Question

Differentiate the following function w.r.t.x.
$\frac{1}{(x^2+3^2)}$

Answer 1

$\frac{d}{dr}\frac{1}{x^2+3^2}$

$=\frac{d}{dr}\frac{1}{x^2+9}$

$=0$

$\frac{d}{dr}\frac{1}{x^2+3^2}=0$

Considering the question to be,

$\frac{d}{dx}\frac{1}{x^2+3^2}$

$=\frac{d}{dx}\frac{1}{x^2+9}$

$=\frac{d}{dx}[x^2+9]\frac{1}{(x^2+9{)}^2}$

$=-\frac{2x+0}{(x^2+9{)}^2}$

$=-\frac{2x}{(x^2+9{)}^2}$

$\frac{d}{dx}\frac{1}{x^2+3^2}=-\frac{2x}{(x^2+3^2{)}^2}$