# Homework Help: F'(x) of a fraction

1. Feb 7, 2012

### cdoss

1. The problem statement, all variables and given/known data
Use the definition of a derivitive to find f'(x).
f(x)=1/(1-4X)

2. Relevant equations
Lim as h approaches 0 [f(x+h)-f(x)]/h

3. The attempt at a solution
I know that the answer is supposed to be -1/(1-4X)2 but I keep getting 4/(1-4X)2. This is what I have done so far (I hope this isn't too hard to understand):

f'(x)= (1/(1-4(x+h))-(1/(1-4x)))/h
= ((1-4x-(1-4(x+h)))/((1-4(x+h))(1-4x)))/h
= ((1-4x-1+4x+4h)/((1-4(x+h))(1-4x)))/h
*cancel the numerator values*
=(4h)/((1-4(x+h))(1-4x))/h
*divide by 1/h and let h=0*
=4/(1-4(x-0))(1-4x)
=4/(1-4X)2

What am I doing wrong?

2. Feb 7, 2012

### Staff: Mentor

Nothing - that's the right answer.

BTW, you don't take "f'(x) of a fraction" as you have in the title. You can take the derivative with respect to x of a fraction (in symbols, d/dx(f(x)) ), but f'(x) already represents the derivative of some function f.

3. Feb 7, 2012

### HallsofIvy

The derivative of 1/u, with respect to u, is -1/u^2. But that "4" in the numerator is from the chain rule. If u is a function of x, the derivative of 1/u, with respect to x is (-1/u^2) du/dx. Here, u= 1- 4x so du/dx= -4. The derivative of 1/(1- 4x), with respect to x, is -1/(1- 4x)^2(-4)= 4/(1- 4x)^2

4. Feb 7, 2012

### cdoss

oh, ok so it's right. I did that problem six times because I thought I was doing it wrong haha thank you! and also thank you for correcting me on the terms! :)

5. Feb 7, 2012

### cdoss

oh yeah the chain rule!! i definitely need to remember that! thanks!