U2 L5 Quotient Rule QUOTIENT RULE UNIT 2 LESSON 5 QUOTIENT RULE

If you thought the product rule was bad... U2 L5 Quotient Rule If you thought the product rule was bad...

Function f is the QUOTIENT of functions g and h U2 L5 Quotient Rule Function f is the QUOTIENT of functions g and h It would be nice if the rule were But, it is NOT!!!

Let’s show that if f(x) = x3 + x2 – 5x and g(x) = – x then Example 1 Let’s show that if f(x) = x3 + x2 – 5x and g(x) = – x then U2 L5 Quotient Rule then = and

If and and so ≠ From previous slide then Example 1 continued U2 L5 Quotient Rule Example 1 continued If and and so ≠ From previous slide then

𝒇 𝒈 ′ 𝒙 = 𝒈 𝒙 𝒇 ′ 𝒙 −𝒇 𝒙 𝒈′(𝒙) 𝒈(𝒙) 𝟐 U2 L5 Quotient Rule Quotient Rule In this section we develop a formula for the derivative of the quotient of two functions. 𝒇 𝒈 ′ 𝒙 = 𝒈 𝒙 𝒇 ′ 𝒙 −𝒇 𝒙 𝒈′(𝒙) 𝒈(𝒙) 𝟐

U2 L5 Quotient Rule Let’s try it in English The bottom (g (x)) times the derivative of the top f ′ (x) minus the top (f (x)) times the derivative of the bottom g ′ (x) all over the bottom g (x) squared.

Example 1 continued Quotient Rule x ≠ 0 FROM BEFORE if U2 L5 Quotient Rule Example 1 continued FROM BEFORE if 𝒇 𝒙 = 𝒙 𝟑 + 𝒙 𝟐 −𝟓𝒙 𝒈 𝒙 =−𝒙 then and Quotient Rule x ≠ 0

If f(x) = x2 + 3x + 2 and g(x) = x + 1 then U2 L5 Quotient Rule EXAMPLE 2 If f(x) = x2 + 3x + 2 and g(x) = x + 1 then

Use f(x) = x2 + 3x + 2 and g(x) = x + 1 to show that Example 2 continued U2 L5 Quotient Rule Use f(x) = x2 + 3x + 2 and g(x) = x + 1 to show that ≠ but from previous slide

Use f(x) = x2 + 3x + 2 and g(x) = x + 1 to show that Example 2 continued U2 L5 Quotient Rule Use f(x) = x2 + 3x + 2 and g(x) = x + 1 to show that and from previous slide =

Example 3 Using the Quotient Rule U2 L5 Quotient Rule Differentiate. State any restrictions on the domain. Restriction on domain x ≠ 4

Example 4 Using the Quotient Rule U2 L5 Quotient Rule Example 4 Using the Quotient Rule Differentiate . State any restrictions on the domain. Since x2 + 1 is always > 0 there are no restrictions

Example 5 Using the Quotient Rule U2 L5 Quotient Rule Example 5 Using the Quotient Rule 𝑦= 𝟐 𝒙 𝟑 +𝟓 𝟑𝒙 𝒅𝒚 𝒅𝒙 = 𝟑𝒙 𝟔 𝒙 𝟐 −(𝟐 𝒙 𝟑 +𝟓)(𝟑) 𝟑𝒙 𝟐 𝒅𝒚 𝒅𝒙 = 𝟏𝟖 𝒙 𝟑 −𝟔 𝒙 𝟑 −𝟏𝟓 𝟗 𝒙 𝟐 𝒅𝒚 𝒅𝒙 = 𝟏𝟐 𝒙 𝟑 −𝟏𝟓 𝟗 𝒙 𝟐 , 𝒙≠𝟎 𝒅𝒚 𝒅𝒙 = 𝟑(𝟒 𝒙 𝟑 −𝟓) 𝟗 𝒙 𝟐 = 𝟒 𝒙 𝟑 −𝟓 𝟑 𝒙 𝟐 , 𝒙≠𝟎

Example 6 Using the Quotient Rule U2 L5 Quotient Rule 𝑦= 3−𝑥 6−5𝑥 Differentiate using the Quotient Rule. 𝒅𝒚 𝒅𝒙 = (𝟔−𝟓𝒙) −𝟏 −(𝟑−𝒙)(−𝟓) 𝟔−𝟓𝒙 𝟐 𝒅𝒚 𝒅𝒙 = −𝟔+𝟓𝒙+𝟏𝟓−𝟓𝒙 𝟔−𝟓𝒙 𝟐 𝒅𝒚 𝒅𝒙 = 𝟗 𝟔−𝟓𝒙 𝟐

Example 6 continued Find the slope of the tangent at P(0, ½ ) U2 L5 Quotient Rule Example 6 continued Find the slope of the tangent at P(0, ½ ) From previous slide 𝒅𝒚 𝒅𝒙 = 𝟗 𝟔−𝟓𝒙 𝟐 𝒅𝒚 𝒅𝒙 = 𝟗 𝟔−𝟓(𝟎 𝟐 ) = 𝟗 𝟔 = 𝟑 𝟐

Example 7 Application x = 0 or x = -5 At what points on the curve U2 L5 Quotient Rule Example 7 Application At what points on the curve is the tangent line horizontal? The tangent line will be horizontal when the derivative = 0 2x(x + 5) = 0 x = 0 or x = -5 Points are (0, 0) and (-5, -5)

Example 7 Application y = 0 y = -5 (0, 0) (- 5, - 5) U2 L5 Quotient Rule Example 7 Application (0, 0) y = 0 y = -5 (- 5, - 5)

Complete Homework Assignment Questions 1-5