Derivative with the quotient rule

Statement

Let $f(x)=\dfrac{2x-1}{x^2+3}$.

• (a) Find $f'(x)$ using the quotient rule.
• (b) Compute $f'(0)$.

Solution

(a) Quotient rule $\left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^2}$ with $u=2x-1$, $v=x^2+3$: $f'(x)=\frac{2(x^2+3)-(2x-1)(2x)}{(x^2+3)^2}=\frac{2x^2+6-4x^2+2x}{(x^2+3)^2}=\frac{-2x^2+2x+6}{(x^2+3)^2}.$

(b) $f'(0)=\dfrac{6}{9}=\dfrac{2}{3}\approx 0{.}667$.

$\boxed{f'(x)=\dfrac{-2x^2+2x+6}{(x^2+3)^2};\qquad f'(0)=\tfrac23}$