Question

How to differentiate ${\left(\frac{kQx}{(x^2+r^2)}\right)}^{3/2}$ with respect to dx when k, Q, r are constants.

Answer 1

$\frac{d}{dx}\left(\frac{kQx}{(x^2+r^2{)}^{3/2}}\right)$
$=kQ\frac{d}{dx}\frac{x}{(x^2+r^2{)}^{3/2}}$
Use Quotient rule
$\frac{d}{dx}\frac{f\left(x\right)}{g\left(x\right)}=\frac{f^{\prime }\left(x\right)g\left(x\right)-f\left(x\right)g^{\prime }\left(x\right)}{g(x{)}^2}$
$=kQ\left(\frac{(x^2+r^2{)}^{\frac{3}{2}}[1]-x\left[\frac{3}{2}\right(x^2+r^2{)}^{\frac{1}{2}}.2x]}{(x^2+r^2{)}^3}\right)$
$=kQ\left[\frac{(x^2+r^2{)}^{\frac{3}{2}}-3x^2[x^2+r^2{]}^{\frac{1}{2}}}{(x^2+r^2{)}^3}\right]$
$=kQ\left[\frac{(x^2+r^2{)}^{\frac{1}{2}}[x^2+r^2-3x^2]}{(x^2+r^2{)}^3}\right]$
$=kQ\frac{[x^2+r^2-2x^2]}{(x^2+r^2{)}^{\frac{5}{2}}}$.