I have a question stating:

Find d/dx f[1/g(x)]

g(x) and f(x) are differentiable. What exactly does this mean?

My solution is f'(1/g(x)*g'(x)/g(x)^2

Using the chain rule there ^

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- Jun 12th 2008, 12:18 AM #1

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- Jun 12th 2008, 12:24 AM #2
Hello,

Yes, this is the answer

Differentiable means that you can differentiate the function. More generally, and in this case, it means that g'(x) and f'(x) exist. Here is an example :

$\displaystyle f(x)=\sqrt{x}$, for $\displaystyle x \in [0, \infty[$

$\displaystyle f'(x)=\frac{1}{2 \sqrt{x}}$

This is defined all over $\displaystyle ]0, \infty[$. But not in 0.

The function is not differentiable in 0.

But here you have to add that g(x) never annulates.

- Jun 12th 2008, 03:35 AM #3