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

**SportfreundeKeaneKent** · Jun 12th 2008, 12:18 AM

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 ^

**Moo** · Jun 12th 2008, 12:24 AM

Hello,

Originally Posted by SportfreundeKeaneKent

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 ^

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

$f'(x)=\frac{1}{2 \sqrt{x}}$
This is defined all over $]0, \infty[$ . But not in 0.
The function is not differentiable in 0.

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

**mr fantastic** · Jun 12th 2008, 03:35 AM

Originally Posted by SportfreundeKeaneKent

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 ^

Originally Posted by Moo

Hello,

Yes, this is the answer

[snip]

The derivative of $\frac{1}{g(x)}$ with respect to x is ${\color{red}-} \frac{g'(x)}{[g(x)]^2}$ so I think there might be a negative in there somewhere, too .....

Thread: d/dx f[1/g(x)] ?

LinkBack

Thread Tools

Display

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

Search Tags