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December 19, 2014

December 19, 2014

Posted by **jessica** on Tuesday, February 20, 2007 at 9:22pm.

f(x) = ((3x-7)/(6x+3))^4

f '(x) =

let u= 3x-7

v= (6x+3)^-4

Then d(uv)= u dv + v du

du= 3 dv= -4(6x+3)^-5 (6)

so what's f'(x)? (thanks for your help)

Goodness. f'(x) is d(uv)

I will be happy to critique your work or thinking.

in the orginal problem, the entire thing is squared to the 4th power not just the denominator. and also, where and how did u change 4 to a -4?

I didn't see that. OK, change u to ((3x-7)^4 that changes du to 12(3x-7)^3

v= 1/(6x+3)^4= (6x+3)^-4

are u sure u change it to 12(3x-7)^3 and not 4(3x-7)^3?? i'm just curious. i'm trying to follow u but my derivative came out different from urs again.

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