Advanced Functions and Introductory Calculus
Examples

Definition of the derivative: [f (x + Δx) - f (x)] ÷ Δx = d[f (x)]/dx Δy/Δx = dy/dx f (x) = - x3 + 3x2 - 2 at the point x = 1 At x = 1 , f (1) = - (1)3 + 3(1)2 - 2 = 0 [(- x3 + 3x2 - 2) - 0] ÷ [x - 1] ... the denominator is zero at the limit x = 1 = [- x2(x - 1) + 2(x + 1)(x - 1)] ÷ (x - 1) ... factoring the numerator = [- x2(x - 1) + 2(x + 1)(x - 1)] ÷ (x - 1) ... cancelling like terms = - x2 + 2(x + 1) = - (1)2 + 2((1) + 1) = -1 + 4 = 3 ... substituting the value of the limit y = - x3 + 3x2 - 2 y + Δy = - (x + Δx)3 + 3(x + Δx)2 - 2 = - x3 - 3x2•Δx - 3x•Δx2 - Δx3 + 3x2 + 6x•Δx + 3Δx2 - 2 ... expanding the right-hand side Δy = [- x3 - 3x2•Δx - 3x•Δx2 - Δx3 + 3x2 + 6x•Δx + 3Δx2 - 2] - [- x3 + 3x2 - 2] = - 3x2•Δx - 3x•Δx2 - Δx3 + 6x•Δx + 3Δx2 ... subtracting y from each side Δy/Δx = [- 3x2•Δx - 3x•Δx2 - Δx3 + 6x•Δx + 3Δx2]/Δx = - 3x2 - 3x•Δx - Δx2 + 6x + 3Δx ... dividing both sides by Δx dy/dx = - 3x2 - 3x•Δx - Δx2 + 6x + 3Δx = - 3x2 + 6x ... passing to the limit At x = 1 , dy/dx = - 3(1)2 + 6(1) = 3 ... substituting for x y = f (x) = - x3 + 3x2 - 2 Let g(x) = - x3 Let h(x) = 3x2 Let k(x) = -2 f (x) = g(x) + h(x) + k(x) g(x) + g(Δx) = - (x + Δx)3 = - x3 - 3x2•Δx - 3x•Δx2 - Δx3 g(Δx) = [- x3 - 3x2•Δx - 3x•Δx2 - Δx3] - [- x3] = - 3x2•Δx - 3x•Δx2 - Δx3 g(Δx)/(Δx) = [- 3x2•Δx - 3x•Δx2 - Δx3]/Δx = - 3x2 - 3x•Δx - Δx2 d[g(x)]/dx = - 3x2 - 3x•Δx - Δx2 = - 3x2 ... passing to the limit h(x) + h(Δx) = 3(x + Δx)2 = 3x2 + 6x•Δx + 3Δx2 h(Δx) = [3x2 + 6x•Δx + 3Δx2] - 3x2 = 6x•Δx + 3Δx2 h(Δx)/Δx = [6x•Δx + 3Δx2]/Δx = 6x + 3Δx d[h(x)]/dx = 6x + 3Δx = 6x ... passing to the limit k(x) + k(Δx) = - 2 + Δc , where Δc, the rate of change of a constant, equals zero k(Δx) = [- 2 + Δc] - [- 2] = Δc k(Δx)/Δx = Δc/Δx = 0 d[k(x)]/dx = = 0 ... passing to the limit dy/dx = d[f (x)]/dx = d[g(x) + h(x) + k(x)]/dx = d[g(x)]/dx + d[h(x)]/dx + d[k(x)]/dx = - 3x2 + 6x At x = 1 , dy/dx = - 3(1)2 + 6(1) = 3 ... substituting for x y = (2 + x)/(2 - 3x) Let 2 + x = u therefore du/dx = 1 Let (2 - 3x) -1 = v therefore dv/dx = d(2 - 3x) -1/d(2 - 3x) • d(2 - 3x)/dx ... Chain Rule dv/dx = (-1)•(2 - 3x) -2•(- 3) y = (2 + x)(2 - 3x) -1 ... changing the division to a multiplication d(uv)/dx = u• d(v)/dx + v• d(u)/dx ... Product Rule dy/dx = (2 + x)•(-1)•(2 - 3x) -2•(- 3) + (2 - 3x) -1•(1) dy/dx = 3(2 + x)/(2 - 3x)2 + 1/(2 - 3x) ... changing the terms to fractions dy/dx = [3(2 + x) + (2 - 3x)]/(2 - 3x)2 ... multiplying the right hand term by (2 - 3x) to create a common denominator dy/dx = (6 + 3x + 2 - 3x)/(2 - 3x)2 dy/dx = 8/(2 - 3x)2 y = (2 + x)/(2 - 3x) Let 2 + x = u therefore du/dx = 1 Let 2 - 3x = v therefore dv/dx = - 3 d(u/v)/dx = [v• d(u)/dx - u• d(v)/dx] ÷ v2 ... Quotient Rule dy/dx = [(2 - 3x)•(1) - (2 + x)•(- 3)] ÷ (2 - 3x)2 dy/dx = (2 - 3x + 6 + 3x)/(2 - 3x)2 dy/dx = 8/(2 - 3x)2 The Product Rule and the Quotient Rule both return the same result: dy/dx = 8/(2 - 3x)2 At P(x, y) = (1, -3) , Slope = dy/dx = 8/(2 - 3(1))2 = 8 Point-Slope form of equation for the tangent: (y - y1) = m(x - x1) (y - (-3)) = 8(x - (1)) ... substituting y = 8x - 11 ... equation of the line tangent at (1, -3) Supplementary: f (x) = - x3 + 3x2 - 2 at the point x = 2 At x = 2 , f (2) = - (2)3 + 3(2)2 - 2 = 2 [(- x3 + 3x2 - 2) - 2] ÷ [x - 2] ... the denominator is zero at the limit x = 2 = [- x3 + 3x2 - 4] ÷ [x - 2] = [- x2(x - 2) + (x + 2)(x - 2)] ÷ (x - 2) ... factoring the numerator = [- x2(x - 2) + (x + 2)(x - 2)] ÷ (x - 2) ... cancelling like terms = - x2 + x + 2 = - (2)2 + 2 + 2 = 0 ... substituting the value of the limit d[f (x)]/dx = - 3x2 + 6x Substituting for x in the derivative: At x = 2 , dy/dx = - 3(2)2 + 6(2) = - 6 + 6 = 0 The tangent is parallel to the x-axis at (2, 2) Supplementary: f (x) = - x3 + 3x2 - 2 at the point x = -2 At x = -2 , f (-2) = - (-2)3 + 3(-2)2 - 2 = -18 [(- x3 + 3x2 - 2) - 18] ÷ [x - (-2)] ... zero divisor at the limit x = -2 = [- x3 + 3x2 - 20] ÷ [x + 2] = [- x2(x + 2) + 5(x - 2)(x + 2)] ÷ (x + 2) ... factoring the numerator = [- x2(x + 2) + 5(x - 2)(x + 2)] ÷ (x + 2) ... cancelling like terms = - x2 + 5x - 10 = - (-2)2 + 5(-2) - 10 = - 4 - 10 - 10 = -24 ... substituting the value of the limit d[f (x)]/dx = - 3x2 + 6x Substituting for x in the derivative: At x = -2 , dy/dx = - 3(-2)2 + 6(-2) = -12 + (-12) = -24