Chapter 3 - Applications of Differentiation - 3.1 Exercises - Page 167: 1

$f'(0)=0.$

Using the quotient rule: $f’(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=x^2; u'(x)=2x$ $v(x)=x^2+4; v'(x)=2x$ $f'(x)=\dfrac{(2x)(x^2+4)-(2x)(x^2)}{(x^2+4)^2}=\dfrac{8x}{(x^2+4)^2}.$ $f'(0)=\dfrac{8(0)}{((0)^2+4)^2}=0.$