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

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

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

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

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