Question

Use log differentiation:

Answer 1

\[\Huge y=(ln x)^{cos7x}\]

Answer 2

So far..
\[\LARGE lny=cosx~ln(lnx)\]
Now differentiation, and product rule..?

Answer 3

@ganeshie8 Right so far?
And that's suppose to be cos7x

Answer 4

two ln's may not be needed

Answer 5

You'll end up with enough logs to build a cabin :-)

Answer 6

Here's the full question..
http://prntscr.com/30chur

Answer 7

\[\Large \ln x^a = a \ln x\]\[\Large (\ln x)^a \ne a \ln x\]

Answer 8

ahh yes i see, cos^7x is to the whole log... so yes luigi is rigiht

Answer 9

\[\LARGE lny=cosx~\ln(lnx) \]just use the product and chain rule now

Answer 10

So I end up with..
\[\LARGE \frac{y'}{y}=cos7x(\frac{1}{lnx}*\frac{1}{x})+-7sin7x(ln(lnx))\]?

Answer 11

looks good, and the answer in ur snap looks right... oly thing is the parenthesis are missing everywhere

Answer 12

once u fix them, i think it wud show checkmark :)

Answer 13

Omg, you were right, it was the parenthesis ._.
Thanks everyone, for everything :)