Differentiate the following w.r.t. x :y=x x x+ x - Vidyadip

Question

Differentiate the following w.r.t. x :$y=\frac{x \log x}{x+\log x}$

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Answer

$y=\frac{x \log x}{x+\log x}$
Differentiating w.r.t. $x$, we get
$\begin{aligned}
& \frac{ d y}{ d x}=\frac{ d }{ d x}\left(\frac{x \log x}{x+\log x}\right) \\
= & \frac{(x+\log x) \frac{ d }{ d x}(x \log x)-x \log x \frac{ d }{ d x}(x+\log x)}{(x+\log x)^2} \\
= & \frac{(x+\log x)\left(x \frac{ d }{ d x} \log x+\log x \frac{ d }{ d x} x\right)-x \log x\left(\frac{ d }{ d x} x+\frac{ d }{ d x} \log x\right)}{(x+\log x)^2} \\
= & \frac{(x+\log x)\left[x\left(\frac{1}{x}\right)+\log x(1)\right]-x \log x\left(1+\frac{1}{x}\right)}{(x+\log x)^2} \\
= & \frac{(x+\log x)(1+\log x)-x \log x\left(1+\frac{1}{x}\right)}{(x+\log x)^2} \\
= & \frac{x+(\log x)^2}{(x+\log x)^2}
\end{aligned}$

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