anonymous
  • anonymous
can someone help me step by step? find the derivative y=ln((e^(7x))/(sqrt(4x-5)))
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
before you begin taking the derivative, use the properties of the log to make this expression easier to differentiate
anonymous
  • anonymous
it is the log of the whole thing right?
anonymous
  • anonymous
yes.

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anonymous
  • anonymous
so first step would be, before beginning to take the derivative, rewrite as \[\ln(e^{7x})-\frac{1}{2}\ln(4x-5)\]
anonymous
  • anonymous
are those steps clear?
anonymous
  • anonymous
why is 1/2 there and not (4x-5)^(1/2)
anonymous
  • anonymous
i used two facts \[\log(\frac{a}{b})=\log(a)-\log(b)\] and \[\log(a^n)=n\log(a)\]
anonymous
  • anonymous
because \[\log(\sqrt{4x-5})=\log((4x-5)^{\frac{1}{2}})=\frac{1}{2}\log(4x-5)\]
anonymous
  • anonymous
but 1/2 is outside the () and not inside ?
anonymous
  • anonymous
You can use the rule like that. As long as it's insside the log.
anonymous
  • anonymous
the one half comes right out front as a multiplier
anonymous
  • anonymous
on other words, \(\log(\sqrt{x})=\frac{1}{2}\log(x)\)
anonymous
  • anonymous
then one more step before differentiating since log and exp are inverse functions, you have \[\log(e^{7x})=7x\]
anonymous
  • anonymous
okay, so do i need to use product rule for 1/2ln(4x-5)?
anonymous
  • anonymous
oh no not at all
anonymous
  • anonymous
\(\frac{1}{2}\) is just a constant , leave it there
anonymous
  • anonymous
okay
anonymous
  • anonymous
Well you could... But it's a waste of time.
anonymous
  • 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
  • anonymous
gotcha. the power rule^
anonymous
  • 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
  • anonymous
typo there, i meant \[7x-\frac{1}{2}\ln(4x-5)\]
anonymous
  • anonymous
okay imma try it/
anonymous
  • anonymous
ok let me know what you get
anonymous
  • anonymous
|dw:1351998238123:dw|
anonymous
  • anonymous
It will be useful when you do the ln part.
anonymous
  • anonymous
(1/4x-5)*4?
anonymous
  • anonymous
Good!
anonymous
  • anonymous
7x-1/2*(1/4x-5)*4
sirm3d
  • sirm3d
oops, you forgot to differentiate 7x
anonymous
  • anonymous
careful of the first term derivative of \(7x\) is just \(7\)
anonymous
  • anonymous
and you might not want to write \[-\frac{4}{2(4x-5)}\] since you can cancel a 2
anonymous
  • anonymous
i thought i was just doing chain rule? i know that that's 7?
anonymous
  • 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
  • anonymous
woah im lost now.
anonymous
  • anonymous
i did the chain rule for ln(4x-5) not 1/2*ln(4x-5)
anonymous
  • anonymous
yea it is right what you wrote is correct
anonymous
  • anonymous
the "minus one half" is just a constant, leave it there like you did
anonymous
  • anonymous
okay, so where do i go from there. that fraction you did confused me.
anonymous
  • 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
  • anonymous
you wrote this 7x-1/2*(1/4x-5)*4
anonymous
  • anonymous
oh i get it now!
anonymous
  • anonymous
the 7x should be 7
anonymous
  • anonymous
and the four is in the numerator, cancels with the 2 in the denominator
anonymous
  • anonymous
derivative of 7x is 7. 1/2 of 4 is 2. 7-2/4x-5
anonymous
  • anonymous
so it should look more like \(7-\frac{2}{4x-5}\)
anonymous
  • anonymous
yup you got it
anonymous
  • anonymous
thanks! im just a little slow sometimes lol

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