f(x)= - (4x^2/3)/((x^2/3)-2)

Question

f'(x)=?

a. -(4/(((x^2/3)-2)^2)
b. -4
c. 16/((3cube rootx)((x^(2/3)-2)^2)
d. ((-16/3cuberootx)+(16/ (3cuberootx)))/((x^(2/3)-2)^2)

Answer 1

f(x)= 4/(1-2x^(-2/3))

check that

f(x)=4(1-2x^(-2/3))^-1

f'= -4(1-2x^(-2/3))^-2 (+4/3x^-5/3)
at x=1
f'=-4(-1)(4/3)= 16/3 making answer c look good, the others wont work.

You probably can manipulate f'to equal answer c, but, I am lazy.