If y = e^x log x,then frac{dy}{dx} is | vedclass

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Question

(C) Given the function $y = e^x \log x$. To find $\frac{dy}{dx}$,we use the product rule of differentiation: $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \

Vedclass · Accepted Answer

$e^x \left( \frac{1}{x} + \log x \right)$

Vedclass · Answer

(C) Given the function $y = e^x \log x$. To find $\frac{dy}{dx}$,we use the product rule of differentiation: $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$. Here,$u = e^x$ and $v = \log x$. Then,$\frac{du}{dx} = e^x$ and $\frac{dv}{dx} = \frac{1}{x}$. Applying the product rule: $\frac{dy}{dx} = e^x \cdot \frac{d}{dx}(\log x) + \log x \cdot \frac{d}{dx}(e^x)$ $\frac{dy}{dx} = e^x \cdot \frac{1}{x} + \log x \cdot e^x$ Factoring out $e^x$,we get: $\frac{dy}{dx} = e^x \left( \frac{1}{x} + \log x \right)$. Thus,the correct option is $C$.

If $y = e^x \log x$,then $\frac{dy}{dx}$ is

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