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Find the gradients of y = 3x^2 − (2/3) x + 1 at x = 0

y = 3x^2 − (2/3) x + 1Derivative of 3x^2 = 6x3 * 2 = 6x^(2-1) = x^1 = xDerivative of -(2/3)x = -2/3-(2/3) * 1 = -2/3x^(1-1) = x^0 = 1dy/dx = 6x - 2/3at x = 0dy/dx = 6(0) - 2/3dy/dx = -2/3

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Find the gradients of y = 3x^2 − (2/3) x + 1 at x = 0

Related Maths A Level answers

Prove the identity: (cos θ + sin θ)/(cosθ-sinθ) ≡ sec 2θ + tan 2θ

using the substitution u=6-x^2 integrate (x^3)/(6-x^2)^1/2 with respect to x, between 1 and 2

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f(x) is defined by f(x) = 3x^3 + 2x^2 - 7*x + 2. Find f(1).