# 1/f differentiation- why am I not getting it

1. Nov 20, 2007

### asif zaidi

What am I missing here

Problem Statement:

Let f be a differentiable at point x and assume f(x) is not equal to 0. Show that function 1/f is differentiable at x=0 and (1/f)' = -f'(x)/( f(x) ^2 )

Solution
1/f = f^-1.
So taking its derivative I get the following: -f^-2. But I fail to see where the f'(x) term comes from. Plz advise.

Thanks

Asif

2. Nov 20, 2007

### Mute

You have to apply the chain rule when you take the derivative:

If u is a function of x, then if you take the derivative of some function g(u) with respect to x,

$$\frac{d}{dx}g(u) = \frac{dg}{du}\frac{du}{dx}$$

In this case, g(u) is your function 1/u, and u = f(x). Does this make sense?

3. Nov 21, 2007

### arildno

Well, you could do it directly like this:
$$\lim_{h\to{0}}\frac{\frac{1}{f(x+h)}-\frac{1}{f(x)}}{h}=\lim_{h\to{0}}\frac{f(x)-f(x+h)}{hf(x+h)f(x)}=-\lim_{h\to{0}}(\frac{f(x+h)-f(x)}{h})*\frac{1}{f(x+h)f(x)}$$
See if you manage to get something out of this.

4. Nov 21, 2007

### arildno

Because you differentiated with respect to f, rather than with respect to x.