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Understanding the Chain Rule and U-Substitution in Calculus

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This lesson focuses on using the chain rule to find derivatives of complex functions, a key concept in calculus. Through various examples, including polynomials and trigonometric functions, learners will understand how to apply the chain rule effectively to calculate derivatives. The session provides a detailed look at both the mathematical process and the underlying principles, reinforcing the relationship between functions and their derivatives. Homework exercises are included to practice the concepts learned.

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Understanding the Chain Rule and U-Substitution in Calculus

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  1. Lesson 3-5 Chain RuleorU-Substitutions

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

  3. Vocabulary • none new

  4. Example 1 Find the derivatives of the following: • g(t) = (5t³ - t + 9)4 • y = (e4x²) (cos 6x) g’(t) = 4 (5t³ - t + 9)³ (15t² - 1) y’(x) = (e4x²) (-sin 6x) (6) + (8x) (e4x²) (cos 6x) = -6 e4x² sin 6x + (8x) (e4x²) (cos 6x)

  5. Example 2 Find the derivatives of the following: d(t) = 360 + 103t – 16t²d’(t³ - 2t² + 1) f(x) = (tan x²) (sin 4x³) 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

  6. Example 3 Find the derivatives of the following: y = 62x-1 f(x) = (tan 2x)³ (3x³ - 4x² + 7x - 9) y’(x) = (ln 6) 62x-1 (2) = (2ln 6) 62x-1 f’(x) = 3(tan 2x)² (2) (3x³ - 4x² + 7x - 9) + (tan 2x)³ (9x² - 8x +7)

  7. 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: If p(x) = xf(x), find p’(2) If q(x) = 3f(x)g(x), find q’(-2) p’(x) = d(xf(x))/dx = (1) f(x) + f’(x) x p’(2) = (-1) + (5) (2) = -1 + 10 = 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

  8. Example 4 cont If r(x) = f(x) / (5g(x)) find r’(0) If s(x) = f(g(x)), find s’(1) 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) [-24 -30] /64 = -54/320 = -0.16875 s’(x) = d(f(g(x)))/dx = f’(x) • g’(x) [chain rule!] s’(1) = (6) (3) = 18

  9. Example 4 cont If t(x) = (2 – f(x)) / g(x) and t’(2) = 4, find g’(2) 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) [-24 -30] /64 = -54/320 = -0.16875

  10. 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

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