0.1 The Chain Rule. db dt = db

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1 0. The Chain Rule A basic illustration of the chain rules comes in thinking about runners in a race. Suppose two brothers, Mark an Brian, hol an annual race to see who is the fastest. Last year Mark won the race, so this year he will be consiere the pace setter. Let M an B be functions that represent the position of each of the brothers at any given time. Since Mark is the pace setter, his position only epens on time, so we write M(t). In contrast, we measure Brian s velocity (an thus position) relative to Mark s, so we write B(M(t)). If Brian is running 2/3 as fast as Mark, then Brian s velocity is simply B t = 2 3 M t, meaning that Brian s position changes at a rate 2/3 of that of Mark s. Knowing Mark s velocity (or instantaneous rate of change of position with time) we can fin Brian s velocity just by multiplying by 2/3, the rate at which Brian s position is changing relative to Mark s. Thus, if Mark is running at 0 mph, it follows Brian is running at 20/ mph. Written symbolically, we observe B t = B M M t. This is calle the chain rule. It states that if we know the rate of change of Brian s position with respect to Mark s an we know the rate of change of Mark s position with respect to time, we can fin the rate of change of Brian s position with respect to time, simply by multiplying them together. If we a a thir runner into the race, Kevin, who is running twice as fast as Brian, then his velocity is given by K t = 2 B T = 2 B M M t. Noting that 2 is just the rate of change of Kevin s position with respect to Brian s, we fin that K t = K t B M M t. In this way we can chain together any number of functions, an fin the overall rate of change by looking at the relative rates of change at each step. Note that here we have the composition K(B(M(t))), so we first ifferentiate K with respect to B, B with respect to M, an finally M with respect to t. Theorem 0.. (The Chain Rule). Suppose f is ifferentiable at g(x) an g is ifferentiable at x. It follows that the composition (f g) is ifferentiable at x, an x (f g)(x) = f (g(x)) g (x). In the composition f(g(x)) we call f the outer function an g the inner function. The basic mechanism of the chain rule is as follows: ifferentiate the outer function holing the inner function as a constant, then multiply the result by the erivative of the inner function. If there is a composition of more than two functions, the above process is simply repeate as many times as necessary. Example. Fin the erivative of e αt (with respect to t), α R. Solution The above function is a composition of two functions, e u an u = αt. Thus, we can apply the chain rule. We take the erivative of the outer function (which is e u ), evaluate the result

2 at the inner function (u = αt), ifferentiate the inner function (yieling α), an then multiply the results. t eαt = e αt α = αe αt. Thus, to fin the erivative of an exponential function where the argument is multiplie by a constant, simply multiply the exponential function by that constant. This is consistent with the erivative of the exponential function, where that constant is simply a. Example 2. Fin the erivative of e x2. Solution Once again we can apply the chain rule, to the composition of e u an u = x 2. We take the erivative of the outer function (which is e u ), evaluate the result at the inner function (u = x 2 ), ifferentiate the inner function (yieling 2x), an then multiply the results. x ex2 = e x2 2x. Example 3. Fin the erivative of a t (with respect to t), where a R +. Solution Using the chain rule we can fin the erivative of a base a exponential using the erivative of the base e exponential. Write a = e ln(a), which can be one as the exponential function an natural logarithm are inverses (as long as a is a positive real number, as we have not efine the natural logarithm for negative numbers). Now we take the erivative using the chain rule, fining t at = t eln(a)t = e ln(a)t ln(a) = ln(a) a t. Example 4. Fin the erivative of +y 2 (with respect to y). Solution In orer to solve this problem it is helpful to first rewrite the function as ( + y 2 ). Then we can see there is a composition of two functions u an u = + y 2. Applying the chain rule we fin y ( + y2 ) = ( + y 2 ) 2 2y = 2y ( + y 2 ) 2. Example 5. Fin the erivative of e e2x+. Solution In this case we re actually looking at a composition of three functions we know how to ifferentiate. How can we apply the chain rule in this case? Let us choose our two functions as e u an u = e 2x+. The chain rule tells us x ee2x+ = u eu x e2x+ = e e2x+ x e2x+. Now we nee to fin the erivative of e 2x+ to complete the problem. Choosing e u an u = 2x + we fin x e2x+ = e 2x+ 2 = 2e 2x+. Substituting this into our first equation we fin x ee2x+ = 2e e2x+ e 2x+. 2

3 The above example illustrates an extremely useful observation we mae earlier. If we have a composition of three functions, to fin the erivative we simply the multiply the erivatives of the three component functions (at each respective step) together. Thus, we fin that f g h f(g(h(x))) = x g h x, where we evaluate f/g at g(h(x)) an g/h at h(x). We can easily exten the formula above to a composition of as many functions as we like. Example 6. Fin the erivative of (x 2 + ) Solution There is nothing specific about the chain rule that tells us how we must choose our inner an outer functions; the only requirement is we choose functions that we know how to ifferentiate (even if we nee to use the chain rule again to o so). Let us choose u an u = x 2 + as our functions, because we know how to ifferentiate both of them. We fin x ((x2 + ) 3 + 2) = 3(x 2 + ) 2 2x. Choosing those two functions as our chain of functions worke well because we knew how to ifferentiate both of them. Nevertheless, another chain of functions woul work just as well. Let us choose u 3 + 2, u = v + an v = x 2. Then we fin x ((x2 + ) 3 + 2) = 3(x 2 + ) 2 2x = 3(x 2 + ) 2 2x, where u/v = an v/x = 2x. This result is exactly the same as our previous result, as it must be. The lesson to be learne here is that the most important thing to consier with the chain rule is choosing functions that are easy to ifferentiate. As long as the composition or chain of functions chosen to represent the original function is equivalent, it oesn t matter how many functions are chosen. Example 7. Fin the erivative of y = ( + a 2t ) 4 with respect to t. Solution Letting our representative functions be u 4, u = + a v an v = 2t we fin y t = 4( + a2t ) 5 ln(a)a 2t 2 = 8 ln(a)a2t ( + a 2t ) 5. As state in the previous section, while it is har to fin an intuitive justification for the quotient rule, it follows simply enough from the chain an prouct rules. We simply begin by writing u(x)/v(x) = u(x) (v(x)), where (v(x)) is raise the negative first power, not an inverse function. Differentiating we fin x (u(x) (v(x)) ) = u (x) (v(x)) + u(x) ( (v(x)) 2 v (x) = (v(x))2 (v(x)) 2 ( u (x) (v(x)) + u(x) ( (v(x)) 2 v (x) = u (x)v(x) u(x)v (x) (v(x)) 2. In fact, one may sometimes fin in practice that is it simply easier to use the prouct an chain rules in conjunction, rather than working with (an remembering) the quotient rule. 3 )

4 There is a final application of the chain rule which is extremely useful for increasing the amount of functions we can ifferentiate. If we have a function f an its inverse f, by efinition (f f)(x) = (f f )(x) = x. Now if two sies of an expression are equal, it follows that they must remain equal after ifferentiation (after all, how can the same function have two ifferent erivatives?). Using this observation along with the chain rule, we fin that (f f ) (x) = f (f (x))(f (x)) =. where the right han sie comes from the fact x/x =. Solving this equation we fin (f ) (x) = f (f (x)). This amazing result is enough to allow us to fin the erivative of a function we on t know, if we can fin the erivative of its inverse function. Through this technique we can fin the erivatives of logarithmic functions, an later will be able to use it to fin the erivatives of inverse trigonometric functions. Theorem 0..2 (Derivative of Inverse Functions). Let f be a function ifferentiable at f, with inverse function f. It follows that (f ) exists, an (f ) (x) = f (f (x)). Example 8. Fin the erivative of ln(x) Solution If f (x) = ln(x) then f(x) = f (x) = e x, so using the above result we fin x ln(x) = e ln(x) = x, because the exponential unoes the action of the logarithm (leaving us with x) in the above equation. Example 9. Fin the erivative of log a (x). Solution If f (x) = log a (x) then f(x) = a x an f (x) = ln(a)a x. Using the above result we fin x log a(x) = ln(a)a log a (x) = ln(a)x, because the base a exponential unoes the action of the logarithm (leaving us with x) in the above equation. Before finishing the current iscussion on the chain rule, it is worth mentioning a subtlety in the chain rule. Given two functions that are ifferentiable, the chain rule tells us how to fin the erivative of their composition. Nevertheless, it may be possible to compose functions which are not ifferentiable, an get a result which is ifferentiable. Let us consier the functions { u Q f(u) = 0 u / Q an u(x) = { 0 u Q u / Q. 4

5 These in an of themselves seem to be very esoteric functions, so let s take a minute to think about what they really are. For the function f, whenever the input is a rational number, we get an output of. Otherwise, for an irrational number, the output is 0. This function is so baly broken up that we cannot raw it, but we can note that it is iscontinuous everywhere (because between any two rationals is an irrational, an between any two irrationals is a rational). The secon function u behaves similarly to f, except that it is 0 for rationals an for irrationals. It is equally baly behave. However, when we look at the composition f(u(x)), something remarkable happens. No matter the input, u outputs a rational number (either 0 or ), so the output of f(u(x)) = for all x. This function is continuous everywhere, as well as ifferentiable (f u) (x) = 0 for all x. There is no inconsistency here with the chain rule, because the component functions o not satisfy the hypothesis of the chain rule; the chain rule tells us nothing about the erivative of a composition of functions which are not ifferentiable. 5