21. Chain rule Chain rule Statement 21.1. Statement Examples The power rule says that d [xn] = nxn−1: dx This rule is valid for any power n, but not for any base other than the simple input variable x. For instance, d (2x + 5)3 6= 3(2x + 5)2: dx The function h(x) = (2x + 5)3 is built up of the two simpler functions g(x) = 2x + 5 and Table of Contents f(x) = x3: h(x) = (2x + 5)3 = (g(x))3 = f(g(x)): JJ II Technically speaking, h is the composition of f and g. The next rule expresses the derivative of such a function in terms of the derivatives of its components. J I Chain rule. For functions f and g Page1 of8 d [f(g(x))] = f 0(g(x)) · g0(x): dx Back Print Version In the composition f(g(x)), we call f the outside function and g the inside function. With this terminology, the rule says that the derivative of the composition of two functions is Home Page the derivative of the outside function evaluated at the inside function times the derivative of the inside function. 21.2. Examples Chain rule d Statement 21.2.1 Example Find the derivative (2x + 5)3. dx Examples Solution We begin by viewing (2x + 5)3 as a composition of functions and identifying the outside function f and the inside function g. The outside function is the last thing you do when computing the expression for a given input x. Here, the outside function is the cubing function: (2x + 5)3 = f(g(x)); where f(x) = x3 and g(x) = 2x + 5: Next, we do the computations required for the chain rule formula: Table of Contents f(x) = x3 g(x) = 2x + 5 f 0(x) = 3x2 g0(x) = 2 JJ II f 0(g(x)) = 3(2x + 5)2 J I Finally, we use the formula: d Page2 of8 [f(g(x))] = f 0(g(x)) · g0(x) dx ### Back d (2x + 5)3 = 3(2x + 5)2 · 2 dx Print Version Home Page d 21.2.2 Example Find the derivative sin(x5). dx Solution Here, the outside function is the sine function: Chain rule sin(x5) = f(g(x)); where f(x) = sin x and g(x) = x5: Statement Examples So f(x) = sin x g(x) = x5 f 0(x) = cos x g0(x) = 5x4 f 0(g(x)) = cos(x5) giving d [f(g(x))] = f 0(g(x)) · g0(x) dx Table of Contents ### d sin(x5) = cos(x5) · 5x4 dx JJ II J I d 21.2.3 Example Find the derivative sin5 x. dx Page3 of8 Solution Recalling that sin5 x means (sin x)5, we see that the outside function is the one Back that raises an input to the fifth power: sin5 x = f(g(x)); where f(x) = x5 and g(x) = sin x: Print Version In order to reduce the number of steps, we go immediately to the chain rule formula and Home Page do the intermediate computations mentally as required: d [f(g(x))] = f 0(g(x)) · g0(x) Chain rule dx ### Statement d sin5 x = 5(sin x)4 · cos x Examples dx d h 2 i 21.2.4 Example Find the derivative 5x −4x+3 . dx Solution Here, the outside function is the exponential function with base 5: 2 5x −4x+3 = f(g(x)); where f(x) = 5x and g(x) = x2 − 4x + 3: Table of Contents Trimming the number of steps a bit more, we omit the formula for the chain rule and just think \Derivative of outside function, evaluated at inside function, times derivative of JJ II inside function": d h 2 i 2 5x −4x+3 = 5x −4x+3 ln 5 · (2x − 4): J I dx Page4 of8 d p 21.2.5 Example Find the derivative 5ex + 4x3 : dx Back Solution The outside function is the square root function: Print Version h i d p x 3 1 x 3 −1=2 x 2 5e + 4x = 2 (5e + 4x ) (5e + 12x ): dx Home Page d h 4 i 21.2.6 Example Find the derivative cos ex . Chain rule dx Statement Solution The outside function is the cosine function: Examples d h 4 i 4 d h 4 i cos ex = − sin ex · ex dx dx 4 4 = − sin ex · ex (4x3): The second step required another use of the chain rule (with outside function the exponen- tial function). ex 21.2.7 Example Find the derivative of f(x) = ee . Table of Contents Solution The chain rule is used twice, each time with outside function the exponential function: JJ II d h ex i f 0(x) = ee J I dx ex d h x i = ee · ee dx Page5 of8 ex x = ee · ee · ex: Back Print Version x3 + 2 21.2.8 Example Find the derivative of f(x) = sin p : 5x − 74x Home Page Solution We have Chain rule d x3 + 2 f 0(x) = sin p Statement dx 5x − 74x Examples x3 + 2 d x3 + 2 = cos p · p 5x − 74x dx 5x − 74x p p 4x d 3 3 d 4x x3 + 2 5x − 7 x + 2 − (x + 2) 5x − 7 = cos p · dx p dx 5x − 74x ( 5x − 74x)2 x3 + 2 = cos p 5x − 74x p 5x − 74x(3x2) − (x3 + 2) 1 (5x − 74x)−1=2(5 − 74x(ln 7)(4)) Table of Contents · 2 : 5x − 74x JJ II J I Page6 of8 Back Print Version Home Page 21 { Exercises Chain rule Statement Examples 21 { 1 Use the chain rule to find the following derivatives. Present your solution just like the solution in Example 21.2.1 (i.e., write the given function as a composition of two functions f and g, compute the quantities required on the right-hand side of the chain rule formula, and finally show the chain rule being applied to get the answer). d (a) (2x4 − 5x)7. dx d Table of Contents (b) [cos(8t − 3)]. dt JJ II 21 { 2 Find the derivatives of the following functions. (You may omit details and proceed as in Example 21.2.5.) J I p (a) f(x) = 3 8 − x3. Page7 of8 (b) f(t) = esin 2t. Back 3 21 { 3 Find the derivative of f(x) = 2x +4x cos(3x). Print Version Home Page e5t 21 { 4 Find the derivative of f(t) = cos . t − t8 Chain rule Statement Examples Table of Contents JJ II J I Page8 of8 Back Print Version Home Page.