# Math Help - bit of trouble with differentiating

1. ## bit of trouble with differentiating

hello, I am having a bit of trouble with taking the derivative of a slighty tricky algebraic equation. I have almost finished it but I may have made a mistake along the way. Here is what I have so far:

1/2
y= (x/3x+1)^ in other words the square root of (x)/(3x+1)

1/2
= ((x)*(3x+1)^-1)^ step 1

1/2 -1/2
= x^ *(3x+1)^ step 2

1/2 -3/2 -1/2 1/2
= ((x^ )*(-1/2*(3x+1)^ )*(3)) + ((3x+1)^ )*(1/(2x^ )

1/2 3/2 1/2 1/2
= (-(3x^ ) / (2*(3x+1)^ )) + ((1) / (2x^ )*(3x+1)^

=

I can understand that this is probably pretty hard to read but I tried to upload a photo and it didn't work.

Whoever helps me it would probably be best if you just copied down the original equation and simplified, used the Product Rule to take the derivative, and then simplified again hopefully correctly because I couldn't.

I am close because I checked the answer sheet but I might have made a mistake with the confusing exponents.

Thank you very much in advance.

[IMG]file:///C:/Users/Robert/AppData/Local/Temp/moz-screenshot.jpg[/IMG][IMG]file:///C:/Users/Robert/AppData/Local/Temp/moz-screenshot-1.jpg[/IMG]

2. $\frac{d}{dx} \, \left(\frac{x}{3x+1}\right)^{\frac{1}{2}} =$

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

$\frac{1}{2}\left(\frac{3x+1}{x}\right)^{\frac{1}{2 }} \cdot \frac{1}{(3x+1)^2} =
$

$\frac{1}{2\sqrt{x(3x+1)^3}}
$

3. ooohh...

So if I understand correctly you simply used the Chain Rule and the derivative of the inside was found by using the Quotient Rule. So I was not taking the best approach to the problem and making the algebra a lot harder for myself.

Thank you very much I appreciate the help.