# Derivative Help

Discussion in 'Community Discussion' started by rhsgolfer33, Sep 16, 2009.

1. ### rhsgolfer33 macrumors 6502a

Joined:
Jan 6, 2006
#1
So, I'm trying to find the derivative of f(x)=3/x. I know, using the power rule, that the derivative is -3x^-2 (or -3/x^2). However, I need to solve the problem using the f'(x)=[f(x+h)-f(x)]/h formula. Here is my work so far:

f'(x)=[(3/x+h)-(3/x)
Make common denominator of x(x+h)
f'(x)=[3x/x(x+h)]-[(3(x+h))/x(x+h)]
Simplify
f'(x)=[3x/x(x+h)]-[(3x+3h)/x(x+h)]
f'(x)=(3x-3x-3h)/x(x+h)
f'(x)=-3h/x(x+h)

That is where I loose it. I see that I can make the denominator (x^2)+hx, but that is about as far as I can get. Any help with this is appreciated. Its probably just my algebra failing me or a dumb mistake. Thanks for the help!

2. ### swiftaw macrumors 603

Joined:
Jan 31, 2005
Location:
Omaha, NE, USA
#2
You forgot the /h part in f'(x)

f'(x)=limit as h-> of [(3/(x+h))-(3/x)]/h

= [3x/x(x+h)]-[(3(x+h))/x(x+h)] / h

= [-3h/x(x+h)] / h = -3 / (x^2 + xh)

Now, as h-> 0, this becomes -3 / x^2

3. ### rhsgolfer33 thread starter macrumors 6502a

Joined:
Jan 6, 2006
#3
Thank you! As I said, probably a dumb mistake, didn't think it would be that dumb!