# I just can't get this

1. Jul 19, 2006

### rockytriton

ok, the reciprocal rule: g(x)' / g(x)^2

problem: -1/x^2

This seems simple, it should be -2/x^3 right?

Am I missing something here?

2. Jul 19, 2006

### Office_Shredder

Staff Emeritus
Ok.... so we have -1/x^2. Now, the reciprocal rule is that [1/g(x)]' = -g'(x)/g2(x).

So (-1/x^2)' = -(-2x/x^4) = 2/x^3

3. Jul 19, 2006

### NateTG

You're off by a sign.

A quick derivation of the 'reciprocal rule':

$$f(x)=\frac{1}{x}$$
$$\frac{d}{dx} f(x)= - \frac{1}{x^2}$$
(This follows from the power rule: $\frac{1}{x}=x^{-1}$)

Now, let's say $g(x)$ is a differentiable non-zero function with non-zero derivative on some domain, then on that domain we have:
$$\frac{d}{dx} f(g(x)) = g'(x) f'(g(x))$$
by the chain rule. Applying what we know about $f(x)$ gives:
$$\frac{d}{dx} \frac{1}{g(x)}=g'(x) \times - \frac{1}{{g(x)}^2}=-\frac{g'(x)}{{g(x)}^2}$$
(With apolgies for sloppy notation.)

Last edited: Jul 19, 2006
4. Jul 19, 2006

### rockytriton

oh ok, thanks guys, my reciprocal rule was wrong, I missed the negative.