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it is the log of the whole thing right?

yes.

are those steps clear?

why is 1/2 there and not (4x-5)^(1/2)

i used two facts
\[\log(\frac{a}{b})=\log(a)-\log(b)\] and
\[\log(a^n)=n\log(a)\]

because
\[\log(\sqrt{4x-5})=\log((4x-5)^{\frac{1}{2}})=\frac{1}{2}\log(4x-5)\]

but 1/2 is outside the () and not inside ?

You can use the rule like that. As long as it's insside the log.

the one half comes right out front as a multiplier

on other words, \(\log(\sqrt{x})=\frac{1}{2}\log(x)\)

okay, so do i need to use product rule for 1/2ln(4x-5)?

oh no not at all

\(\frac{1}{2}\) is just a constant , leave it there

okay

Well you could... But it's a waste of time.

gotcha. the power rule^

typo there, i meant
\[7x-\frac{1}{2}\ln(4x-5)\]

okay imma try it/

ok let me know what you get

|dw:1351998238123:dw|

It will be useful when you do the ln part.

(1/4x-5)*4?

Good!

7x-1/2*(1/4x-5)*4

oops, you forgot to differentiate 7x

careful of the first term
derivative of \(7x\) is just \(7\)

and you might not want to write
\[-\frac{4}{2(4x-5)}\] since you can cancel a 2

i thought i was just doing chain rule? i know that that's 7?

woah im lost now.

i did the chain rule for ln(4x-5) not 1/2*ln(4x-5)

yea it is right
what you wrote is correct

the "minus one half" is just a constant, leave it there like you did

okay, so where do i go from there. that fraction you did confused me.

you wrote this
7x-1/2*(1/4x-5)*4

oh i get it now!

the 7x should be 7

and the four is in the numerator, cancels with the 2 in the denominator

derivative of 7x is 7.
1/2 of 4 is 2.
7-2/4x-5

so it should look more like \(7-\frac{2}{4x-5}\)

yup you got it

thanks! im just a little slow sometimes lol