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f(x) = In((3x-4)3)

f(x) = In((3x-4)*(3x-4)*(3x-4))

f(x) = In(3x-4) + In(3x-4) + In(3x-4)

f'(x) = 1/(3x-4) + 1/(3x-4) + 1/(3x-4)

f'(x) = 3/(3x-4)

 

I'd assume this was correct, but it isn't. I know how to work it out, but I don't understand why this method doesn't work. 

 

Someone explain.

 

Thanks.

Guest Aug 4, 2017
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2+0 Answers

 #1
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Your mistake is in the second last line as the derivative of

log(3x-4) is 3/(3x-4)

Guest Aug 4, 2017
 #2
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I see now. Thanks.

Guest Aug 4, 2017

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