# F'(x) of a fraction

## Homework Statement

Use the definition of a derivitive to find f'(x).
f(x)=1/(1-4X)

## Homework Equations

Lim as h approaches 0 [f(x+h)-f(x)]/h

## 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?

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## Homework Statement

Use the definition of a derivitive to find f'(x).
f(x)=1/(1-4X)

## Homework Equations

Lim as h approaches 0 [f(x+h)-f(x)]/h

## 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?
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.

HallsofIvy