what is the derivative of [(x^3)/2-x]^4/3; d/dx(f'(x))?

Question

thnxs

Answer 1

(4/3)[x^3/(2-x)]^(1/3)
[(2-x)(3)x^2-x^3(-1)]/
(2-x)^2=

4x[1/(2-x)]^(1/3)
[3x^2(2-x)+x^3]/
3(2-x)^2

Answer 2

the derivative is = 4/3 * ((x^3)/2 - x)^4/3 * (3/2*x^2 -1)

Answer 3

F'(X) = (24 X^5 - 8 X^6 ) / (6 - 3X)(2 - X)^2

This is the most Simplified answer if u wish u can open the square by a^2 + b^2 - 2ab

Integral

Answer 4

deivative of u^n = n u^(n-1) * du/dx here u = (x^3)/2-x

du/dx =3/2 x^2 -1 and u^(n-1) = ((x-3)/2-x)^1/3

final result first derivative 1derivative = (1/2) x^2 (( (x-3)/2-x))^1/3

second derivative
1/2 (2x * ((x-3))/2 -x)^1/3 + x^2 (((x-3)/2-x)^-2/3)*3/2 x^2-1)

Answer 5

y=[(x^3)/2-x]^4/3

dy/dx=4/3 *[(x^3)/2-x]^(4/3-1) *derv(x^3/2-x) [chain rule,derv(x^n)=n*x^n-1]

dy/dx=4/3 * [(x^3)/(2-x)]^(1/3) * [ { (2-x)(3 x^ 2) } - { (x^3)(-1) } ] / (2-x)^2 [Division rule]

dy/dx=4/3 * [xcube/(2-x)]raiseto(1/3) * [ x^4 - 5x^3 +6x^2] / (2-x)^2

Answer 6

let f(x)= [g(x)/h(x)]^r
then f'(x) = {r[g(x)/h(x)]^r-1 }*{g'(x)/h(x) - g(x)h'(x)}/{h(x)}^2}

r=4/3, r-1 = 1/3

g(x) = x^3
g'(x) =3x^2

h(x) = (2-x)
h'(x)= -1
{h(x)}^2= (2-x)^2

combining all:
f'(x) = (4/3){(x^3/(2-x))^1/3}{{3x^2/(2-x) }+{x^3/(2-x)^2}}

similarly find f''(x)

Answer 7

[(x^3)/2-x]^4/3;
f'(x)=4/3*((x^3)/2-x)^1/3*(3(x^2)/2-1)
d/dx(f'(x))=4/3*1/3*((x^3)/2-x)^(-2/3)*(3(x^2)/2-1)^2+4/3*((x^3)/2-x)^1/3*(6x/2)
d/dx(f'(x))=4/3((x^3)/2-x)^1/3{[(3x^2/2-1)/(x^3/2-x)]+6x/2}