Can you help w/ der. of y = ln(sqrt((4x-7)/(x^2+2x))

[yummymummy1713]

This involves the natural log and that really confuses me. Here is what I have so far:

. . .y = ln(sqrt[(4x-7)/(x^2 +2x)])

I know this is the same as:

. . .y = ln [(4x-7)/(x^2 +2x)]^1/2

Then I can:

. . .y= 1/2 [ln(4x-7) - ln(x^2+ 2x)]

What next? Thank you!

 

[stapel]

Why not just take the derivative of the function as it stands, rather than rearranging things first? Does your book require that you rearrange the log first...?

Eliz.

 

[yummymummy1713]

ni i just dont know how to do it that way. i am not very good with logs

thanks

 

[galactus]

This problem is mostly lotsa algebra. Don't forget the chain rule a couple of times.


\(\displaystyle \L\\f(x)=\ln(\sqrt{\frac{4x-7}{x^{2}+2x}})\)


Use the quotient rule and chain rule to find the derivative of the radical and don't forget the derivative of ln(x) is 1/x

\(\displaystyle \L\\\underbrace{(\frac{4x-7}{x^{2}+2x})^{\frac{-1}{2}}}_{\text{derivative\\of ln(x)}}\underbrace{\frac{1}{2}(\frac{4x-7}{x^{2}+2x})^{\frac{-1}{2}}}_{\text{chain rule}}\underbrace{\left(\frac{(x^{2}+2x)(4)-(4x-7)(2x+2)}{(x^{2}+2x)^{2}}\right)}_{\text{quotient rule\\inside radical}}\)

\(\displaystyle \L\\\frac{x^{2}+2x}{2(4x-7)}\cdot\frac{-2(2x^{2}-7x-7)}{(x^{2}+2x)^{2}}\)

You can finish simplifying. Okey-doke?.

 

[yummymummy1713]

I'm sorry, but I still can't get an answer. I don't understand. Can you please explain?

 

[galactus]

exponent law:

-1/2+(-1/2)=-1

\(\displaystyle \L\\\frac{1}{2}\left(\frac{4x-7}{x^{2}+2x}\right)^{\frac{-1}{2}}\left(\frac{4x-7}{x^{2}+2x}\right)^{\frac{-1}{2}}=\frac{1}{2}\left(\frac{4x-7}{x^{2}+2x}\right)^{-1}=\frac{x^{2}+2x}{2(4x-7)}\)