Thread: 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:




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.

Thanks in advance.

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:




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 ... ... you know better, don't you?

if you're trying to find , just use the chain rule. expanding that binomial is just going to make a bigger mess.

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)