If $y=\frac{e^x \log x}{x^2}$, then $\frac{d y}{d x}=$
If $y=\frac{e^x \log x}{x^2}$, then $\frac{d y}{d x}=$
- $\frac{e^x\{1+(x+2) \log x\}}{x^3}$
- $\frac{e^x\{1-(x-2) \log x\}}{x^4}$
- $\frac{e^x\{1-(x-2) \log x\}}{x^3}$
- $\frac{e^x\{1+(x-2) \log x\}}{x^3}$
Solution
It is given that,
$
y=\frac{e^x \log x}{x^2}
$
On applying logarithm both sides, we get
$
\log _e y=x+\log _e(\log x)-2 \log x
$
Now, on differentiating both sides w.r.t ' $x$ ', we get
$
\begin{aligned}
\frac{1}{y}\left(\frac{d y}{d x}\right) & =1+\frac{1}{x \log x}-\frac{2}{x} \\
\Rightarrow \quad \frac{d y}{d x} & =y\left(1+\frac{1}{x \log x}-\frac{2}{x}\right)
\end{aligned}
$
$
\begin{aligned}
& =\frac{e^x \log x}{x^2}\left(\frac{x \log x+1-2 \log x}{x \log x}\right) \\
& =\frac{e^x}{x^3}[1+(x-2) \log x]
\end{aligned}
$
Hence, option (4) is correct
Asked in: AP EAMCET 2020 (22 Sep Shift 1)
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