Q.Differentiate the function $x^{\sin x} + (\sin x)^{\cos x}$ OR If $y = 3\cos(\log x) + 4\sin(\log x)$, show that : $x^2 \dfrac{d^2y}{dx^2} + x\dfrac{dy}{dx} + y = 0$

Use logarithmic differentiation separately on each variable-exponent term, then add.

Main part. Let $y = x^{\sin x} + (\sin x)^{\cos x} = u+v$.

For $u=x^{\sin x}$: $\ln u = \sin x\ln x$. Differentiate: $\dfrac{1}{u}\dfrac{du}{dx} = \cos x\ln x + \dfrac{\sin x}{x}$, so $\dfrac{du}{dx} = x^{\sin x}\left[\cos x\ln x + \dfrac{\sin x}{x}\right]$.

For $v=(\sin x)^{\cos x}$: $\ln v = \cos x\ln(\sin x)$. Differentiate: $\dfrac{1}{v}\dfrac{dv}{dx} = -\sin x\ln(\sin x) + \cos x\cdot\dfrac{\cos x}{\sin x}$, so $\dfrac{dv}{dx} = (\sin x)^{\cos x}\left[\dfrac{\cos^2 x}{\sin x} - \sin x\ln(\sin x)\right]$.

$\dfrac{dy}{dx} = \dfrac{du}{dx}+\dfrac{dv}{dx}$.

OR (alternative part). $y=3\cos(\log x)+4\sin(\log x)$.

$y' = 3\cdot(-\sin(\log x))\cdot\dfrac1x + 4\cos(\log x)\cdot\dfrac1x = \dfrac1x\left[4\cos(\log x)-3\sin(\log x)\right]$

So $xy' = 4\cos(\log x) - 3\sin(\log x)$. Differentiate both sides w.r.t. $x$ (product rule on LHS):

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