Problem with the inverse function

**jenkki** · November 3rd 2010, 09:01 AM

Given that g is the inverse function of f, where:

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

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

I have to show that:

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

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

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

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

**Opalg** · November 3rd 2010, 12:49 PM

Originally Posted by jenkki

Given that g is the inverse function of f, where:

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

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

I have to show that:

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

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

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

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

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