PDA

View Full Version : I just can't get this


rockytriton
Jul19-06, 09:59 AM
ok, the reciprocal rule: g(x)' / g(x)^2

In this book I'm reading:

problem: -1/x^2

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

But the answer says: 2x^(-3)

Am I missing something here?

Office_Shredder
Jul19-06, 10:14 AM
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

NateTG
Jul19-06, 10:15 AM
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.)

rockytriton
Jul19-06, 10:44 AM
oh ok, thanks guys, my reciprocal rule was wrong, I missed the negative.