If y = x^n log x + x(log x)^n,then frac{dy}{d | vedclass

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(A) Given $y = x^n \log x + x(\log x)^n$. Applying the product rule $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$: $\frac{dy}{dx} = \frac{d}{

Vedclass · Accepted Answer

$x^{n-1}(1 + n \log x) + (\log x)^{n-1}(n + \log x)$

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(A) Given $y = x^n \log x + x(\log x)^n$. Applying the product rule $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$: $\frac{dy}{dx} = \frac{d}{dx}(x^n \log x) + \frac{d}{dx}(x(\log x)^n)$ $\frac{dy}{dx} = [x^n \cdot \frac{1}{x} + (\log x) \cdot nx^{n-1}] + [x \cdot n(\log x)^{n-1} \cdot \frac{1}{x} + (\log x)^n \cdot 1]$ $\frac{dy}{dx} = [x^{n-1} + nx^{n-1} \log x] + [n(\log x)^{n-1} + (\log x)^n]$ $\frac{dy}{dx} = x^{n-1}(1 + n \log x) + (\log x)^{n-1}(n + \log x)$.

If $y = x^n \log x + x(\log x)^n$,then $\frac{dy}{dx} = $

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