Find frac{dy}{dx} for the function (cos x)^{y | vedclass

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Question

(C) Given the function $(\cos x)^{y}=(\cos y)^{x}$.Taking the natural logarithm on both sides:$y \log \cos x = x \log \cos y$Differentiating both side

Vedclass · Accepted Answer

$\frac{y 	an x + \log \cos y}{x 	an y + \log \cos x}$

Vedclass · Answer

(C) Given the function $(\cos x)^{y}=(\cos y)^{x}$.Taking the natural logarithm on both sides:$y \log \cos x = x \log \cos y$Differentiating both sides with respect to $x$ using the product rule:$\frac{d}{dx}(y \log \cos x) = \frac{d}{dx}(x \log \cos y)$$\frac{dy}{dx} \log \cos x + y \cdot \frac{1}{\cos x} \cdot (-\sin x) = 1 \cdot \log \cos y + x \cdot \frac{1}{\cos y} \cdot (-\sin y) \cdot \frac{dy}{dx}$$\frac{dy}{dx} \log \cos x - y 	an x = \log \cos y - x 	an y \frac{dy}{dx}$Rearranging the terms to isolate $\frac{dy}{dx}$:$\frac{dy}{dx} \log \cos x + x 	an y \frac{dy}{dx} = \log \cos y + y 	an x$$\frac{dy}{dx} (\log \cos x + x 	an y) = y 	an x + \log \cos y$Therefore,$\frac{dy}{dx} = \frac{y 	an x + \log \cos y}{x 	an y + \log \cos x}$.

Find $\frac{dy}{dx}$ for the function $(\cos x)^{y}=(\cos y)^{x}$.

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