Consider the function f, defined as f(x)=x^5+x^3 +3x

Determine (f^-1)'(5)

The answer should be 1/11

You want to find x such that f(x) = 5. By observation, f(1) = 5. Now use the fact that

d/dx f^(-1)(a) = 1/f '(b)

provided f '(b) ≠ 0 and f(b) = a.

f '(x) = 5x^4 + 3x² + 3 ==> f '(1) = 5 + 3 + 3 = 11.

So

d/dx f^(-1)(5) = 1/f '(1) = 1/11.

First note that
f(1) = 1+1+3 = 5 ==> f ˉ¹ (5) = 1

Also
f ' (x) = 5x⁴ + 3x² +3

==> f ' (1) = 5+3+3= 11
-------------------------------------

f(f ˉ¹ (x)) =x

Take derivative of both sides (use chain rule for the left-hand side)

[(f ˉ¹) ' (x)] * [f ' (fˉ¹(x))] = 1

Plug x =5

[(f ˉ¹) ' (5)] * [f ' (fˉ¹(5))] = 1

==> [(f ˉ¹) ' (5)] = 1/ f ' (fˉ¹(5)) = 1/ f ' (1) = 1/11