Over the last several sections, we have been steadily refining a model for the motion of a weight hanging from a spring. In Section 2.5, we found that the cosine function could be used to describe the oscillations of the weight. In Section 2.6, we found that if we combined this with an exponential function, we could describe an oscillatory motion that gradually dies out. However, the model u(t ) = e-t cos t  represents only one possible motion (specifically, one where the motion dies out quite rapidly).  If we wanted to change the rate at which the motion dies out, we could change the exponential function used. If we wanted to change the period of the oscillations, we could change the trigonometric function used. For example, u(t ) = e-t /6cos 4t  describes a somewhat different motion (see Figure 2.38).  This looks much more like the gradual dying out of a spring's motion that you might expect, compared to the motion shown in Figure 2.36. Unfortunately, we do not presently have any rules for computing the derivative of the composition function e-t /6. Figure 2.38 u = e-t /6 cos 4t  and u = e-t  cos t  Recall that in Example 5.2, we used the product rule to show that Using this result and the product rule, notice that   = cos xsin2 x+sin x(2sin xcos x)   = 3sin2 xcos x. We leave it as a straightforward exercise to extend this result to You should notice that in each case, we have brought the exponent down, lowered the power by one and then multiplied by the derivative of sin x. Notice that we can write sin 4x as the composite function  sin 4x = (sin x) 4 = f (g(x)), where g(x) = sin x and f (x) = x4. Differentiating f and g, we have f '(x) = 4x3 and g'(x) = cos x. It's now easy to see that 4sin3x = f '(g(x)). Finally, observe that the derivative of the composite function is This is an example of the chain rule, which has the following general form.

 

7.1    Chain Rule   If g is differentiable at x and f is differentiable at g(x), then   At this point, we can only prove the special case where g'(x) 0. Let F(x) = f (g(x)). Then,           Since F(x) = f (g(x)).           Multiply numerator and denominator by g(x+h)-g(x).           Regroup terms.                     = f ' (g(x))g'(x) , where the next to the last line is valid since as h 0, g(x+h) g(x), by the continuity of g. (Recall that since g is differentiable, it is also continuous.) You will be asked in the exercises to fill in some of the gaps in this argument. In particular, you should identify why we need g' (x) 0 in this proof.

 

It is often helpful to think of the chain rule in Leibniz notation. If y = f (u) and u = g(x), then y = f (g(x)) and the chain rule says that (7.1)

 

7.1    Using the Chain Rule   Differentiate y = (x3+x-1)5.     For u = x3+x-1, note that y = u5. From (7.1), we have Since y = u5.     = 5(x3+x-1)4(3x2+1). The last term is the derivative of the expression inside the parentheses (often referred to as the derivative of the inside).

 

7.2    Using the Chain Rule for an Exponential Function   Find     Let u = -t /6. Then, from (7.1), Notice that here, the derivative of the inside is the derivative of the exponent.

 

7.3    Finding the Velocity of a Hanging Weight   Suppose the position of a weight hanging from a spring is given by u(t ) = e-t /6cos 4t. Find the velocity of the weight at any time t.     Using the product rule and the chain rule, we have v(t ) By the product rule.   By the chain rule.     = -e-t /6cos 4t-4e-t /6sin 4t.

 

You are now in a position to calculate the derivative of virtually every function you can write down, by using the rules we have at our disposal. In many cases, the chain rule must be combined with other differentiation rules.

 

7.4    The Derivatives of Some Similar Trigonometric Functions   Compute the derivative of f (x) = cos x3, g(x) = cos3x, and h(x) = cos 3x.     Note the differences in these three functions. Using the implied parentheses we normally do not bother to include, we have f (x) = cos (x3), g(x) = (cos x)3, and h(x) = cos (3x). For the first function, we have Next, we have g' (x)   = 3(cos x)2 (- sin x) = -3sin xcos2x. Finally, we have

 

7.5    A Derivative Involving a Chain Rule and a Quotient Rule   Find the derivative of     We have f '(x) By the chain rule.   By the quotient rule.

 

7.6    A Derivative Involving a Quotient Rule and Several Chain Rules   Find the derivative of     We have f '(x) By the quotient rule.   By the chain rule.

 

7.7    A Derivative Involving Repeated Chain Rules   Find the derivative of     You can think of the chain rule here as differentiating from the outside-in, where we illustrate the process one step at a time, as follows:

 

Next, we consider a simple application.

 

7.8    Finding the Maximum Concentration of a Chemical   The concentration x of a certain chemical after t seconds of an autocatalytic reaction is given by Show that x'(t ) > 0 and use this information to determine that the concentration of the chemical never exceeds 10.     Before computing the derivative, look carefully at the function x(t ). The independent variable is t and the only term involving t is in the denominator. So, we don't need to use the quotient rule. Instead, first rewrite the function as x(t ) = 10(9e-20 t +1)-1 and use the chain rule. We get x'(t )   = -10(9e-20 t +1)-2(-180e-20t )   = 1800e-20t (9e-20t +1)-2   Notice that since e-20 t  > 0 for any t, both the numerator and denominator are positive, so that x'(t ) > 0. Since all of the tangent lines have positive slope, the graph of y = x(t ) rises from left to right, as shown in Figure 2.39. Since the concentration increases for all time, the concentration is always less than the limiting value which is easily computed to be Figure 2.39 Chemical concentration

 

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