Question
Find $\frac{d y}{d x}$ if : $\mathrm{y}=\frac{(\log x+1)}{x}$

Answer

$
y=\frac{(\log x+1)}{x}
$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
\frac{\mathrm{d} y}{\mathrm{~d} x} & =\frac{\mathrm{d}}{\mathrm{d} x}\left[\frac{\log x+1}{x}\right] \\
& =\frac{x \frac{\mathrm{d}}{\mathrm{d} x}(\log x+1)-(\log x+1) \frac{\mathrm{d}}{\mathrm{d} x}(x)}{x^2} \\
& =\frac{x\left(\frac{1}{x}+0\right)-(\log x+1)(1)}{x^2} \\
& =\frac{1-\log x-1}{x^2} \text { } \\
& =\frac{-\log x}{x^2}
\end{aligned}
$

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