Question

If $y=\left[\frac{x^2+1}{x+1}\right]$, then find $\frac{dy}{dx}$.

Answer 1

Here, $u\left(x\right)=x^2+1,v\left(x\right)=x+1$.
Using quotient rule, we get
$\frac{dy}{dx}=\frac{(x+1)\frac{d(x^2+1)}{dx}-(x^2+1)\frac{d(x+1)}{dx}}{(x+1{)}^2}$
$=\frac{(x+1)2x-(x^2+1)1}{(x+1{)}^2}$
$=\frac{2x^2+2x-x^2-1}{(x+1{)}^2}=\frac{x^2+2x-1}{(x+1{)}^2}$.