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6.2 Rates of Change for Composite Functions - The Chain Rule. Given: y = (4 – 3 x ) 3. then y = u 3. let u = 4 – 3 x. y is a function of u and u is a function of x is a composite function.
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6.2 Rates of Change for Composite Functions - The Chain Rule Given: y = (4 – 3x)3 then y = u3 let u = 4 – 3x y is a function of u and u is a function of x is a composite function. Therefore if h(x) = (fog)(x), then h´(x) = f ´(g(x))·g´(x)
Proof of the Chain Rule multiply by Du/Du provided Du ¹0 if both limits exist (def. of derivative)
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.
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) = (2s3 – 5)4 4(2s3 – 5)3 (6s2) f ´(s) = derivative of outer · derivative of inner f ´(s) = 24s2(2s3 – 5)3
Example 3: Determine
Differentiating a quotient with constant numerator. Example 4: Determine f ´(t) f(t) = 5(4t + 3)–3 f ´(t) = (–3)(5)(4t + 3)–3–1 (4) derivative of outer · derivative of inner f ´(t) = (–60)(4t + 3)–4
Example 5: Determine y´ This is the composition of y = u3 and
Example 6: A cylindrical barrel has a radius of 1.2 m. If water is pouring in at a rate of 3 m3/min, at what rate is the height of water in the barrel increasing? 1.2 m V = pr2h 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.