f (θ) = Sin2θ + Cos3(θ), find f ' (π/6) and f " (π/6) y = 2x + 3x^2, explain and simplify x^2 (d^y/dx^2) - 2x(dy/dx) + 2y f (x) = x^2 - 1/x^3, find f ' (-1) and f " (-1)
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f '(θ) = 2 cos 2θ - 3 sin 3θ f ''(θ) = -4 sin 2θ - 9 cos 3θ So f ' (π/6) = 1 - 3 = - 2 And f " (π/6) = -4(sqrt3/2) - 0 = -4(sqrt3/2)
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