Can you guys please find the derivative of this function?
f(x)= x^(2/3)(2x-5)
You have to use the product rule
2006-06-10 04:34:21 · 5 answers · asked by A 2 in Science & Mathematics ➔ Mathematics
df(x)/dx = (x)^(2/3)*(2) -2/3(x)^(-1/3)(2x-5) [using product rule]
df(x)/dx = 2(x)^(2/3) - 4/3 (x)^(2/3) - 10/3(x)^(-1/3) [simplifying]
df(x)/dx = (2/3)(x)^(2/3) - (10/3)(x)^(-1/3) [ after more simplification]
2006-06-10 04:37:37 · answer #1 · answered by organicchem 5 · 1⤊ 1⤋
use the product rule:
= 2/3 (x)^(2/3-1) + x^(2/3) * (2 - 0)
= 2/3 * x^(-1/3) + x^ (2/3) * 2
arrange the sequence,
= (2/3) * x^(-1/3) + 2x^(2/3)
2006-06-10 11:45:48 · answer #2 · answered by kcool 2 · 0⤊ 0⤋
f(x)= x^(2/3)(2x-5) I read this as f(x)= x^{(2/3)(2x-5)}
f'(x) = (2/3)(2x-5) X x^{(2/3)(2x-5)-1} x (2/3) X 2
When you read it as f(x)= {x^(2/3)} X (2x-5) then
f'(x) = {x^(2/3)} X 2 + {(2/3) X x^(2/3 - 1)} X (2x-5)
Be happy!
2006-06-10 11:42:38 · answer #3 · answered by Thermo 6 · 0⤊ 0⤋
f'(x)={(2/3) .x^(-1/3)}(2x-5)+(x^2/3)(2)
2006-06-10 11:45:51 · answer #4 · answered by minee 2 · 0⤊ 0⤋
2/3x^(-1/3)(2x-5)+2x^(2/3)
2006-06-10 11:42:42 · answer #5 · answered by Eulercrosser 4 · 0⤊ 0⤋