# Problem with the inverse function

• Nov 3rd 2010, 09:01 AM
jenkki
Problem with the inverse function
Given that g is the inverse function of f, where:

$\displaystyle f(x) = \int^x_0 \frac {1}{\sqrt{t^3+1}} dt$

for $\displaystyle f: [0,\infty) \rightarrow \mathbb{R}$

I have to show that:

$\displaystyle g''(x) = \frac{3}{2} (g(x))^2$

At first I thought that all I had to do was to use the relation:

$\displaystyle g'(y) = \frac{1}{f'(x)}$ where $\displaystyle f'(x) = \frac {1}{\sqrt{x^3+1}}$

however, this expression gives me g'(y) and g''(y) (not x). What can I do to find the g(x)'s ?
• Nov 3rd 2010, 12:49 PM
Opalg
Quote:

Originally Posted by jenkki
Given that g is the inverse function of f, where:

$\displaystyle f(x) = \int^x_0 \frac {1}{\sqrt{t^3+1}} dt$

for $\displaystyle f: [0,\infty) \rightarrow \mathbb{R}$

I have to show that:

$\displaystyle g''(x) = \frac{3}{2} (g(x))^2$

At first I thought that all I had to do was to use the relation:

$\displaystyle g'(y) = \frac{1}{f'(x)}$ where $\displaystyle f'(x) = \frac {1}{\sqrt{x^3+1}}$

however, this expression gives me g'(y) and g''(y) (not x). What can I do to find the g(x)'s ?

So far, you have found that $\displaystyle g'(y) = \sqrt{x^3+1}$. Now you need to differentiate again to find $\displaystyle g''(y)$. But remember that this means the second derivative of g with respect to y. So you need to differentiate $\displaystyle \sqrt{x^3+1}$ with respect to y. Use the chain rule for that: differentiate it with respect to x and then multiply the result by $\displaystyle \frac{dx}{dy}$. You should find that $\displaystyle g''(y) = \frac{3x^2}{2\sqrt{x^3+1}}\,\frac{dx}{dy} = \frac32x^2$ (because $\displaystyle \frac{dx}{dy} = 1\big/\frac{dy}{dx}$). But $\displaystyle x = g(y)$ (because g is the inverse of f), so you get $\displaystyle g''(y) = \frac32\bigl(g(y)\bigr)^2$. All that remains is to trade the variable y for an x.