SDSU

 

Math 121 - Calculus for Biology I
Spring Semester, 2001
Chain Rule - Examples

 © 2001, All Rights Reserved, SDSU & Joseph M. Mahaffy
San Diego State University -- This page last updated 16-May-01

Chain Rule - Examples

  1. Chain Rule of Differentiation
  2. Graphing a Composite Function
  3. Hassell's Model

Several examples are presented below to demonstrate the chain rule. Examples show the basic differentiation and applications to extrema in graphing. We also present a couple of applications, including the stability analysis of Hassell's population model and a composite function for studying the effects of energy consumption with increasing population.

Example 1: (Differentiation using the Chain Rule) Differentiate the following functions:

f(x) = (x3 - 4x2 + e-2x)6 and g(x) = (4x + ln(x4 + 4x2 + 1))4

Solution: The function f(x) can be considered a composite of the function f1(u) = u6 and the function f2(x) = x3 - 4x2 + e-2x. It is easy to find the derivatives of both f1 and f2. We have f1'(u) = 6u5 u' and f2'(x) = 3x2 - 8x - 2e-2x. From the chain rule, we see that

f '(x) = 6(f2(x))5f2'(x) = 6(x3 - 4x2 + e-2x)5(3x2 - 8x - 2e-2x).

To analyze the function g(x), we have a couple of composite functions to consider. Let the function g1(u) = u4 and the function g2(x) = 4x + ln(h2(x)). Then the chain rule first gives g1'(u) = 4u3 u' or

g'(x) = 4(g2(x))3g2'(x) = 4(4x + ln(x4 + 4x2 + 1))3 g2'(x).

To find the derivative of g2(x), we need to differentiate the composite h(x) = h1(h2(x)), where h1(v) = ln(v) with h1'(v) = (1/ v)v' and h2(x) = x4 + 4x2 + 1 with h2'(x) = 4x3 + 8x, so

But g2'(x) = 4 + h'(x). So combining the results above, we see that

Example 2: (Graphing a Composite Function) An important function in statistics is the normal distribution function, which classically gives the Bell curve. A normal distribution function is given by

 

where a is a normalizing factor and s is the standard deviation. We would like to find the points of inflection for this curve and determine its significance. Also, plot this function for several values of s.

Solution: To find the points of inflection, we need to take the derivative twice. Its clear that this is an even function, and that its maximum occurs at x = 0 with N(0) = a/s. We take the first derivative of N(x) and obtain

 

Note that N '(x) = 0 at x = 0, as expected.

The second derivative requires the product rule along with the chain rule. The result is given by

The points of inflection occur when N "(x) = 0, which is easily seen to be when either x = -s or x = s. Thus, the points of inflection occur one standard deviation out the normal distribution function. It turns out that 68% of the area under the normal distribution occurs in the interval -s < x < s.

Below we show a graph of the normal distribution with s = 1, 2, 3, and 4.

Example 3: ( Hassell's Model) Suppose that a study shows that a population, Pn, of butterflies satisfies the dynamic model given by the following equation:

 

where n is measured in weeks. Let P0 = 200, then find P1 and P2. Find the intercepts, all extrema of H(P), and any asymptotes for P > 0.

Determine the equilibria and analyze the behavior of the solution near the equilibria.

Solution: The iterations are found in the standard way with

P1= H(200) = 16200/(1.4)4 = 4271.

Similarly, P2= H(4271) = 43. Thus, we see dramatic population swings with this model, suggesting an instability.

Analyzing H(P), we find that the only intercept is (0, 0) and that there is a horizontal asymptote with H = 0. To find where the maximum occurs, we differentiate the function.

Solving H '(P) = 0, we have 1 - 0.006P = 0 or P = 500/3 = 166.7. With H(500/3) = 4271.5, the maximum occurs at (166.7, 4271.5). A graph of the updating function with the identity function is shown below.

 

As always, the equilibria are found by solving Pe = H(Pe), which is equivalent to solving

Pe(1 + 0.002Pe)4 = 81Pe.

So either Pe = 0 or (1 + 0.002Pe)4 = 81, which gives 1 + 0.002Pe= 3 or Pe= 1000. The stability of these equilibria can be determined by examining the derivative at the equilibria. At Pe = 0, H '(0) = 81, which implies from our rules that the solutions monotonically grow away from 0. At Pe= 1000, H '(1000) = 81(-5)/243 = -5/3. This implies that the solution near this equilibrium oscillates and goes away from the equilibrium. In fact, this model produces a period 4 solution with the solution asymptotically oscillating from 163 to 4271 to 42 to 2453. A simulation of this model is shown below.