Compute Inverse Derivative for f(x) = x^3/(x^2+1) | Algebraic Help"

[physicsernaw]

Homework Statement

compute the inverse derivative of f(x) = x^3/(x^2+1)

Homework Equations

n/a

The Attempt at a Solution

My issue is a purely algebraic one... getting the inverse function!
Attempting to solve for x yields no results!

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??

 

[Curious3141]

Homework Statement

compute the inverse derivative of f(x) = x^3/(x^2+1)

Homework Equations

n/a

The Attempt at a Solution

My issue is a purely algebraic one... getting the inverse function!
Attempting to solve for x yields no results!

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??

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

Hint: apply integration by parts twice.

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

 

[physicsernaw]

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

Hint: apply integration by parts twice.

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

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

 

[Curious3141]

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

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.

 

[SteamKing]

The inverse derivative can be found using the following procedure:

http://www.analyzemath.com/calculus/Differentiation/derivative_inverse.html

 

[Curious3141]

The inverse derivative can be found using the following procedure:

http://www.analyzemath.com/calculus/Differentiation/derivative_inverse.html

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 $\frac{1}{2}x^{-\frac{1}{2}}$.

 

[Curious3141]

The inverse derivative can be found using the following procedure:

http://www.analyzemath.com/calculus/Differentiation/derivative_inverse.html

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).

 

[SteamKing]

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.

 

[Curious3141]

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.

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.

 

[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

 

[Curious3141]

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

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?