The Chain Rule

The Chain Rule Suppose we have two functions, y = f (u) and u = g(x), and we know that y changes at a rate dy = 3 and 3 times as fast as u, and that u changes at a rate 2 times as fast as x (ie. f 0 (u) = du 0 (x), the derivative of the composition = 2). What if anything can we say about (f ◦ g) g 0 (x) = du dx (f ◦ g)(x)? It stands to reason that y should change at a rate 6 times as fast as x, so we would dy du find (f ◦ g)0 (x) = du · dx = 6. The chain rule asserts that our intuition is correct, and provides us with a means of calculating the derivative of a composition of functions, using the derivatives of the functions in the composition. The Chain Rule Suppose f (u) is differentiable at u = g(x), and g(x) is differentiable at x. It follows that the composite function (f ◦ g)(x) is differentiable at x, and that (f ◦ g)0 (x) = f 0 (g(x)) · g 0 (x) If y = f (u) and u = g(x), then dy du dy = · dx du dx where dy du is evaluated at u = g(x). Many undergraduate calculus texts assert the following proof of the chain rule. Purported Proof of the Chain Rule: Recall that f (u + ∆u) − f (u) dy = f 0 (u) = lim ∆u→0 du ∆u and let u = g(x) and ∆u = ∆g = g(x + ∆x) − g(x). It follows that f [g(x) + ∆g] − f [g(x)] ∆g→0 ∆g f [g(x + ∆x)] − f [g(x)] = lim ∆x→0 g(x + ∆x) − g(x) f 0 [g(x)] = = = lim f [g(x+∆x)]−f [g(x)] ∆x g(x+∆x)−g(x) ∆x→0 ∆x (f ◦ g)0 (x) lim g 0 (x) We are able to replace lim∆g→0 with lim∆x→0 as ∆x → 0 implies ∆g → 0. If we multiply both sides of the equation with g 0 (x), then the proof follows. This proof feels very intuitive, and does arrive to the conclusion of the chain rule. However, there are two fatal flaws with this proof. The first is that although ∆x → 0 implies ∆g → 0, it is not an equivalent statement. It is very possible for ∆g → 0 while ∆x does not approach 0. This leads us to the second flaw with the proof. Let g(x) be a constant function. It follows that for every x and ∆x, g(x + ∆x) − g(x) = 0. This means that the above derivation included division by 0, which is clearly not permitted by the rules of mathematics. For those reasons we will have to discard the above proof, but we currently do not possess the tools to properly prove the chain rule. In the composition f (g(x)) we refer to f as the outer function, and g as the inner function. We can describe the basic mechanism of the chain rule as follows: differentiate the outer function holding the inner function as a constant. Then, multiply the result by the derivative of the inner function. If there is a composition of more than two functions, the above process is simply repeated as many times as necessary. Consider the following examples. Example 1 Find the derivative of eαt (with respect to t), α ∈ R. Solution The above function is a composition of two functions, et and αt. Thus, we can apply the chain rule. We take the derivative of the outer function (which is et ), evaluate the result at the inner function (αt), differentiate the inner function (yielding α), and then multiply the results. d αt e = eαt · α = αeαt dt Thus, to find the derivative of an exponential function where the argument is multiplied by a constant, simply multiply the exponential function by that constant. This is consistent with the derivative of the exponential function, where that constant is simply a 1. 2 Example 2 Find the derivative of ex . Solution Once again we can apply the chain rule, to the composition of ex and x2 . We take the derivative of the outer function (which is ex ), evaluate the result at the inner function (x2 ), differentiate the inner function (yielding 2x), and then multiply the results. d x2 2 e = ex · 2x dx 1 Example 3 Find the derivative of 1+y 2 (with respect to y). Solution In order to solve this problem it is helpful to first rewrite the function as (1 + y 2 )−1 . Then we can see there is a composition of two functions y −1 and 1 + y 2 . Applying the chain rule we find −2y d (1 + y 2 )−1 = −(1 + y 2 )−2 · 2y = dy (1 + y 2 )2 Example 4 Find the derivative of ln[(1 − x)4 + 1]. Solution In this case we have a composition of three functions (ln(x), x4 +1, and 1−x). Fortunately it is simple to repeatedly apply the chain rule, and doing so we find that we only need to multiply by the derivative of each composed function to find the derivative of the entire composition. 1 −4(1 − x)3 d 3 ln[(1 − x)4 + 1] = · 4(1 − x) · (−1) = dx (1 − x)4 + 1 (1 − x)4 + 1 Example 5 Find the derivative of 2t (with respect to t) using the chain rule. Solution We previously calculated this derivative using the definition of the limit, but we can more easily calculate it using the chain rule. Write 2 = eln(2) , which can be done as the exponential function and natural logarithm are inverses. Now take the derivative using the chain rule d t d 2 = eln(2)t = eln(2)t · ln(2) dt dt