# Inverse derivative help

1. Mar 5, 2013

### physicsernaw

1. The problem statement, all variables and given/known data
compute the inverse derivative of f(x) = x^3/(x^2+1)

2. Relevant equations
n/a

3. The attempt at a solution
My issue is a purely algebraic one... getting the inverse function!

y = x^3/(x^2+1)
y(x^2+1) = x^3
yx^2 + y = x^3
y = x^3 - yx^2
y = x^2(x - y)
y/(x-y) = x

cannot isolate x!!! help??

2. Mar 5, 2013

### Curious3141

You're asked for the inverse derivative (which means indefinite integral, I presume). Not the inverse function.

Hint: apply integration by parts twice.

Hint: $x^3 = x^2.x$

3. Mar 5, 2013

### physicsernaw

Thanks for the reply but it's a first semester Calculus course. We haven't touched the topic of integration yet.

4. Mar 5, 2013

### Curious3141

Well, if it's really the inverse *function* that you want, it's going to be *extremely difficult* to work out. You need to solve the cubic $x^3 - yx^2 - y = 0$ giving x in terms of y, and that's anything but trivial. You'd better check the question again, and write it out exactly as it's stated.

5. Mar 5, 2013

### SteamKing

Staff Emeritus
6. Mar 5, 2013

### Curious3141

Oh, so *that's* what they meant by "inverse derivative"?

Can't they just say derivative of the inverse function?

Is "inverse derivative" even correct usage? Because technically, I would assume the inverse derivative to be the inverse of the derivative as opposed to the derivative of the inverse. Quite different things, in general.

Of course, even assuming the first meaning, there's bound to be confusion. In my first post, I assumed they meant the inverse operation of the derivative, which is the integral. Whereas the inverse function of a derivative is different, e.g. the inverse function of the derivative of $f(x) = x^2$ (for x > 0) is $\frac{1}{2}x$, whereas the inverse operation of the derivative of $x^2$ is simply the integral, $\frac{1}{3}x^3 + c$. And the derivative of the inverse is another thing altogether, in this case it's $\frac{1}{2}x^{-\frac{1}{2}}$.

Last edited: Mar 5, 2013
7. Mar 5, 2013

### Curious3141

Even doing it "implicitly", I don't think it really solves his issue. There's still a step where one needs to basically find the inverse function to get everything in terms of a single independent variable. It's still a very ugly expression (I checked with Wolfram).

8. Mar 5, 2013

### SteamKing

Staff Emeritus
I think you should take a look at Method 2 from the attachment. It is not required to obtain the inverse of the function directly to obtain the inverse derivative.

9. Mar 5, 2013

### Curious3141

I understand what they're trying to do. What I'm saying is that in the final expression, it does become necessary to find the inverse function in terms of a single variable, to get an acceptable answer. Let me illustrate a basically equivalent approach (which I use) and highlight the problem:

$y = f(x)$
$x = f^{-1}(y)$
$\frac{dx}{dy} = \frac{d}{dy}f^{-1}(y)$
$\frac{d}{dy}f^{-1}(y) = \frac{1}{\frac{dy}{dx}} = \frac{1}{\frac{d}{dx}(f(x))} = \frac{1}{\frac{d}{dx}f(f^{-1}(y))}$
$(f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))}$
$(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$

You can see that, in order to express everything in terms of a single independent variable (x), you still need to work out $f^{-1}(x)$.

I don't like the formulation written in the link, because Method 2 uses y to indicate the inverse function (whereas Method 1 started by using y to indicate the function itself), and there's a great potential for confusion. Nevertheless, it's essentially the same as what I've derived here. The function they used to illustrate the example was a very simple linear function, which is why no problem was apparent, but with a more complex function, like in this problem, the issues will become obvious.

Note that it's still possible to get an answer, it's just that it's very ugly and tedious. Try it if you don't believe me.

Last edited: Mar 5, 2013
10. Mar 5, 2013

### physicsernaw

My apologies, the question asks to find the g'(-1/2) where g(x) is the inverse of f(x) = x^3/(x^2+1). So the problem is actually really easy. Busted my *** for no reason trying to find the inverse function XD

11. Mar 5, 2013

### Curious3141

Ah, now that is MUCH more tractable. And you can use what I wrote in my last post to give you a quick answer.

To be proper, you need to define $f(x) = \frac{x^3}{x^2+1}, f:\mathbb{R} \rightarrow \mathbb{R}$ to avoid ambiguity.

Anyway, what answer did you get?