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

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 ^

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 :
$\displaystyle f(x)=\sqrt{x}$, 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.

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