dg/dx = dg/df * df/dx = 0
since f'(2) = 0
If f′ (x)=((x−2)^3(x^2−4))/16 and g (x)= f(x^2−1) , what is g′ (2) ?
a. 2
b. 5/4
c. 5/8
d. 5/16
e. 0
I don't know how I would find f(x) from f'(x). I was thinking maybe it was the chain rule? Because that one has multiplication in it but I don't know where the divided by 16 would come from
1 answer