Question

Let f(x)=x^3+x. If h is the inverse function of f, then h'(2) =

The answer is 1/4 but I have no idea how. This is what i thought to do, and if someone could correct me it would be greatly appreciated.

F(x)=x^3+x
X=y^3+y
1=3y^2(y')+y'
1=y'(3y^2+1)
1/(3y^2+1)=y' where y=(x^3+x)

Answer 1

sorry... i didn't mean that...

Answer 2

ok... can you calculate f(1)...

Answer 3

so f(1)=1^3+1 = 2

Answer 4

ok
since f(1)=2, that means \(\large f^{-1}(2)=1 \)

Answer 5

now... what's f'(1)

Answer 6

f'(1)=3(1)^2+1 = 4

Answer 7

so since \(\large f'(1)=4 \) then \(\large f'(2)=\frac{1}{4} \)

Answer 8

because the derivative is just the reciprocal...