Find the derivative of f(x)=[x^2/3](2x-5)^2
F prime(x)=?
2006-10-07 14:35:15 · 6 answers · asked by Zidane 3 in Science & Mathematics ➔ Mathematics
nice try kelly but if the answer and calculus was that wasy, i wouldn't have asked the question
2006-10-07 15:21:03 · update #1
*easy instead of wasy
2006-10-07 15:21:41 · update #2
*easy instead of wasy
2006-10-07 15:21:42 · update #3
=Find the answer and get back with us
2006-10-07 14:36:34 · answer #1 · answered by Anonymous · 0⤊ 1⤋
Here it is..
F prime(x) = (2x)/3*(2x-5)^2+4/3*x^2*(2x-5)
You need to break the original equation down using the chain rule.
F(x) =u*v
F'(x) = u*dV + v*du
Set
u = x^2/3
v = (2x-5)^2
and go from there. Hint. You may need to use the chain rule a second time on the v term by itself in order to calculate -its- derivative, dV
2006-10-07 21:40:02 · answer #2 · answered by Guru 6 · 0⤊ 0⤋
Hmm..so I got
f'(x) = [(2/3)x^(-1/3) * (2x-5)^2] + [4x^(2/3) * (2x-5)]
just be using the chain rule.
2006-10-07 23:57:37 · answer #3 · answered by mathstinks 1 · 0⤊ 0⤋
2 / 3 = 0.666667 (or 2/3)
2006-10-07 22:11:40 · answer #4 · answered by SunniGirl 2 · 0⤊ 1⤋
f'(x)=[x^2/3](2x-5)^2
=x^2/3(4x^2-20x+25)
=[4x^4-20x^3+25x^2]/3
=[16x^3-60x^2+50x]/3
2006-10-07 21:47:53 · answer #5 · answered by tul b 3 · 0⤊ 0⤋
Tried could not sorry!
2006-10-07 21:37:35 · answer #6 · answered by Anonymous · 0⤊ 0⤋