Question

If $y=\frac{(x+1{)}^2\sqrt{x-1}}{(x+4{)}^2{e}^x}$, then show that $\frac{dy}{dx}=\frac{x+1^2\sqrt{(x-1)}}{(x+4)e^x}[\frac{5x-3}{2(x^2-1)}+\frac{x+6}{x+4}]$

Answer 1

$y=\frac{(x+1{)}^2\sqrt{x-1}}{(x+4{)}^2{e}^x}$
Taking log on both sides, we get

$\log y=\log \left[\right(x+1{)}^2\sqrt{x-1}]-\log \left[\right(x+4{)}^2{e}^x]$

$\log y=2\log (1+x)+\frac{1}{2}\log (x-1)-2\log (x+4)-x$
Differentiating both sides w.r.t. $x$
$\frac{1}{y}\frac{dy}{dx}=\frac{2}{1+x}+\frac{1}{2(x-1)}-\frac{2}{x+4}-1$

$\therefore \frac{dy}{dx}=\frac{(x+1{)}^2\sqrt{(x-1)}}{(x+4)e^x}[\frac{5x-3}{2(x^2-1)}-\frac{x+6}{x+4}]$