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Mathematics 22 Online
OpenStudy (anonymous):

f(x) = [ln x ]^(7) find f'(x) and f'(e^4)

OpenStudy (anonymous):

Remember to use the chain rule

OpenStudy (campbell_st):

this is the chain rule let u = ln(x) u' = 1/x y = u^7 y' = 7u^6 the f'(x) = 7(ln(x))^6 * 1/x

OpenStudy (anonymous):

oh ok does it matter which one if u and v

OpenStudy (anonymous):

No he just used it to make it look more simple and less intimidating

OpenStudy (anonymous):

It can easily be done without doing a u substitution

OpenStudy (campbell_st):

then substitute x = e^4

OpenStudy (anonymous):

is this what im suppose to get 28672/e^4

OpenStudy (campbell_st):

well I would have thought you would have gotten \[\frac{7 \times((\ln(e^4))^6}{e^4} = \frac{7 \times 4^6}{e^4} = \frac{7 \times 4096}{e^4}\] which gives the answer you have.

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