monroe17
can someone help me step by step?
find the derivative
y=ln((e^(7x))/(sqrt(4x-5)))
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anonymous
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before you begin taking the derivative, use the properties of the log to make this expression easier to differentiate
anonymous
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it is the log of the whole thing right?
monroe17
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yes.
anonymous
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so first step would be, before beginning to take the derivative, rewrite as
\[\ln(e^{7x})-\frac{1}{2}\ln(4x-5)\]
anonymous
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are those steps clear?
monroe17
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why is 1/2 there and not (4x-5)^(1/2)
anonymous
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i used two facts
\[\log(\frac{a}{b})=\log(a)-\log(b)\] and
\[\log(a^n)=n\log(a)\]
anonymous
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because
\[\log(\sqrt{4x-5})=\log((4x-5)^{\frac{1}{2}})=\frac{1}{2}\log(4x-5)\]
monroe17
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but 1/2 is outside the () and not inside ?
Dido525
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You can use the rule like that. As long as it's insside the log.
anonymous
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the one half comes right out front as a multiplier
anonymous
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on other words, \(\log(\sqrt{x})=\frac{1}{2}\log(x)\)
anonymous
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then one more step before differentiating
since log and exp are inverse functions, you have
\[\log(e^{7x})=7x\]
monroe17
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okay, so do i need to use product rule for 1/2ln(4x-5)?
anonymous
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oh no not at all
anonymous
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\(\frac{1}{2}\) is just a constant , leave it there
monroe17
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okay
Dido525
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Well you could... But it's a waste of time.
anonymous
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for example if you wanted the derivative of \(\frac{1}{2}x^3\) you do not use the product rule, you just say \(\frac{3}{2}x^2\)
monroe17
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gotcha. the power rule^
anonymous
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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
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typo there, i meant
\[7x-\frac{1}{2}\ln(4x-5)\]
monroe17
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okay imma try it/
anonymous
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ok let me know what you get
Dido525
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|dw:1351998238123:dw|
Dido525
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It will be useful when you do the ln part.
monroe17
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(1/4x-5)*4?
Dido525
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Good!
monroe17
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7x-1/2*(1/4x-5)*4
sirm3d
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oops, you forgot to differentiate 7x
anonymous
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careful of the first term
derivative of \(7x\) is just \(7\)
anonymous
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and you might not want to write
\[-\frac{4}{2(4x-5)}\] since you can cancel a 2
monroe17
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i thought i was just doing chain rule? i know that that's 7?
anonymous
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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
monroe17
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woah im lost now.
monroe17
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i did the chain rule for ln(4x-5) not 1/2*ln(4x-5)
anonymous
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yea it is right
what you wrote is correct
anonymous
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the "minus one half" is just a constant, leave it there like you did
monroe17
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okay, so where do i go from there. that fraction you did confused me.
anonymous
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the only mistake in your answer was that you left \(7x\) there, when the derivative of \(7x\) is \(7\)
everything else was right
anonymous
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you wrote this
7x-1/2*(1/4x-5)*4
monroe17
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oh i get it now!
anonymous
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the 7x should be 7
anonymous
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and the four is in the numerator, cancels with the 2 in the denominator
monroe17
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derivative of 7x is 7.
1/2 of 4 is 2.
7-2/4x-5
anonymous
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so it should look more like \(7-\frac{2}{4x-5}\)
anonymous
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yup you got it
monroe17
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thanks! im just a little slow sometimes lol