Question

Let y = (3 - x^2)^(3/2). Find the first and second derivatives of y.

(dy)/(dx) = A x (3 - x^2)^B

Answer 1

dy/dx= 3/2(3-x^2)^(1/2) * 2x= 3x(3-x^2)^(1/2); A=3, B=1/2 don't forget chain rule
d2y/dx2= 3x^2(3-x/2)^(-1/2)