Derivative of inverse at specific x value

Question

I am having trouble with finding the derivative of an inverse function at a specific x value. I know these things:

$$f(x)=x^3-2$$

$$f^{-1}(x)=\sqrt[3]{x+2}$$

if

$$x=25$$

$$f^{-1}(25)=3$$

However, I am trying to find $f^{'}(f^{-1}(25))$ and getting nowhere.

I plug in 3 for the inverse function at 25 and all I get for the first step after this is $f^{'}(3)$ which is clearly not right. I am supposed to get $3(something)^2$ after the $f^{'}$ step and simplify and the online homework software is asking me to put in an exact number at every step. I assume that means that whatever I do, I shouldn't put some function of x as the answer to any of the steps. So I am stuck here. If I can't use the variable x in my answer, how am I going to find the derivative at that specific point where x=25?

Answer 1

You seem to be confused about a few things, so I'll try to go go through them.

Method to find Derivative of $f^{-1}$ at $x=25$

If $f(x)=x^3-2$ then you're correct in saying $f^{-1}(x)=\sqrt[3]{x+2}$.

I'm going to ignore the $f'$ notation in favor of the derivative operator $\left(\frac{\mathrm{d}}{\mathrm{d}x}\right)$, to remove some confusion.

Using the chain rule, we find that the derivative of the inverse function is: $\frac{\mathrm{d}}{\mathrm{d}x}f^{-1}(x)=\frac{\mathrm{d}}{\mathrm{d}x}\sqrt[3]{x+2}=\frac{\mathrm{d}}{\mathrm{d}x}\left(x+2\right)^{1/3}=\frac{1}{3}\left(x+2\right)^{(1/3)-1}\cdot\frac{\mathrm{d}}{\mathrm{d}x}(x+2)=\frac{1}{3}\left(x+2\right)^{-2/3}$.

To find the derivative at a certain point, say $x=25$, you'd substitute $x=25$ into the equation and calculate $\frac{1}{3}\left(25+2\right)^{-2/3}$. When you simplify that number, it's the "derivative of $f^{-1}(x)$ at $x=25$". Nothing else is required.

Unecessary Step $1$: Computing $f^{-1}(x)$

If you're given that $x=25$ then you don't need to plug $25$ into the inverse function. We're not interested in what $f^{-1}(x)$ is doing at $x=25$; we want to know what $\frac{\mathrm{d}}{\mathrm{d}x}f^{-1}(x)$ is doing there. These are different things.

Unecessary Step $2$: Computing $f'(f^{-1}(x))$

This might just be a mistake in your notation but $\frac{\mathrm{d}}{\mathrm{d}x}f^{-1}(x)$ is not the same thing as $f'(f^{-1}(x))$. The first formula, $\frac{\mathrm{d}}{\mathrm{d}x}f^{-1}(x)$, says that we're differentiating the inverse function. But the second formula, $f'(f^{-1}(x))$, says that we're plugging the inverse function into $f'(x)$, which is the derivative of the ordinary function not what we want (we want the derivative of the inverse function). Furthermore, you mention "$3\,\text{something}^2$" which is $f'(x)$ (not what we want).

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I don't mean to discourage you by pointing out mistakes but its best to learn how to avoid them. Also, I'd initially answered the wrong question, since I'd misunderstood what you were asking. As Raffaele points out, there is a possible method you can use to find the answer, from the Inverse Function Theorem, with $\frac{\mathrm{d}}{\mathrm{d}x}f^{-1}(x)\big|_{x=25}=\frac{1}{f'(x)}\big|_{x=f^{-1}(25)}$ but I think it'd be better for you to be more comfortable with the basics before using that.

Answer 2

You have

$ f(x)=x^3 -2 $ and $f^{-1}(x)=\sqrt[3]{x+2}$,

so:

$f'(x)=3x^2$ and $\left(f^{-1}\right)'(x)=\frac{1}{3}\frac{1}{\sqrt[3]{(x+2)^2}}$

Frome the inverse function derivative formula we have:

$$ \left(f^{-1}\right)'(25)=\frac{1}{f'\left(f^{-1}(25)\right)} $$ so: $$ \left(f^{-1}\right)'(25)=\frac{1}{f'(3)}=\frac{1}{27} $$

and deriving the inverse function we have the same result: $$ \left(f^{-1}\right)'(25)=\frac{1}{3} \frac{1}{\sqrt[3]{27^2}}=\frac{1}{27} $$