1. Lesson 3-5 Chain Rule or U-Substitutions

2. Objectives Use the chain rule to find derivatives of complex functions

3. Vocabulary none new

4. Differentiation Chain Rule If f and g are both differentiable and F = f○g is the composite function defined by F(x) = f(g(x)) d dy dy du ---- [F(x)] = F’(x) = f’(g(x)) g’(x) or ---- = ---- ---- dx dx du dx In words: the derivative of a composite function is equal to the derivation of the outer function times the derivative of the inner function. The notation on the right is Leibniz notation and is often referred to as u substitution: y = f(u) and u = g(x) so y’(x) = f’(u) g’(x)

5. Differentiation Chain Rule Cont. Combine with the Power Rule: d du d ---- [u n ] = nu n-1 ---- or ----- [g(x)] n = n[g(x)] n-1 g’(x) dx dx dx Example: y = (x 6 – 2x + 1) 8 let u(x) = (x 6 – 2x + 1) u’(x) = 6x 5 – 2 dy dy du du then y = u 8 ------ = ----- ---- = 8 u 7 ----- dx du dx dx = 8 (x 6 – 2x + 1) 7 (6x 5 – 2)

6. Specific Applications of the Chain Rule Differentiation of exponential functions: d du d ---- [e u ] = e u ----- ---- [a x ] = a x ln a dx dx dx Example: y = e 2x + 1 dy dy du du let u = (2x + 1) then y = e u ------ = ----- ---- = e u ----- dx du dx dx = e 2x + 1 (2) = 2 e 2x + 1

7. Specific Applications of the Chain Rule cont. Greater length Chains: y = e cos(3x + 1) let u = cos(3x + 1) and v = 3x + 1 then y = e u dy dy du dv ------ = ----- ---- ----- = e u (-sin(v)) (3) dx du dv dx = e cos(3x + 1) (-sin(3x+1)) (3) = -3sin(3x+1) e cos(3x + 1)

8. Example 1 1.y = (2x² - 4x + 1) 60 2.f(x) = 1 / (2x 5 – 7)³ Find the derivatives of the following: y’(x) = 60 (2x² - 4x + 1) 59 (4x – 4) f(x) = (2x 5 – 7) -3 f’(x) = (-3) (2x 5 – 7) -4 (10x 4 )

9. Example 2 3.y = cos (x²) 4.y =  (1 + (1/x) Find the derivatives of the following: y’(x) = -sin (x²) (2x) = -2x sin (x²) y’(x) = (1 + (1/x)) ½ = ½ (1 + (1/x)) -½ (-1/x²)

10. Example 3 5.y = ( 3 + (x³ - 2x) 5 ) 8 6.y = tan³ (  x) Find the derivatives of the following: If y = v 8, with v(x) = 3 + (x³ - 2x) 5 and, v(u) = (3 + u 5 ) with u(x) = (x³ - 2x) So dy/dx = dy/dv dv/du du/dx = 8v 7 (0 + 5u 4 ) (3x² - 2) y’(x) = 8(3 + (x³ - 2x) 5 ) 7 (5(x³ - 2x) 4 ) (3x² - 2) y’(x) = 3 tan² (  x) (½ x -½ )

11. Example 4 7.y =  x³ + 6x 8. y = sec x² Find the derivatives of the following: y’(x) = (1 + (1/x)) ½ = ½ (1 + (1/x)) -½ (-1/x²) y’(x) = 3 tan² (  x) (½ x -½ )

12. Example 5 9. y = sec² x 10. y = cos³ (x²) Find the derivatives of the following: y’(x) = 2 sec x (sec x tan x) = 2 sec² x tan x y’(x) = 3 cos² (x²) (2x) = 6x cos² (x²)

13. Example 6 9. y = (sin 2x) (cos 3x) 10. y = x² cos (e x ) Find the derivatives of the following: y’(x) = 2 (cos 2x) (cos 3x) – (sin 2x) (-sin 3x) (3) = 2 (cos 2x) (cos 3x) + 3 (sin 2x) (sin 3x) y’(x) = 2x (cos e x ) + (x²) (-sin e x ) (e x ) = 2x (cos e x ) – x² e x (sin e x )

14. Summary & Homework Summary: –Chain rule allows derivatives of more complex functions –Chain rule is also known as u-substitution Homework: –pg 224 - 227: 3, 4, 7, 8, 11, 14, 15, 22, 29, 32, 43, 67