Question

y=xlogx/x+log x..differentiathe the following w. r. t. x

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Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{y=\dfrac{x\,log\,x}{x+log\,x}}$

$\underline{\textbf{To find:}}$

$\textsf{Derivate of}\;\mathsf{y=\dfrac{x\,log\,x}{x+log\,x}}$

$\underline{\textbf{Quotient rule:}}$

$\mathsf{\dfrac{d\left(\frac{u}{v}\right)}{dx}=\dfrac{v\,\dfrac{du}{dx}-u\,\dfrac{dv}{dx}}{v^2}}$

$\textsf{Consider}$

$\mathsf{y=\dfrac{x\,log\,x}{x+log\,x}}$

$\textsf{Differentiate with respect to "x"}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{(x+log\,x)\left(x.\dfrac{1}{x}+log\,x.1\right)-(x\,log\,x)\left(1+\dfrac{1}{x}\right)}{(x+log\,x)^2}}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{(x+log\,x)(1+log\,x)-(x\,log\,x)\left(1+\dfrac{1}{x}\right)}{(x+log\,x)^2}}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{x+x\,logx+logx+(logx)^2-x\,logx-logx}{(x+log\,x)^2}}$

$\implies\boxed{\mathsf{\dfrac{dy}{dx}=\dfrac{x+(logx)^2}{(x+log\,x)^2}}}$