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

Mathematics
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before you begin taking the derivative, use the properties of the log to make this expression easier to differentiate
it is the log of the whole thing right?
yes.

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Other answers:

so first step would be, before beginning to take the derivative, rewrite as \[\ln(e^{7x})-\frac{1}{2}\ln(4x-5)\]
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, \(\log(\sqrt{x})=\frac{1}{2}\log(x)\)
then one more step before differentiating since log and exp are inverse functions, you have \[\log(e^{7x})=7x\]
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.
for example if you wanted the derivative of \(\frac{1}{2}x^3\) you do not use the product rule, you just say \(\frac{3}{2}x^2\)
gotcha. the power rule^
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
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?
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
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.
the only mistake in your answer was that you left \(7x\) there, when the derivative of \(7x\) is \(7\) everything else was right
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

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