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Drill

Drill. Let f(x) = sin(x) and g(x) = 1+ x 2 and h(x) = 7x f(g(x)) = g(h(x)) = f(g(h(x)) h(g(x)). Chain Rule. Lesson 3.6. Objectives. Students will be able to differentiate composite functions using the Chain Rule. find slopes of parameterized curves. Relating Derivatives.

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Drill

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  1. Drill • Let f(x) = sin(x) and g(x) = 1+ x2 and h(x) = 7x • f(g(x)) = • g(h(x)) = • f(g(h(x)) • h(g(x))

  2. Chain Rule Lesson 3.6

  3. Objectives • Students will be able to • differentiate composite functions using the Chain Rule. • find slopes of parameterized curves.

  4. Relating Derivatives. The function y = 6x -10 = 2(3x – 5) It is the composite function of y = 2u and u = 3x -5 y(u) = y(3x – 5) = 2 (3x – 5) = 6x – 10 How are the derivatives of these three functions related? dy/dx = 6 dy/du = 2 du/dx = 3 dy/dx = (dy/du)(du/dx) =6

  5. Relating Derivatives The polynomial y = 9x4 + 6x2 + 1 = (3x2 + 1)2 is the composite of y = u2 and u = 3x2 + 1. Calculate each of the following : dy/dx dy/du du/dx (dy/du)(du/dx)

  6. Chain Rule If y = f (u) and u = g(x) are both differentiable functions, then so is y = f (g(x)) and

  7. Chain Rule The derivative of the composition of two differentiable functions f and g is

  8. Examples

  9. Examples

  10. Examples

  11. Examples

  12. Homework • Page 153: 1-14

  13. Example: #16, p. 153 • y = x3 (2x – 5)4 • We are going to have to use the product rule AND the chain rule. • u = x3 , so du/dx = 3x2 • v = (2x – 5)4 • Let v = w4 , w = 2x – 5 • dv/dw = 4w3 and dw/dx = 2 • So, (dv/dw)(dw/dx) = dv/dx = 8w3 = 8(2x-5)3

  14. Example: #16, p. 153 • The product rule: • u (dv/dx) + v (du/dx) • u = x3 , so du/dx = 3x2 • v = (2x – 5)4 , dv/ dx 8(2x-5)3 • x3 (8(2x-5)3) + (2x – 5)4 (3x2) • 8x3(2x – 5)3 + 3x2(2x -5)4 • x2 (2x -5)3 [8x + 3(2x -5)] • x2 (2x -5)3 [8x + 6x – 15] • x2 (2x -5)3 [14x– 15]

  15. Example: #33, p. 153 • Find the value of (f◦g)’ at the given value of x. • f(u) = u 5+ 1 and u = g(x) = x1/2 • at x = -1 • f’(u) = 5u4 and g’(x) = ½x-1/2 • 5(x1/2 )4 • ½x-1/2 • 5x2 • ½x-1/2 • 5(1) • ½(1)-1/2 = 5/2

  16. Differentiating with Parameterized Curves A parameterized curve (x(t), y(t)) is differentiable at t is x and y are differentiable at t. If this is true, then It follows then that

  17. Example with Parametrized Curve Find the equation of the line tangent to the curve at the given point value of t.

  18. Example with Parametrized Curve Find the equation of the line tangent to the curve at the given point value of t.

  19. Homework • Page 153: 18-38 even, 42, 46 • #36 is BONUS! PLEASE PUT ON A SEPARATE PIECE OF PAPER!

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