Question

Find the derivative of y with the respect to x:
y=sin‾²(x³)
y'=cos‾²(x³)(3x²)
y'=3x²cos‾²(x³)

How come my answer is incorrect? Thanks in advance.

Answer 1

let z = x^3 then dz/dx = 3 x^2

y = sin^-2 z

dy/dz = (-2 )(sin^-3 z)(cos z)

chain dy/dx = (dy/dz)(dz/dx)
so
dy/dx = (-2 )(sin^-3 z)(cos z)(3x^2)

= -6 x^2 (sin^-3 z)(cos z)