Question
If $y = e ^{1+\log x}$ then find $\frac{ d y}{ d x}$
✓
Answer
$ y = e ^{1+\log x}$
$= e \cdot e ^{\log x}$
$= e \cdot x$
$\therefore \frac{ d y}{ d x}= e \cdot 1= e$
$\text { OR }$
$y = e ^{1+\log x}$
$\frac{ d y}{ d x}=\frac{ d }{ d x}\left( e ^{1+\log x}\right)$
$= e ^{1+\log x} \cdot \frac{ d }{ d x}(1+\log x)$
$= e ^{1+\log x} \cdot\left(0+\frac{1}{x}\right)$
$=\frac{ e ^{1+\log x}}{x}$
$= e \cdot e ^{\log x}$
$= e \cdot x$
$\therefore \frac{ d y}{ d x}= e \cdot 1= e$
$\text { OR }$
$y = e ^{1+\log x}$
$\frac{ d y}{ d x}=\frac{ d }{ d x}\left( e ^{1+\log x}\right)$
$= e ^{1+\log x} \cdot \frac{ d }{ d x}(1+\log x)$
$= e ^{1+\log x} \cdot\left(0+\frac{1}{x}\right)$
$=\frac{ e ^{1+\log x}}{x}$
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