can someone help me step by step?
find the derivative
y=ln((e^(7x))/(sqrt(4x-5)))

- anonymous

can someone help me step by step?
find the derivative
y=ln((e^(7x))/(sqrt(4x-5)))

- katieb

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- anonymous

before you begin taking the derivative, use the properties of the log to make this expression easier to differentiate

- anonymous

it is the log of the whole thing right?

- anonymous

yes.

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## More answers

- anonymous

so first step would be, before beginning to take the derivative, rewrite as
\[\ln(e^{7x})-\frac{1}{2}\ln(4x-5)\]

- anonymous

are those steps clear?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

the one half comes right out front as a multiplier

- anonymous

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

- anonymous

then one more step before differentiating
since log and exp are inverse functions, you have
\[\log(e^{7x})=7x\]

- anonymous

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

- anonymous

oh no not at all

- anonymous

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

- anonymous

okay

- anonymous

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

- anonymous

for example if you wanted the derivative of \(\frac{1}{2}x^3\) you do not use the product rule, you just say \(\frac{3}{2}x^2\)

- anonymous

gotcha. the power rule^

- anonymous

so now you have
\[7x+\frac{1}{2}\ln(4x-5)\] so the only rule you need now is the chain rule for the second part, and also knowing what the derivative of the log is

- anonymous

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

- anonymous

okay imma try it/

- anonymous

ok let me know what you get

- anonymous

|dw:1351998238123:dw|

- anonymous

It will be useful when you do the ln part.

- anonymous

(1/4x-5)*4?

- anonymous

Good!

- anonymous

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

- sirm3d

oops, you forgot to differentiate 7x

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

chain rule for \(-\frac{1}{2}\ln(4x-5)\) because it is a composition , the log of something
for \(7x\) that is just a line with slope 7, derivative is 7

- anonymous

woah im lost now.

- anonymous

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

- anonymous

yea it is right
what you wrote is correct

- anonymous

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

- anonymous

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

- anonymous

the only mistake in your answer was that you left \(7x\) there, when the derivative of \(7x\) is \(7\)
everything else was right

- anonymous

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

- anonymous

oh i get it now!

- anonymous

the 7x should be 7

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

yup you got it

- anonymous

thanks! im just a little slow sometimes lol

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