If y = log x cdot e^{(tan x + x^2)},then frac | vedclass

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(C) Given $y = \log x \cdot e^{(	an x + x^2)}$.Using the product rule $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$:$\frac{dy}{dx} = \log x

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$e^{(	an x + x^2)} \left[ \frac{1}{x} + (\sec^2 x + 2x) \log x ight]$

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(C) Given $y = \log x \cdot e^{(	an x + x^2)}$.Using the product rule $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$:$\frac{dy}{dx} = \log x \cdot \frac{d}{dx}(e^{(	an x + x^2)}) + e^{(	an x + x^2)} \cdot \frac{d}{dx}(\log x)$.Applying the chain rule for the exponential term:$\frac{dy}{dx} = \log x \cdot e^{(	an x + x^2)} \cdot (\sec^2 x + 2x) + e^{(	an x + x^2)} \cdot \frac{1}{x}$.Factoring out $e^{(	an x + x^2)}$:$\frac{dy}{dx} = e^{(	an x + x^2)} \left[ \frac{1}{x} + (\sec^2 x + 2x) \log x ight]$.

If $y = \log x \cdot e^{(\tan x + x^2)}$,then $\frac{dy}{dx} = $

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