarithm on both sides: $\cos x \log y = \sin y \log x$ Differentiating both sides with respect to

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

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(A) Given equation: $y^{\cos x}=x^{\sin y}$ Taking logarithm on both sides: $\cos x \log y = \sin y \log x$ Differentiating both sides with respect to $x$ using the product rule: $\frac{d}{dx}(\cos x \log y) = \frac{d}{dx}(\sin y \log x)$ $(\cos x) \cdot \frac{1}{y} \frac{dy}{dx} + (\log y) \cdot (-\sin x) = (\sin y) \cdot \frac{1}{x} + (\log x) \cdot (\cos y) \frac{dy}{dx}$ Rearranging the terms to group $\frac{dy}{dx}$: $\frac{dy}{dx} \left( \frac{\cos x}{y} - \cos y \log x \right) = \frac{\sin y}{x} + \sin x \log y$ $\frac{dy}{dx} \left( \frac{\cos x - y \cos y \log x}{y} \right) = \frac{\sin y + x \sin x \log y}{x}$ $\frac{dy}{dx} = \frac{y(\sin y + x \sin x \log y)}{x(\cos x - y \cos y \log x)}$

If $y^{\cos x}=x^{\sin y}$, then $\frac{d y}{d x}=$

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