Lecture : The Chain Rule MTH 124

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1 At this point you may wonder if these derivative rules are some sort of Sisyphean task. Fear not, by the end of the next few lectures we will complete all the differentiation rules needed for this course. At which point you can easily impress friends and family with your savant-esque deriving abilities. Before any such tomfoolery, let s review the derivative rules we have learned thus far. At this point you should be comfortable finding the derivatives of Power Functions Polynomials Constant Functions Linear Functions Quadratic Functions General Polynomials Moebius Strip II by M.C. Escher. Products of Functions Quotients of Functions 1. For homework review examples of each of the cases above and make sure you can apply the rules we have discussed to determine these derivatives. Today we will learn how to take derivatives of compositions of functions. Compositions of Functions The rule below gives us a way to determine derivatives of compositions of functions. Chain Rule: Derivative of a composition of functions If fx) and gx) are differentiable functions, then the composite function hx) = fgx)) is differentiable, and h x) = [fgx))] = f gx))g x). Using Leibniz notation we can formulate this as dh dx = df dg dg dx. It may not be immediately clear what this rule gives us. The basic idea is that by viewing a function as a composition of two or more simpler functions we can apply the rule above to determine the derivative of a more complex function. The hard part here is to train yourself to view functions in this different way and getting comfortable with the notation. Fortunately this comes with practice. Let s start by considering the example on the next page 63

2 Example 1 Determine the derivative of hx) = x 2 + 3) 5. We could proceed by expanding the function, then using our existing rules. However this is impractical, and as we ll see, the rule above allows us to differentiate this function much quicker. Notice that this looks very similar to problems we ve encountered before. We already know how to differentiate the inside expression x We also know how to differentiate outside expression x 5. We can think of hx) as the composition of Put another way, fx) = x 5, and gx) = x hx) = fgx)), = fx 2 + 3), = x ) 5. By viewing this function as a composition of two functions we can apply the chain rule to find the derivative. According to the chain rule we have that h x) = fgx) ) = f gx))g x), So we need to determine f gx)) and g x). The term f gx)) looks a bit gnarly, but it simply represents the expression that comes from plugging gx) into the derivative of fx). So before we can determine f gx)) we first need to determine f x). Taking the derivative of fx) and gx) we have the following. f x) = 5x 4, and g x) = 2x. To determine f gx)) we evaluate f x) at gx), which gives us f gx)) = f x 2 + 3), = 5x 2 + 3) 4. Combining our results give us a final answer of h x) = f gx) )) g x) ) = 5x 2 + 3) 4) 2x) = 10xx 2 + 3) Determine the derivative of fx) = t + 1) 4.Hint: Use the chain rule!) 64

3 In the previous lecture we saw the way we naturally wanted to take the derivatives of products and quotients of functions was incorrect. The same holds for the chain rule. In other words f gx) )) f g x) ). Example 2 To illustrate this example let us consider the function fx) = x + 3) 2. It s tempting to think that the derivative is given by x + 3) 2 ) 21) 1 = 2. However notice that expanding fx) gives us that fx) = x 2 + 6x + 9. Using our previous derivative rules we get f x) = 2x + 6. Since these answers don t agree, and we assume the previous derivative rules to be true we can see why f gx) )) f g x) ). 3. Use the chain rule to determine the derivative of each of the following functions. 1 a) ht) = t 3 + t) 100 b) wr) = 52r + 3) 5 c) At) = 1 1 t 1 If you are having trouble applying the chain rule, write the function as a composite of two functions then differentiate according to the chain rule. 65

4 d) mx) = 1 2 x 2 4. Suppose hx) = fgx)) and kx) = gfx)). Use the figure below to answer the following questions. If a derivative does not exist write DNE. a) Determine h 2). b) Determine k 2). 66

5 Example 3 The percentage y, of total personal consumption, an individual spends on food is approximately y = 35x 0.25 percentage points 5.6 x 17.5), where x is the percentage the individual spends on education. An individual finds that she is spending x = t percent of her personal consumption on education, where t is time in months since January 1 st. Estimate how fast the percentage she spends on food is changing on November 1 st. Be sure to specify units. 5. The amount of gas, G, in gallons, consumed by a car depends on the distance traveled, s, in miles, and s depends on the time, t, in hours. If 0.05 gallons of gas is consumed for each mile traveled, and the car is traveling at 30 miles/hr, how fast is gas being consumed? Write the amount of gas consumed as a function of time t as a composition of s and t, then interpret your answer in terms of the chain rule.) 67

6 A Composition of More Than Two Functions We can apply the chain rule iteratively to handle the composition of multiple functions. 1 Example 4 Determine the derivative of P x) = x 4 + 1) First notice that by exponent rules we can rewrite this function as P x) = x 4 + 1) 5 +7 ) 1. From here one can view the function as a composition of the three simpler functions In other words fx) = x 1, gx) = x 5 + 7, and hx) = x P x) = fghx))) = x 4 + 1) ) 1. To take the derivative we ll apply the chain rule twice. Applying the chain rule once gives us P x) = f ghx)) ) ghx)) ) We can then apply the chain rule a second time, to the term ghx)) ), to get ghx)) ) = g hx))h x). Putting these two facts together gives us P x) = f ghx)) ) g hx))h x). Admittedly, this looks expression looks rather terrible. However we have now successfully reduced our bigger problem to a series of simpler problems. Thus to finish this problem we need to determine the two factors f ghx)) ), and g hx)). Using derivative rules we have that This gives us The second term is given by f x) = x 2, g x) = 5x 4, and h x) = 4x 3. Putting everything together we have f ghx)) = f gx 4 + 1)), = f x 4 + 1) 5 + 7), = x 4 + 1) 5 + 7) 2. g hx)) = g x 4 + 1) = 5x 4 + 1) 4 P x) = x 4 + 1) 5 + 7) 2) 5x 4 + 1) 4) 4x 3). Put in another way, the chain rule is a telescoping derivative process where we start by differentiating the outermost function while leaving the inner function intact, then we multiply the resulting expression by the derivative of next outermost function. This process is then repeated until we have differentiated to the inner function. 68

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