Lesson 3-5 Chain Rule or U-Substitutions

Objectives Use the chain rule to find derivatives of complex functions

Vocabulary none new

Example 1 1.g(t) = (5t³ - t + 9) 4 2.y = (e 4x² ) (cos 6x) Find the derivatives of the following: g’(t) = 4 (5t³ - t + 9)³ (15t² - 1) y’(x) = (e 4x² ) (-sin 6x) (6) + (8x) (e 4x² ) (cos 6x) = -6 e 4x² sin 6x + (8x) (e 4x² ) (cos 6x)

Example 2 3.d(t) = t – 16t² d’(t³ - 2t² + 1) 4.f(x) = (tan x²) (sin 4x³) Find the derivatives of the following: d’(t) = 103 – 32t d’(t³ - 2t² + 1) = 103 – 32(t³ - 2t² + 1) = -32t³ + 64t² + 71 f’(x) = tan x² (cos 4x³) (12x²) + (sec² x²) (2x) sin 4x³

Example 3 5.y = 6 2x-1 6.f(x) = (tan 2x)³ (3x³ - 4x² + 7x - 9) Find the derivatives of the following: f’(x) = 3(tan 2x) ² (2) (3x³ - 4x² + 7x - 9) + (tan 2x)³ (9x² - 8x +7) y’(x) = (ln 6) 6 2x-1 (2) = (2ln 6) 6 2x-1

Example 4 Assume that f(x) and g(x) are differentiable functions about which we know information about a few discrete data points. The information we know is summarized in the table below: 1.If p(x) = xf(x), find p’(2) 2.If q(x) = 3f(x)g(x), find q’(-2) x f(x)f’(x)g(x)g’(x) ? p’(x) = d(xf(x))/dx = (1) f(x) + f’(x) x p’(2) = (-1) + (5) (2) = = 9 q’(x) = d(3f(x)g(x))/dx = 3[g’(x) f(x) + f’(x) g(x)] q’(-2) = 3[(6)(4) + (-1)(5)] = 3[24 – 5] = 57

Example 4 cont 3.If r(x) = f(x) / (5g(x)) find r’(0) 4.If s(x) = f(g(x)), find s’(1) x f(x)f’(x)g(x)g’(x) ? r’(x) = d(f(x)/(5g(x)))/dx = (1/5) [g(x) f’(x) - f(x) g’(x)] / g(x)² r’(0) = (1/5)[(8)(-3) - (-6) (-5)] / (8)² = (1/5) [ ] /64 = -54/320 = s’(x) = d(f(g(x)))/dx = f’(x) g’(x) [chain rule!] s’(1) = (6) (3) = 18

Example 4 cont 5.If t(x) = (2 – f(x)) / g(x) and t’(2) = 4, find g’(2) x f(x)f’(x)g(x)g’(x) ? t’(x) = d((2-f(x)) / g(x))/dx = [g(x) (-f’(x)) – (2-f(x)) g’(x)] / g(x)² t’(2) = 4 = [(1)(-5) - (2-(-1)) (x)] / (1)² = (1/5) [ ] /64 = -54/320 =

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