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Mathematics 19 Online
OpenStudy (anonymous):

I need to find dy/dx of 3cos^2 2x - 3sin^2 2x

OpenStudy (anonymous):

d/dx cos²(x³) ..................... use the chain rule 3 times as follows = d/dz cos²(z) dz/dx ......... z = x³ = d/du u² du/dz dz/dx ...... u = cos(z) = (2u) (-sin(z)) (3x²) ......... du/dz = -sin(z) , d/du u² = 2u, dz/dx = 3x² = -6x² cos(x³) sin(x³) Answer: -6x² cos(x³) sin(x³)

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