Question

g(x)=rad(16-x^2) h(x)=f(g(x)) h'(0) if g is differentiable at 4

Answer 1

h'(x)=f'(rad(16-x^2)(1/2)(rad(16-x^2))(-2x)

Answer 2

if you plug in 0, you get f'(rad(16-x^2)(0/4)=0