# Finding the derivative

• Oct 14th 2009, 03:58 PM
dark-ryder341
Finding the derivative
Hi guys...I'm having some troubles with this one question and it's really starting to bug me cause I keep getting it wrong.

The question is:

http://www.webassign.net/cgi-bin/sym...x%29%29%29%5E2

So, I tried simplifying it...

v = (9*x^1/2 + 5/(3*x^1/3))^2
v = (9x^1/2 + 5/(3x^1/3))^2
I then distributed the 2 from outside the brackets into the brackets, so:
v = (81x + 25/9x^2/3)
Then I took the derivative...
v(der) = 81 + 25/6x^-1/3

Should I have done something with the 25 on the top of that fraction? I'm really not sure.

• Oct 14th 2009, 04:33 PM
skeeter
Quote:

Originally Posted by dark-ryder341
Hi guys...I'm having some troubles with this one question and it's really starting to bug me cause I keep getting it wrong.

The question is:

http://www.webassign.net/cgi-bin/sym...x%29%29%29%5E2

So, I tried simplifying it...

v = (9*x^1/2 + 5/(3*x^1/3))^2
v = (9x^1/2 + 5/(3x^1/3))^2
I then distributed the 2 from outside the brackets into the brackets, so: v = (81x + 25/9x^2/3)

can't do that ... $\displaystyle \textcolor{red}{(a+b)^2 \ne a^2 + b^2}$ ... you know better, don't you?

if you're trying to find $\displaystyle \frac{dv}{dx}$, just use the chain rule. expanding that binomial is just going to make a bigger mess.
• Oct 14th 2009, 07:52 PM
PMatt
I really doubt you're expected to have to multiply that out. So just use the chain rule as the previous poster suggested.

n(a + b)^n-1*(da/dx + db/dx)