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4-3 Derivative of a Quotient of Two Functions

4-3 Derivative of a Quotient of Two Functions. Pages 136- 141 Katie Warmbold Jillian Defrietas. Introduction. This chapter allows you to solve “catching up problems” in which one moving object wants to reach another moving object. Topics of Discussion.

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4-3 Derivative of a Quotient of Two Functions

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  1. 4-3Derivative of a Quotient of Two Functions Pages 136- 141 Katie Warmbold Jillian Defrietas

  2. Introduction • This chapter allows you to solve “catching up problems” in which one moving object wants to reach another moving object.

  3. Topics of Discussion • Property: Derivative of a Quotient of Two Functions: • If y = u/v, where u and v are differentiable functions, and v doesn’t equal 0, then u’v – uv’ y’ = v^2 Words – Derivative of top times bottom, minus top times derivative of bottom, all divided by bottom squared

  4. Relation to Earlier Lessons • In order to differentiate, or find the derivative, some functions, its important to remember the Property: Derivatives of Sine and Cosine Functions. • d/dx(sinx) = cosX • d/dx(cosx) = -sinX • The chain rule, explained in 3-7, is applied in numerous problems when finding the derivative.

  5. Typical Problem • Differentiate: (5x – 2)^7 y = (4x +9)^3 • Differentiate: d ( 5 ) dx ( 7x^3) • Differentiate: sin5x f(x) = 8x - 3

  6. Detailed Procedure • Problem # 1: (5x – 2)7 y= (4x +9)3 Use the Property to solve problem (HINT:Derivative of toptimesbottom, minustoptimesderivative of bottom, all divided bybottom squared.) The chain rule, as in 4-2, also is applied when finding derivatives y’ = 7(5x-2)^6(5) * (4x+9)^3 - (5x - 2)^7 * 3(4x + 9)^2(4) (4x + 9)^6 = (5x - 2)^6 (80x + 339) (4x + 9)^6

  7. Solution • Differentiate: • sin5x • f(x) = 8x - 3 • 1.) cos5x(5)(8x - 3) - sin5x(8) • f’x = (8x - 3)^2 • 2.) 5cos5x(8x - 3) - 8sin5x • f’x = (8x - 3)^2

  8. Real Life Example • Black Hole Problem : • Ann Astronaut’s spaceship gets trapped in the gravitational field of a black hole! Her velocity, v(t) miles per hour is given: • 1000 • v(t) = 3 - t • a.) How fast is she going when t=1? When t = 2? When t = 3? • B.) Write an equation for the acceleration function, a(t). • C.) What is her acceleration when t=1? When t = 2? When t = 3?

  9. Real Life Example Solution • A.) a(1) = 250 mph/hr • a(2) = 1000 mph/hr • a(3) = 1000/0 mph/hr = no answer • B.) 1000 • a(t) = (3 - t)^2 • C.) a(1) = 250 mph/hr • a(2) = 1000 mph/hr • a(3) = 1000/0 mph/hr = no answer

  10. Conclusion • This section taught you how to find an equation for the derivative function in one step when the function contains a quotient of two other functions. • The Property; Derivative of a Quotient of Two Functions was discussed and applied.

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