If y(cos x)^{sin x}=(sin x)^{sin x}, then the | vedclass

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(C) Given equation: $y(\cos x)^{\sin x} = (\sin x)^{\sin x}$ Dividing both sides by $(\cos x)^{\sin x}$, we get: $y = \frac{(\sin x)^{\sin x}}{(\cos x

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$\sqrt{2}$

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(C) Given equation: $y(\cos x)^{\sin x} = (\sin x)^{\sin x}$ Dividing both sides by $(\cos x)^{\sin x}$, we get: $y = \frac{(\sin x)^{\sin x}}{(\cos x)^{\sin x}} = (	an x)^{\sin x}$ Taking natural logarithm on both sides: $\ln y = \sin x \cdot \ln(	an x)$ Differentiating with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = \cos x \cdot \ln(	an x) + \sin x \cdot \frac{1}{	an x} \cdot \sec^2 x$ Simplifying the second term: $\sin x \cdot \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\cos x} = \sec x$ So, $\frac{dy}{dx} = y [\cos x \cdot \ln(	an x) + \sec x]$ At $x = \frac{\pi}{4}$, $	an x = 1$, $\sin x = \frac{1}{\sqrt{2}}$, $\cos x = \frac{1}{\sqrt{2}}$, $\sec x = \sqrt{2}$, and $y = (1)^{1/\sqrt{2}} = 1$ Substituting these values: $\left. \frac{dy}{dx} ight|_{x=\frac{\pi}{4}} = 1 \cdot [\frac{1}{\sqrt{2}} \cdot \ln(1) + \sqrt{2}] = 1 \cdot [0 + \sqrt{2}] = \sqrt{2}$

If $y(\cos x)^{\sin x}=(\sin x)^{\sin x}$, then the value of $\frac{dy}{dx}$ at $x=\frac{\pi}{4}$ is

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