Differentiation Product Rule By Mr Porter.

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Presentation transcript:

Differentiation Product Rule By Mr Porter

Important Information. To help explain the Product Rule used in differentiation, I will be using the following derivatives. Exponential Function: Trigonometic Sine Function:

Definition of the Product Rule. If u = f(x) and v = g(x), then Using the dash notation: If y = uv, then Example: Use the Product Rule to differentiate The easy way to do this problem would be to simplify the algebra before differentiating.

Example 1 Differentiate the following, given that Solution Solution Let u = x3 and v = ex Let u = 4x -2 and v = ex then u’ = 3x2 and v’ = ex then u’ = -8x -3 and v’ = ex y’ = uv’ + vu’ y’ = uv’ + vu’ y’ = 4x -2 (ex) +ex (-8x -3) y’ = x3 (ex) +ex (3x2) y’ = x3 ex + 3x2ex factorise y’ = 4x -2 ex – 8x -3ex factorise y’ = x2 ex (x + 3) y’ = 4x -2 ex (1 – 2x -1)

Example 2 Differentiate the following, given that Solution Let u = 4x 3 and v = sin x Solution Let u = -5x and v = sin x then u’ = -5 and v’ = cos x y’ = uv’ + vu’ then u’ = 12x 2 and v’ = cos x y’ = uv’ + vu’ y’ = -5x (cos x) + (-5) sin x y’ = 4x 3 (cos x) + (12x 2) sin x y’ = -5x cos x – 5sin x factorise y’ = 4x 3 cos x + 12x2 sin x factorise y’ = -5(x cos x + sin x) y’ = 4x 2(x cos x + 3 sin x)

Example 3: Combination of Product Rule and Chain Rule. Chain Rule is the Function of a Function Rule. Differentiate y = x2 (3x + 1)4 Now y’ = uv’ + vu’ y’ = x2 x 12(3x + 1)3 +2x (3x + 1)4 Solution Let u = x2 and v = (3x + 1)4 y’ = 12x2(3x + 1)3 + 2x (3x + 1)4 factorise y’ = 2x(3x + 1)3[6x + (3x + 1)] y’ = 2x(3x + 1)3[6x + 3x + 1] Differentiate the brackets, multiply by the derivative of what’s inside! y’ = 2x(3x + 1)3(9x + 1)

Example 4: Combination of Product Rule and Chain Rule. Differentiate y = -2x4 (5 – x2)3 Solution Let u = -2x4 and v = (5 - x2)3 Now y’ = uv’ + vu’ y’ = -2x4 x -6x(5 – x2)2 + -8x3 (5 - x2)3 y’ = 12x5(5 – x2)2 – 8x3(5 - x2)3 factorise Differentiate the brackets, multiply by the derivative of what’s inside! y’ = 4x3(5 - x2)2[3x2 – 2(5 – x2)] y’ = 4x3(5 - x2)2 [3x2 – 10 + 2x2] y’ = 4x3(5 - x2)2 (5x2 – 10) factorise y’ = 20x3(5 - x2)2 (x2 – 2)