y=sin‾²(x³) y'=cos‾²(x³)(3x²) y'=3x²cos‾²(x³) How come my answer is incorrect? Thanks in advance.
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)
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