# Derivative of Inverse Function

1. Feb 16, 2017

1. The problem statement, all variables and given/known data
Suppose $f(x) = x^5 + 2x + 1$ and $f^{-1}$ is the inverse of function f. Evaluate $f^{-1}(4)$

solution: 1/7

2. Relevant equations
$(f^{-1}(x))=\frac{1}{f'(f^{-1}(x))}$

3. The attempt at a solution
I attempted to use my calculator's solve function to get the solution of $4 = 5x^4 + 2$, so the value of x when the derivative is equal to 4. I then took the inverse of this. Sort of still grasping in the dark right now tho.

2. Feb 16, 2017

### Simon Bridge

Your problem statement says you have to:
Note: $f^{-1}(f(x))=x \implies f^{-1}(4) = x:f(x)=4$
Why were you computing a derivative?

3. Feb 16, 2017

### BvU

There's something fishy in the combination problem statement $\leftrightarrow$ solution 1/7 . James ?

4. Feb 16, 2017

My mistake guys,

The problem statement asks for $(f^{-1})^{'}(4)$

5. Feb 16, 2017

### PeroK

The relevant equation you posted is not that useful, as it involves $f'$. You need something similar involving $(f^{-1})'$.

6. Feb 16, 2017

### BvU

So you don't want to solve for 'derivative = 4' but you want to evaluate the derivative at the point where the function value is 4 !

 perok: he had a $\$ ' $\$ missing in
$(f^{-1}(x))=\frac{1}{f'(f^{-1}(x))}$ ; should've been $$(f^{-1}(x))'=\frac{1}{f'(f^{-1}(x))}$$

7. Feb 16, 2017

Hmmm. $f^{'} = 5x^4+2$ and $f^{'}(4)= 1282$

So I have the derivative of of the function itself at a certain x value. But I'm not sure how I can translate that into finding the derivative of the inverse at the same x value.

8. Feb 16, 2017

### BvU

Again: you don't want to evaluate the derivative at the point x = 4 but at the point f-1(4)

9. Feb 16, 2017

### BvU

Do you understand the relationship $$(f^{-1}(x))'=\frac{1}{f'(f^{-1}(x))}\ \ \rm ?$$

10. Feb 16, 2017

I understand this relationship, but I do not know how I can evaluate $f^{'}(4)$ since I can't figure out how to get the inverse of the original function. Plugging in for $y=x^5 + 2x + 1$, I can't figure out how I would be able to solve for x in terms of y.

11. Feb 16, 2017

### PeroK

You could always just take a guess!

12. Feb 16, 2017

### BvU

Try a few numbers, or do bisection ( f(0) = 1, f(2) >> 4 so...)

And again (again) :
You don't want that. You want $f'(f^{-1}(4))$ !

13. Feb 16, 2017

### Ray Vickson

Plot the graph of $y = x^5+2x+1$ over some modest $x$-range; that will give you a good feeling for approximately where you have the solution of $f(x)=4.$ Alternately you can try using the "rational root theorem" on the equation $x^5 + 2x - 3 = 0.$ (Once you get the solution you will wonder why you did not see it right away!)

14. Feb 17, 2017

Thanks guys, I guess I just really needed to sleep on it, judging by my sloppiness. Coming back today, I can see where you all were going:

$(f^{-1}(x))^{'} = \frac{1}{f^{'}(f^{-1}(x))}$

First thing, we solve the interior function on the RHS, or $4 = x^5 + 2x +1$. I used the numerical solver on the calculator, but I see now that 1 could be obvious if you take a step back for a second. After that, we're no longer working with inverses, the derivative function $f^{'}(x) = 5x^4 + 2$ taken at x = 1. One over that and boom.