Inverse function

1. Feb 8, 2005

trap

I have no idea how to do this question, can anyone provide some help?

Suppose f is differentiable with derivative f '(x) = (1+x^3) ^(1/2).
If g = f^-1, show that g ''(x) = 3/2 g(x)^2.

2. Feb 8, 2005

Crosson

These sort of problems are fun, inverse calculus.

Write f(x) = int[0,x, (1+t^3) ^(1/2),dt]

Set the integral equal to u so that you have:

f(x) = u

Take the inverse of both sides:

x = g(u)

Differentiate wrt x, using the chain rule:

(dg/du)(du/dx) = 1

Using FTC, du/dx = (1+x^3) ^(1/2)

Then dg/du = (1+x^3) ^(-1/2)

but remember x =g(u)

So g' = (1+g^3) ^(-1/2)

differentiate again (remember to use the chain rule):

g'' = (-1/2)(3g)(g')(1+g^3) ^(-3/2)

g'' = (-1/2)(3g)(1+g^3) ^(-1/2))(1+g^3) ^(-3/2)

Which simplifies to your expression. I am sorry to have deprived you of solving this wonderful problem, but it is pretty difficult to get started on a problem like this if you do not know where to begin.

I invented this method to to derive an expression for the integral int[0,x, 1/t, dt] .

3. Feb 8, 2005

trap

wow, thank you so much, you are amazing

4. Feb 8, 2005

mathwonk

if f '(x) = dy/dx = (1+x^3) ^(1/2). where x = g(y).
If g = f^-1, show that g ''(x) = 3/2 g(x)^2. \\

this seems incorrect, as shown by the previous post, which also seems however slightly incorrect.

i.e. dx/dy = 1/(dy/dx). so dx/dy = (1+x^3)^(-1/2). so d^2x/dy^2, by chain rule,

= (-1/2)(1+x^3)^(-3/2) (3x^2) dx/dy = (-1/2)(1+x^3)^(-3/2) (3x^2)(1+x^3)^(-1/2)

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

= (-3/2)[1+g(y)^3)^(-2)] (g(y)^2). (the previous post omitted the power 2.)

in particular all the integration in the previous post is entirely superfluous.

now i may easily be missing something here, but my problem is that now I do not see how this simplifies to (3/2) [g(y)]^2, since it does not appear to me that

[1+g(y)^3)^(-2)] = -1, i.e. that

1 + x^3 = -1, i.e.that x = cuberoot(-2), for all x.

what am i missing?

5. Feb 9, 2005

matt grime

You are correct, Mathwonk, as far as I'm concerned: I independently reached exactly the same answer as you.

6. Feb 9, 2005

mathwonk

thanks, Matt; I guess you are kindly giving me credit for what I meant to write, even if one of my braces is misplaced in the last line.

I probably meant [1+g(y)^3)]^(-2) = -1, so the left side is a square while the right side is negative.

the point is the problem was probably miscopied from the source.

7. Feb 9, 2005

matt grime

we had the same issue, and it roughly stated all x satisfied some polynomial relation, i didn't check the details - perhaps i should stop using the word exactly when i don't mean it...

8. Feb 9, 2005

mathwonk

I assumed you quit reading my details as soon as you saw we raised the same point and found the same result.