# Thread: Differentiating logarithm question

1. ## Differentiating logarithm question

I'd really appreciate it if somebody could show me the working for this question. Try as I might I can't seem to get the correct answer to come up. The question is as follows:

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By writing the function y = ((2x - 5) / (3x + 2))^0.5 in logarithmic form, show that dy/dx = 19/(2((2x-5)^0.5)((3x + 2)^1.5)

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I've used ^ to mean 'to the power of'.

I've spent a while on this question and I'd really like to understand how it's done, it's annoying me! Please help

2. Hi !

Originally Posted by wardj
I'd really appreciate it if somebody could show me the working for this question. Try as I might I can't seem to get the correct answer to come up. The question is as follows:

-------------------------------

By writing the function y = ((2x - 5) / (3x + 2))^0.5 in logarithmic form, show that dy/dx = 19/(2((2x-5)^0.5)((3x + 2)^1.5)

-------------------------------
I've used ^ to mean 'to the power of'.

I've spent a while on this question and I'd really like to understand how it's done, it's annoying me! Please help
$\displaystyle y={\color{red}\frac{2x-5}{(3x+2)^{.5}}}$

Taking the logarithms :

$\displaystyle \ln y=\ln (2x-5)-\ln ((3x+2)^{.5})$, because $\displaystyle \ln \frac ab=\ln a-\ln b$

Plus, $\displaystyle \ln a^n=n \ln a$

Therefore :

$\displaystyle \ln y=\ln (2x-5)-.5 \cdot \ln (3x+2)$

The derivative of $\displaystyle \ln y$ is $\displaystyle \frac{\frac{dy}{dx}}{y}=\frac{\frac{dy}{dx}}{\colo r{red}\dots}$

Can you continue ?

3. Originally Posted by wardj
I'd really appreciate it if somebody could show me the working for this question. Try as I might I can't seem to get the correct answer to come up. The question is as follows:

-------------------------------

By writing the function y = ((2x - 5) / (3x + 2))^0.5 in logarithmic form, show that dy/dx = 19/(2((2x-5)^0.5)((3x + 2)^1.5)

-------------------------------
I've used ^ to mean 'to the power of'.

I've spent a while on this question and I'd really like to understand how it's done, it's annoying me! Please help

$\displaystyle y=\left( \frac{2x-5}{3x+2}\right)^{\frac{1}{2}}$

Taking the natrual log of both sides gives

$\displaystyle \ln(y)=\ln\left( \frac{2x-5}{3x+2}\right)^{\frac{1}{2}}$

Using log properties we get

$\displaystyle \ln(y)=\frac{1}{2}\ln(2x-5)-\frac{1}{2}\ln(3x+2)$

Now taking the implict derivative of the equation gives

$\displaystyle \frac{1}{y}\frac{dy}{dx}=\frac{1}{2}\left( \frac{2}{2x-5} \right)-\frac{1}{2}\left( \frac{3}{3x+2}\right)=\frac{1}{2} \left( \frac{(2(3x+2)-3(2x-5)}{(2x-5)(3x+2)}\right)=$

$\displaystyle \frac{1}{y}\frac{dy}{dx}=\frac{1}{2} \left( \frac{19}{(2x-5)(3x+2)}\right) \iff \frac{dy}{dx}=y\frac{1}{2} \left( \frac{19}{(2x-5)(3x+2)}\right)$

Now replace y with what it is equal to and multiply.

$\displaystyle \frac{dy}{dx}=\left( \frac{2x-5}{3x+2}\right)^{\frac{1}{2}}\frac{1}{2} \left( \frac{19}{(2x-5)(3x+2)}\right)=\frac{1}{2} \left( \frac{19}{(2x-5)^{\frac{1}{2}}(3x+2)^{\frac{3}{2}}}\right)$

Yeah

4. Thanks for the reply, however the top part of the fraction is also square rooted, i.e (2x - 5)^0.5 / (3x + 2)^0.5

I do get to:

0.5 ln (2x - 5) - 0.5 ln (3x + 2)

Even though I have tried to find the derivative with the rule, I can not get to the answer so I would be very grateful if you could outline every step from 0.5 ln (2x - 5) - 0.5 ln (3x + 2) all the way.

5. I see the empty set has already posted!

Thanks a lot for the help guys, that's exactly what I was looking for. Now to learn how to do it!

Thanks.

6. Originally Posted by wardj
I see the empty set has already posted!

Thanks a lot for the help guys, that's exactly what I was looking for. Now to learn how to do it!

Thanks.
Sorry for misreading your post
... It's so rare that people respect like that the parenthesis rules and... I just happen to misread... :'(

Well, good night