1. ## Chain rule problem?

f(x)= 1/2x - 1/3x^2
=(2x)^-1 -(3x)^-2
f ' =(-2x)^-2 + (6x)^-3
=-2/x^2 + 6/x^-3
This is as far as I can get, the finished answer should be 4-3x/6x^3

Can you show me how to finish it please? Thanks

2. Originally Posted by wolfhound

f(x)= 1/2x - 1/3x^2
=(2x)^-1 -(3x)^-2
f ' =(-2x)^-2 + (6x)^-3
=-2/x^2 + 6/x^-3
This is as far as I can get, the finished answer should be 4-3x/6x^3

Can you show me how to finish it please? Thanks
For clarification: Do you mean

f(x)= 1/2*x - 1/3*x^2

or

f(x)= 1/(2x) - 1/(3x^2)

or

f(x)= 1/(2x) - 1/((3x)^2)

3. Originally Posted by earboth
For clarification: Do you mean

or

f(x)= 1/(2x) - 1/((3x)^2)

4. $f(x) = (2x)^{-1} - (3x)^{-2}$

$f'(x) = -(2x)^{-2} \cdot 2 + 2(3x)^{-3} \cdot 3$

$f'(x) = -\frac{2}{(2x)^2} + \frac{6}{(3x)^3}$

you can finish the algebra grunt work

5. Originally Posted by skeeter
$f(x) = (2x)^{-1} - (3x)^{-2}$

$f'(x) = -(2x)^{-2} \cdot 2 + 2(3x)^{-3} \cdot 3$

$f'(x) = -\frac{2}{(2x)^2} + \frac{6}{(3x)^3}$

you can finish the algebra grunt work
I dont know how to finish it!

6. Originally Posted by wolfhound
I dont know how to finish it!
Take skeeter's result:

1. Expand the terms at the RHS. Cancel if possible.

2. Determine a common denominator. Transform the fractions such that they have the same denominator. Collect like terms in the numerator.

First step:

$
f'(x) = -\frac{2}{(2x)^2} + \frac{6}{(3x)^3} = -\frac{1}{2x^2}+\frac{2}{9x^3}
$

Next steps for you.