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September 29th, 2017, 03:05 AM   #1
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derivatives and natural logarithms


I don't really understand (b) and (c).
Thanks beforehand.

Last edited by skipjack; September 29th, 2017 at 07:54 AM.
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September 29th, 2017, 03:47 AM   #2
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September 29th, 2017, 04:43 AM   #3
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Originally Posted by vizzy22 View Post

I don't really understand (b) and (c).
Thanks beforehand.
Okay, do you know what the "laws of logarithms" are?

$\displaystyle \ln(ab)= \ln(a)+ \ln(b)$
$\displaystyle \ln(a/b)= \ln(a)- \ln(b)$ and
$\displaystyle \ln(a^b)= b \ln(a)$

For example, (c) can be written as $\displaystyle y= \ln\left(\left(\frac{x+ 1}{x- 1}\right)^{1/2}\right)= \frac{1}{2} \ln\left(\frac{x+ 1}{x- 1}\right)= \frac{1}{2}\left(\ln(x+ 1)- \ln(x- 1)\right)$.

Now, can you differentiate that?

You try the others.

Last edited by skipjack; September 29th, 2017 at 07:56 AM.
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September 30th, 2017, 03:49 AM   #4
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Okay, do you know what the "laws of logarithms" are?

$\displaystyle \ln(ab)= \ln(a)+ \ln(b)$
$\displaystyle \ln(a/b)= \ln(a)- \ln(b)$ and
$\displaystyle \ln(a^b)= b \ln(a)$

For example, (c) can be written as $\displaystyle y= \ln\left(\left(\frac{x+ 1}{x- 1}\right)^{1/2}\right)= \frac{1}{2} \ln\left(\frac{x+ 1}{x- 1}\right)= \frac{1}{2}\left(\ln(x+ 1)- \ln(x- 1)\right)$.

Now, can you differentiate that?

You try the others.
I differenced that, but the thing I don't really understand is - how did you take ^1/2 and made it into 1/2ln.
I tried expanding B, but did not do it correctly, as the answers don't match.
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September 30th, 2017, 05:17 AM   #5
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Then you need to go back and review logarithms!

One of the basic "laws of logarithms", as I said before is "$\displaystyle \ln(a^b)= b \ln(a)$". If b = 1/2, that says that $\displaystyle \ln(\sqrt{a})= \ln(a^{1/2})= (1/2) \ln(a)$.

Problem (b) is $\displaystyle \ln\left(x\sqrt{3x- 1}\right)$. That can be written as $\displaystyle \ln\left(x(3x- 1)^{1/2}\right)$ and, by the "laws of logarithms", $\displaystyle \ln(x)+ \ln((3x- 1)^{1/2})= \ln(x)+ (1/2)\ln(3x- 1)$.

Of course to do problems like this, you have to know that the derivative of ln(x) is 1/x.

And you need to know the "chain rule": the derivative of ln(3x- 1) is $\displaystyle \frac{1}{3x- 1}$ times the derivative of 3x- 1 which is 3.

Putting all of those together the derivative of $\displaystyle \ln\left(x\sqrt{3x-1}\right)$, which is the same as the derivative of $\displaystyle \ln(x)+ (1/2)\ln(3x- 1)$, is $\displaystyle \frac{1}{x}+ \frac{3}{2(3x- 1)}$.

Last edited by skipjack; September 30th, 2017 at 07:24 AM.
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