1. ## Exponent/Log Functions

You told me to post a couple more on here to make sure I had gotten this...
This one is going to prove more difficult for me.

f(x) = the 4th root of (2x+1/1-3x)....the whole fraction under the 4th root.

u(x) = the 4th root of u or u^(-1/4)
u'(x) = -1/4u^(-5/4)
v(x) = [2x+1/1-3x]
v'(x) = (1-3x)(2) - (2x+1)(-3)/(1-3x)^2
v'(x) = (2-6x) - (-6x-3)/(1-3x)^2
v'(x) = -1/(1-3x)^2

If that is correct, I will move on with
u(x) * v'(x) - v'(x) * u(x)/u(x)^2

2. ## I must not be getting this....

Wasnt asking for an answer, just if the steps were correct so far.

3. Originally Posted by becky
Wasnt asking for an answer, just if the steps were correct so far.
the fourthroot of u is not u^(-1/4), so the first one is wrong. it's just u^(1/4)

the second one is incorrect also. you made a mistake when expanding the brackets. you will have in the top 2 - - 3 = 2 + 3 = 5, you forgot that there was a double negative

4. Originally Posted by becky
If that is correct, I will move on with
u(x) * v'(x) - v'(x) * u(x)/u(x)^2
what's this for? shouldn't you be doing the chain rule? you have u(v(x)) so

[u(v(x))]' = u'(v(x))*v'(x)

Re: