thevisualclassroom.com 6.2 Rates of Change for Composite Functions - The Chain Rule Therefore if h(x) = (f o g)(x), then h ´ (x) = f ´ (g(x)) · g ´ (x) Given: y = (4 – 3x) 3 let u = 4 – 3x then y = u 3 y is a function of u and u is a function of x is a composite function.

thevisualclassroom.com Proof of the Chain Rule multiply by  u/  u provided  u  0 if both limits exist (def. of derivative)

thevisualclassroom.com Example 1: Determine if y = (4x + 6) 3 = 3(4x + 6) 2 (4) = 12(4x + 6) 2 If we had expanded y = (4x + 6) 3 and found the derivative of the product, the result would have been the same.

thevisualclassroom.com The Chain Rule with the Power Rule = n[g(x)] n–1 ·g´(x), where n is a constant Take the derivative of the ‘outer’ function multiplied by the derivative of the ‘inner’ function. Example 2: Determine f ´(s) if f(s) = (2s 3 – 5) 4 f ´(s) = derivative of outer · derivative of inner 4(2s 3 – 5) 3 (6s 2 ) f ´(s) = 24s 2 (2s 3 – 5) 3

thevisualclassroom.com Determine Example 3:

thevisualclassroom.com Determine f ´(t) Example 4: f(t) = 5(4t + 3) –3 f ´(t) = (–3)(5)(4t + 3) –3–1 derivative of outer · derivative of inner f ´(t) = (–60)(4t + 3) –4 Differentiating a quotient with constant numerator. (4)

thevisualclassroom.com Determine y´ Example 5: This is the composition of y = u 3 and

thevisualclassroom.com A cylindrical barrel has a radius of 1.2 m. Example 6: If water is pouring in at a rate of 3 m 3 /min, at what rate is the height of water in the barrel increasing? 1.2 m V =  r 2 h Since r = 1.2 m, Using the chain rule we have: The height of water is changing at a rate of 0.66 m/min.