Quotient Rule of Derivatives Chain Rule Quotient Rule

Home , Quotient rule

fundamental rules of derivatives fundamental rules of derivatives Chain Rule MCV4U: Calculus & Vectors Recap Determine the derivative of f (x) = 12√3 2x5 7x. − The inner function is h(x) = 2x5 7x, so h (x) = 10x4 7. − 0 − 1 2 Quotient Rule of Derivatives The outer function is g(h) = 12h 3 , so g 0(x) = 4h− 3 .

5 2 4 f 0(x) = 4(2x 7x)− 3 (10x 7x) J. Garvin − − 40x4 28 or − 2 (x(2x4 7)) 3 −

J. Garvin— Quotient Rule of Derivatives Slide 1/13 Slide 2/13

fundamental rules of derivatives fundamental rules of derivatives Quotient Rule Quotient Rule

p(x) 1 Sometimes a function can be written as the quotient of two Rewrite f (x) = as f (x) = p(x)[q(x)]− and use the other functions. q(x) chain rule to diﬀerentiate. To determine the derivative of a function with the form p(x) 1 2 f 0(x) = p0(x)[q(x)]− + p(x)( 1)[q(x)]− q0(x) f (x) = , use the Quotient Rule. − q(x) 2 1 2 [q(x)] = p0(x)[q(x)]− p(x)[q(x)]− q0(x) Quotient Rule − × [q(x)]2 p(x) p (x)q(x) p(x)q (x) p0(x)q(x) p(x)q0(x) If f (x) = , then f (x) = 0 − 0 , = − q(x) 0 [q(x)]2 [q(x)]2 provided that q(x) = 0. 6 v du u dv If y = u , then dy = dx − dx , v = 0. v dx v 2 6

Note the similarity to the product rule in the numerator.

J. Garvin— Quotient Rule of Derivatives J. Garvin— Quotient Rule of Derivatives Slide 3/13 Slide 4/13

fundamental rules of derivatives fundamental rules of derivatives Quotient Rule Quotient Rule Example Example 3x2 1 4x2 6 Determine the derivative of f (x) = − . Determine the derivative of y = − . 2x + 3 √x

2 2 1 Let p(x) = 3x 1 and q(x) = 2x + 3. Let u = 4x 6 and v = √x = x 2 . − − du dv 1 Then p0(x) = 6x and q0(x) = 2. Then dx = 8x and dx = 2√x . 2 (6x)(2x + 3) (3x 1)(2) 1 f 0(x) = − − (8x)(√x) (4x2 6) (2x + 3)2 dy − − 2√x dx = 2 12x2 + 18x 6x2 + 2 [√x] = − 2 6x2 + 3 (2x + 3) or 6x2 + 18x + 2 √x3 = (2x + 3)2

J. Garvin— Quotient Rule of Derivatives J. Garvin— Quotient Rule of Derivatives Slide 5/13 Slide 6/13 fundamental rules of derivatives fundamental rules of derivatives Quotient Rule An Alternative to the Quotient Rule Example Since we derived the general formula for the quotient rule The concentration, C (g/L), of a chemical released into a using the chain rule, an alternative to the quotient rule is to 6t use both the product rule and the chain rule instead. stream after t days is given by C(t) = 2 . At what rate 2t + 9 This may be useful, because it is not necessary to memorize is the concentration changing after 4 days? a separate formula for quotients. Let p(t) = 6t and q(t) = 2t2 + 9. Derivatives derived using each technique may appear

Then p0(t) = 6 and q0(t) = 4t. diﬀerent, but should be the same after simpliﬁcation. 6(2t2 + 9) 6t(4t) f 0(x) = − (2t2 + 9)2 6(2(4)2 + 9) 6(4)(4(4)) = − (2(4)2 + 9)2 138 = −1681

J. Garvin— Quotient Rule of Derivatives J. Garvin— Quotient Rule of Derivatives Slide 7/13 Slide 8/13

fundamental rules of derivatives fundamental rules of derivatives An Alternative to the Quotient Rule An Alternative to the Quotient Rule Example Example Use the product and chain rules to show that the derivative 3x 4 At what point(s), if any, is the tangent to y = − 3x2 1 6x2 + 18x + 2 x2 5x of f (x) = − is f (x) = . 2x + 3 0 (2x + 3)2 horizontal? −

2 1 2 1 Rewrite the function as y = (3x 4)(x 5x)− . Rewrite the function as f (x) = (3x 1)(2x + 3)− . − − − The slope of the tangent is given by dy . 1 2 2 dx f 0(x) = (6x)(2x + 3)− + (3x 1)( 1)(2x + 3)− (2) − − 1 2 2 dy 2 1 2 2 = (6x)(2x + 3)− 2(3x 1)(2x + 3)− dx = 3(x 5x)− + (3x 4)( 1)(x 5x)− (2x 5) − − − − − − − 2 3(x2 5x) (3x 4)(2x 5) (6x)(2x + 3) 6x + 2 = − − − − = −2 (x2 5x)2 (2x + 3) − 2 3x2 + 8x 20 6x + 18x + 2 = − − = 2 (x2 5x)2 (2x + 3) −

J. Garvin— Quotient Rule of Derivatives J. Garvin— Quotient Rule of Derivatives Slide 9/13 Slide 10/13

fundamental rules of derivatives fundamental rules of derivatives An Alternative to the Quotient Rule An Alternative to the Quotient Rule dy Since the slope of a horizontal line is 0, set dx = 0 and solve for x. 3x2 + 8x 20 0 = − − (x2 5x)2 − = 3x2 + 8x 20 − − Since 82 4( 3)( 20) < 0, there are no real solutions to − − − this equation. 3x 4 Therefore, there are no points on y = − where the x2 5x tangent is horizontal. −

J. Garvin— Quotient Rule of Derivatives J. Garvin— Quotient Rule of Derivatives Slide 11/13 Slide 12/13 fundamental rules of derivatives Questions?

J. Garvin— Quotient Rule of Derivatives Slide 13/13