3.3 Chain Ruleap Calculus

Chain Rule in Derivatives: The Chain rule is a rule in calculus for differentiating the compositions of two or more functions. All functions are functions of real numbers that return real values. Find Derivatives Using Chain Rules: The Chain rule states that the derivative of f(g(x)) is f'(g(x)).g'(x). It helps to differentiate composite functions. This video explains about Topic 3.3 for subject BUM2123 Applied Calculus which will be used in Substitute Blended Learning.

• 3.3 Chain Ruleap Calculus 2nd Edition
• 3.3 Chain Ruleap Calculus 14th Edition
• 3.3 Chain Ruleap Calculus Calculator

Learning Objectives

• State the chain rule for the composition of two functions.
• Apply the chain rule together with the power rule.
• Apply the chain rule and the product/quotient rules correctly in combination when both are necessary.
• Recognize the chain rule for a composition of three or more functions.
• Describe the proof of the chain rule.

We have seen the techniques for differentiating basic functions ([latex]x^n, , sin x, , cos x[/latex], etc.) as well as sums, differences, products, quotients, and constant multiples of these functions. However, these techniques do not allow us to differentiate compositions of functions, such as [latex]h(x)= sin (x^3)[/latex] or [latex]k(x)=sqrt{3x^2+1}[/latex]. In this section, we study the rule for finding the derivative of the composition of two or more functions.

State the chain rule for the composition of two functions. Apply the chain rule together with the power rule. Apply the chain rule and the product/quotient rules correctly in combination when both are necessary. Recognize the chain rule for a composition of three or more functions. Describe the proof of the chain rule.

When we have a function that is a composition of two or more functions, we could use all of the techniques we have already learned to differentiate it. However, using all of those techniques to break down a function into simpler parts that we are able to differentiate can get cumbersome. Instead, we use the chain rule, which states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.

To put this rule into context, let’s take a look at an example: [latex]h(x)= sin (x^3)[/latex]. We can think of the derivative of this function with respect to [latex]x[/latex] as the rate of change of [latex]sin (x^3)[/latex] relative to the change in [latex]x[/latex]. Consequently, we want to know how [latex]sin (x^3)[/latex] changes as [latex]x[/latex] changes. We can think of this event as a chain reaction: As [latex]x[/latex] changes, [latex]x^3[/latex] changes, which leads to a change in [latex]sin (x^3)[/latex]. This chain reaction gives us hints as to what is involved in computing the derivative of [latex]sin (x^3)[/latex]. First of all, a change in [latex]x[/latex] forcing a change in [latex]x^3[/latex] suggests that somehow the derivative of [latex]x^3[/latex] is involved. In addition, the change in [latex]x^3[/latex] forcing a change in [latex]sin (x^3)[/latex] suggests that the derivative of [latex]sin (u)[/latex] with respect to [latex]u[/latex], where [latex]u=x^3[/latex], is also part of the final derivative.

We can take a more formal look at the derivative of [latex]h(x)= sin (x^3)[/latex] by setting up the limit that would give us the derivative at a specific value [latex]a[/latex] in the domain of [latex]h(x)= sin (x^3)[/latex].

This expression does not seem particularly helpful; however, we can modify it by multiplying and dividing by the expression [latex]x^3-a^3[/latex] to obtain

From the definition of the derivative, we can see that the second factor is the derivative of [latex]x^3[/latex] at [latex]x=a[/latex]. That is,

However, it might be a little more challenging to recognize that the first term is also a derivative. We can see this by letting [latex]u=x^3[/latex] and observing that as [latex]xto a, , uto a^3[/latex]:

Thus, [latex]h^{prime}(a)= cos (a^3) cdot 3a^2[/latex].

In other words, if [latex]h(x)= sin (x^3)[/latex], then [latex]h^{prime}(x)= cos (x^3) cdot 3x^2[/latex]. Thus, if we think of [latex]h(x)= sin (x^3)[/latex] as the composition [latex](fcirc g)(x)=f(g(x))[/latex] where [latex]f(x)=sin x[/latex] and [latex]g(x)=x^3[/latex], then the derivative of [latex]h(x)= sin (x^3)[/latex] is the product of the derivative of [latex]g(x)=x^3[/latex] and the derivative of the function [latex]f(x)= sin x[/latex] evaluated at the function [latex]g(x)=x^3[/latex]. At this point, we anticipate that for [latex]h(x)= sin (g(x))[/latex], it is quite likely that [latex]h^{prime}(x)= cos (g(x))g^{prime}(x)[/latex]. As we determined above, this is the case for [latex]h(x)= sin (x^3)[/latex].

Now that we have derived a special case of the chain rule, we state the general case and then apply it in a general form to other composite functions. An informal proof is provided at the end of the section.

Rule: The Chain Rule

Let [latex]f[/latex] and [latex]g[/latex] be functions. For all [latex]x[/latex] in the domain of [latex]g[/latex] for which [latex]g[/latex] is differentiable at [latex]x[/latex] and [latex]f[/latex] is differentiable at [latex]g(x)[/latex], the derivative of the composite function

is given by

Alternatively, if [latex]y[/latex] is a function of [latex]u[/latex], and [latex]u[/latex] is a function of [latex]x[/latex], then

Problem-Solving Strategy: Applying the Chain Rule

1. To differentiate [latex]h(x)=f(g(x))[/latex], begin by identifying [latex]f(x)[/latex] and [latex]g(x)[/latex].
2. Find [latex]f^{prime}(x)[/latex] and evaluate it at [latex]g(x)[/latex] to obtain [latex]f^{prime}(g(x))[/latex].
3. Find [latex]g^{prime}(x)[/latex].
4. Write [latex]h^{prime}(x)=f^{prime}(g(x)) cdot g^{prime}(x)[/latex].

Note: When applying the chain rule to the composition of two or more functions, keep in mind that we work our way from the outside function in. It is also useful to remember that the derivative of the composition of two functions can be thought of as having two parts; the derivative of the composition of three functions has three parts; and so on. Also, remember that we never evaluate a derivative at a derivative.

We can now apply the chain rule to composite functions, but note that we often need to use it with other rules. For example, to find derivatives of functions of the form [latex]h(x)=(g(x))^n[/latex], we need to use the chain rule combined with the power rule. To do so, we can think of [latex]h(x)=(g(x))^n[/latex] as [latex]f(g(x))[/latex] where [latex]f(x)=x^n[/latex]. Then [latex]f^{prime}(x)=nx^{n-1}[/latex]. Thus, [latex]f^{prime}(g(x))=n(g(x))^{n-1}[/latex]. This leads us to the derivative of a power function using the chain rule,

Rule: Power Rule for Composition of Functions

For all values of [latex]x[/latex] for which the derivative is defined, if

Then

Using the Chain and Power Rules

Find the derivative of [latex]h(x)=frac{1}{(3x^2+1)^2}[/latex].

First, rewrite [latex]h(x)=frac{1}{(3x^2+1)^2}=(3x^2+1)^{-2}[/latex].

Applying the power rule with [latex]g(x)=3x^2+1[/latex], we have

Rewriting back to the original form gives us

Find the derivative of [latex]h(x)=(2x^3+2x-1)^4[/latex].

[latex]h^{prime}(x)=4(2x^3+2x-1)^3(6x+2)=8(3x+1)(2x^3+2x-1)^3[/latex]

Hint

Use (Figure) with [latex]g(x)=2x^3+2x-1[/latex]

Using the Chain and Power Rules with a Trigonometric Function

Find the derivative of [latex]h(x)=sin^3 x[/latex].

First recall that [latex]sin^3 x=(sin x)^3[/latex], so we can rewrite [latex]h(x)= sin^3 x[/latex] as [latex]h(x)=(sin x)^3[/latex].

Applying the power rule with [latex]g(x)= sin x[/latex], we obtain

Finding the Equation of a Tangent Line

Find the equation of a line tangent to the graph of [latex]h(x)=frac{1}{(3x-5)^2}[/latex] at [latex]x=2[/latex].

Because we are finding an equation of a line, we need a point. The [latex]x[/latex]-coordinate of the point is 2. To find the [latex]y[/latex]-coordinate, substitute 2 into [latex]h(x)[/latex]. Since [latex]h(2)=frac{1}{(3(2)-5)^2}=1[/latex], the point is [latex](2,1)[/latex].

For the slope, we need [latex]h^{prime}(2)[/latex]. To find [latex]h^{prime}(x)[/latex], first we rewrite [latex]h(x)=(3x-5)^{-2}[/latex] and apply the power rule to obtain

By substituting, we have [latex]h^{prime}(2)=-6(3(2)-5)^{-3}=-6[/latex]. Therefore, the line has equation [latex]y-1=-6(x-2)[/latex]. Rewriting, the equation of the line is [latex]y=-6x+13[/latex].

Find the equation of the line tangent to the graph of [latex]f(x)=(x^2-2)^3[/latex] at [latex]x=-2[/latex].

Hint

Use the preceding example as a guide.

Now that we can combine the chain rule and the power rule, we examine how to combine the chain rule with the other rules we have learned. In particular, we can use it with the formulas for the derivatives of trigonometric functions or with the product rule.

Using the Chain Rule on a General Cosine Function

Find the derivative of [latex]h(x)= cos (g(x))[/latex].

Thus, the derivative of [latex]h(x)= cos (g(x))[/latex] is given by [latex]h^{prime}(x)=−sin (g(x))g^{prime}(x)[/latex].

In the following example we apply the rule that we have just derived.

Using the Chain Rule on a Cosine Function

Find the derivative of [latex]h(x)= cos (5x^2)[/latex].

Using the Chain Rule on Another Trigonometric Function

Find the derivative of [latex]h(x)= sec (4x^5+2x)[/latex].

Apply the chain rule to [latex]h(x)= sec (g(x))[/latex] to obtain

In this problem, [latex]g(x)=4x^5+2x[/latex], so we have [latex]g^{prime}(x)=20x^4+2[/latex]. Therefore, we obtain

Find the derivative of [latex]h(x)= sin (7x+2)[/latex].

Hint

Apply the chain rule to [latex]h(x)= sin g(x)[/latex] first and then use [latex]g(x)=7x+2[/latex].

At this point we provide a list of derivative formulas that may be obtained by applying the chain rule in conjunction with the formulas for derivatives of trigonometric functions. Their derivations are similar to those used in (Figure) and (Figure). For convenience, formulas are also given in Leibniz’s notation, which some students find easier to remember. (We discuss the chain rule using Leibniz’s notation at the end of this section.) It is not absolutely necessary to memorize these as separate formulas as they are all applications of the chain rule to previously learned formulas.

Using the Chain Rule with Trigonometric Functions

For all values of [latex]x[/latex] for which the derivative is defined,

Combining the Chain Rule with the Product Rule

Find the derivative of [latex]h(x)=(2x+1)^5(3x-2)^7[/latex].

First apply the product rule, then apply the chain rule to each term of the product.

Find the derivative of [latex]h(x)=frac{x}{(2x+3)^3}[/latex].

[latex]h^{prime}(x)=frac{3-4x}{(2x+3)^4}[/latex]

Hint

Start out by applying the quotient rule. Remember to use the chain rule to differentiate the denominator.

We can now combine the chain rule with other rules for differentiating functions, but when we are differentiating the composition of three or more functions, we need to apply the chain rule more than once. If we look at this situation in general terms, we can generate a formula, but we do not need to remember it, as we can simply apply the chain rule multiple times.

In general terms, first we let

Then, applying the chain rule once we obtain

Applying the chain rule again, we obtain

Rule: Chain Rule for a Composition of Three Functions

For all values of [latex]x[/latex] for which the function is differentiable, if

then

In other words, we are applying the chain rule twice.

Notice that the derivative of the composition of three functions has three parts. (Similarly, the derivative of the composition of four functions has four parts, and so on.) Also, remember, we always work from the outside in, taking one derivative at a time.

Differentiating a Composite of Three Functions

Find the derivative of [latex]k(x)=cos^4 (7x^2+1)[/latex].

First, rewrite [latex]k(x)[/latex] as

Then apply the chain rule several times.

Find the derivative of [latex]h(x)=sin^6 (x^3)[/latex].

[latex]h^{prime}(x)=18x^2 sin^5 (x^3) cos (x^3)[/latex]

Hint

Rewrite [latex]h(x)=sin^6 (x^3)=(sin(x^3))^6[/latex] and use (Figure) as a guide.

Using the Chain Rule in a Velocity Problem

A particle moves along a coordinate axis. Its position at time [latex]t[/latex] is given by [latex]s(t)= sin (2t)+ cos (3t)[/latex]. What is the velocity of the particle at time [latex]t=frac{pi}{6}[/latex]?

To find [latex]v(t)[/latex], the velocity of the particle at time [latex]t[/latex], we must differentiate [latex]s(t)[/latex]. Thus,

Substituting [latex]t=frac{pi}{6}[/latex] into [latex]v(t)[/latex], we obtain [latex]v(frac{pi}{6})=-2[/latex].

A particle moves along a coordinate axis. Its position at time [latex]t[/latex] is given by [latex]s(t)= sin (4t)[/latex]. Find its acceleration at time [latex]t[/latex].

Hint

Acceleration is the second derivative of position.

Proof of the Chain Rule

At this point, we present a very informal proof of the chain rule. For simplicity’s sake we ignore certain issues: For example, we assume that [latex]g(x)ne g(a)[/latex] for [latex]xne a[/latex] in some open interval containing [latex]a[/latex]. We begin by applying the limit definition of the derivative to the function [latex]h(x)[/latex] to obtain [latex]h^{prime}(a)[/latex]:

Rewriting, we obtain

Although it is clear that

it is not obvious that

To see that this is true, first recall that since [latex]g[/latex] is differentiable at [latex]a, , g[/latex] is also continuous at [latex]a[/latex]. Thus,

Next, make the substitution [latex]y=g(x)[/latex] and [latex]b=g(a)[/latex] and use change of variables in the limit to obtain

Finally,

Using the Chain Rule with Functional Values

Let [latex]h(x)=f(g(x))[/latex]. If [latex]g(1)=4, , g^{prime}(1)=3[/latex], and [latex]f^{prime}(4)=7[/latex], find [latex]h^{prime}(1)[/latex].

Use the chain rule, then substitute.

Let [latex]h(x)=f(g(x))[/latex]. If [latex]g(2)=-3, , g^{prime}(2)=4[/latex], and [latex]f^{prime}(-3)=7[/latex], find [latex]h^{prime}(2)[/latex].

Hint

Follow (Figure).

As with other derivatives that we have seen, we can express the chain rule using Leibniz’s notation. This notation for the chain rule is used heavily in physics applications.

For [latex]h(x)=f(g(x))[/latex], let [latex]u=g(x)[/latex] and [latex]y=h(x)=g(u)[/latex]. Thus,

Consequently,

Rule: Chain Rule Using Leibniz’s Notation

If [latex]y[/latex] is a function of [latex]u[/latex], and [latex]u[/latex] is a function of [latex]x[/latex], then

Taking a Derivative Using Leibniz’s Notation, Example 1

Find the derivative of [latex]y=big(frac{x}{3x+2}big)^5[/latex].

First, let [latex]u=frac{x}{3x+2}[/latex]. Thus, [latex]y=u^5[/latex]. Next, find [latex]frac{du}{dx}[/latex] and [latex]frac{dy}{du}[/latex]. Using the quotient rule,

and

Finally, we put it all together.

It is important to remember that, when using the Leibniz form of the chain rule, the final answer must be expressed entirely in terms of the original variable given in the problem.

Taking a Derivative Using Leibniz’s Notation, Example 2

Find the derivative of [latex]y= tan (4x^2-3x+1)[/latex].

First, let [latex]u=4x^2-3x+1[/latex]. Then [latex]y= tan u[/latex]. Next, find [latex]frac{du}{dx}[/latex] and [latex]frac{dy}{du}[/latex]:

Finally, we put it all together.

Use Leibniz’s notation to find the derivative of [latex]y= cos (x^3)[/latex]. Make sure that the final answer is expressed entirely in terms of the variable [latex]x[/latex].

Hint

Let [latex]u=x^3[/latex].

Key Concepts

• The chain rule allows us to differentiate compositions of two or more functions. It states that for [latex]h(x)=f(g(x))[/latex],
  [latex]h^{prime}(x)=f^{prime}(g(x))g^{prime}(x)[/latex].

  In Leibniz’s notation this rule takes the form

  [latex]frac{dy}{dx}=frac{dy}{du} cdot frac{du}{dx}[/latex].
• We can use the chain rule with other rules that we have learned, and we can derive formulas for some of them.
• The chain rule combines with the power rule to form a new rule:
  If [latex]h(x)=(g(x))^n[/latex], then [latex]h^{prime}(x)=n(g(x))^{n-1}g^{prime}(x)[/latex].
• When applied to the composition of three functions, the chain rule can be expressed as follows: If [latex]h(x)=f(g(k(x)))[/latex], then [latex]h^{prime}(x)=f^{prime}(g(k(x)))g^{prime}(k(x))k^{prime}(x)[/latex].

• The chain rule
  [latex]frac{d}{dx}(f(g(x)))=f^{prime}(g(x))g^{prime}(x)[/latex]
• The power rule for functions
  [latex]frac{d}{dx}((g(x)^n)=n(g(x))^{n-1}g^{prime}(x)[/latex]

For the following exercises, given [latex]y=f(u)[/latex] and [latex]u=g(x)[/latex], find [latex]frac{dy}{dx}[/latex] by using Leibniz’s notation for the chain rule: [latex]frac{dy}{dx}=frac{dy}{du}frac{du}{dx}[/latex].

[latex]frac{dy}{dx} = 18u^2 cdot 7=18(7x-4)^2 cdot 7[/latex]

[latex]frac{dy}{dx} = −sin u cdot frac{-1}{8}=−sin (frac{−x}{8}) cdot frac{-1}{8}[/latex]

[latex]frac{dy}{dx} = frac{8x-24}{2sqrt{4u+3}}=frac{4x-12}{sqrt{4x^2-24x+3}}[/latex]

For each of the following exercises,

1. decompose each function in the form [latex]y=f(u)[/latex] and [latex]u=g(x)[/latex], and
2. find [latex]frac{dy}{dx}[/latex] as a function of [latex]x[/latex].

a. [latex]f(u) = u^3, , u=3x^2+1[/latex];
b. [latex]frac{dy}{dx} = 18x(3x^2+1)^2[/latex]

a. [latex]f(u)=u^7, , u=frac{x}{7}+frac{7}{x}[/latex];
b. [latex]frac{dy}{dx} = 7(frac{x}{7}+frac{7}{x})^6 cdot (frac{1}{7}-frac{7}{x^2})[/latex]

a. [latex]f(u)= csc u, , u=pi x+1[/latex];
b. [latex]frac{dy}{dx} = −pi csc (pi x+1) cdot cot (pi x+1)[/latex]

a. [latex]f(u)=-6u^{-3}, , u= sin x[/latex];
b. [latex]frac{dy}{dx} = 18 sin^{-4} x cdot cos x[/latex]

For the following exercises, find [latex]frac{dy}{dx}[/latex] for each function.

3.3 Chain Ruleap Calculus 2nd Edition

[latex]frac{dy}{dx} = 6(2x^3-x^2+6x+1)^2(3x^2-x+3)[/latex]

[latex]frac{dy}{dx} = -3(tan x+ sin x)^{-4} cdot (sec^2 x+ cos x)[/latex]

[latex]frac{dy}{dx} = -7 cos (cos 7x) cdot sin 7x[/latex]

[latex]frac{dy}{dx} = -12 cot^2 (4x+1) cdot csc^2 (4x+1)[/latex]

25. Let [latex]y=(f(x))^3[/latex] and suppose that [latex]f^{prime}(1)=4[/latex] and [latex]frac{dy}{dx}=10[/latex] for [latex]x=1[/latex]. Find [latex]f(1)[/latex].

26. Let [latex]y=(f(x)+5x^2)^4[/latex] and suppose that [latex]f(-1)=-4[/latex] and [latex]frac{dy}{dx}=3[/latex] when [latex]x=-1[/latex]. Find [latex]f^{prime}(-1)[/latex]

27. Let [latex]y=(f(u)+3x)^2[/latex] and [latex]u=x^3-2x[/latex]. If [latex]f(4)=6[/latex] and [latex]frac{dy}{dx}=18[/latex] when [latex]x=2[/latex], find [latex]f^{prime}(4)[/latex].

28. [T] Find the equation of the tangent line to [latex]y=−sin (frac{x}{2})[/latex] at the origin. Use a calculator to graph the function and the tangent line together.

29. [T] Find the equation of the tangent line to [latex]y=(3x+frac{1}{x})^2[/latex] at the point [latex](1,16)[/latex]. Use a calculator to graph the function and the tangent line together.

30. Find the [latex]x[/latex]-coordinates at which the tangent line to [latex]y=(x-frac{6}{x})^8[/latex] is horizontal.

31. [T] Find an equation of the line that is normal to [latex]g(theta)= sin^2 (pi theta)[/latex] at the point [latex](frac{1}{4},frac{1}{2})[/latex]. Use a calculator to graph the function and the normal line together.

For the following exercises, use the information in the following table to find [latex]h^{prime}(a)[/latex] at the given value for [latex]a[/latex].

35. [latex]h(x)=(frac{f(x)}{g(x)})^2; , a=3[/latex]

40. [T] The position function of a freight train is given by [latex]s(t)=100(t+1)^{-2}[/latex], with [latex]s[/latex] in meters and [latex]t[/latex] in seconds. At time [latex]t=6[/latex] s, find the train’s

1. velocity and
2. acceleration.
3. Using a. and b. is the train speeding up or slowing down?

a. [latex]-frac{200}{343}[/latex] m/s;
b. [latex]frac{600}{2401} , text{m/s}^2[/latex];
c. The train is slowing down since velocity and acceleration have opposite signs.

41. [T] A mass hanging from a vertical spring is in simple harmonic motion as given by the following position function, where [latex]t[/latex] is measured in seconds and [latex]s[/latex] is in inches:

[latex]s(t)=-3 cos (pi t+frac{pi}{4})[/latex].

1. Determine the position of the spring at [latex]t=1.5[/latex] s.
2. Find the velocity of the spring at [latex]t=1.5[/latex] s.

42. [T] The total cost to produce [latex]x[/latex] boxes of Thin Mint Girl Scout cookies is [latex]C[/latex] dollars, where [latex]C=0.0001x^3-0.02x^2+3x+300[/latex]. In [latex]t[/latex] weeks production is estimated to be [latex]x=1600+100t[/latex] boxes.

1. Find the marginal cost [latex]C^{prime}(x)[/latex].
2. Use Leibniz’s notation for the chain rule, [latex]frac{dC}{dt}=frac{dC}{dx} cdot frac{dx}{dt}[/latex], to find the rate with respect to time [latex]t[/latex] that the cost is changing.
3. Use b. to determine how fast costs are increasing when [latex]t=2[/latex] weeks. Include units with the answer.

a. [latex]C^{prime}(x)=0.0003x^2-0.04x+3[/latex]
b. [latex]frac{dC}{dt}=100 cdot (0.0003x^2-0.04x+3)[/latex]
c. Approximately $90,300 per week

43. [T] The formula for the area of a circle is [latex]A=pi r^2[/latex], where [latex]r[/latex] is the radius of the circle. Suppose a circle is expanding, meaning that both the area [latex]A[/latex] and the radius [latex]r[/latex] (in inches) are expanding.

1. Suppose [latex]r=2-frac{100}{(t+7)^2}[/latex] where [latex]t[/latex] is time in seconds. Use the chain rule [latex]frac{dA}{dt}=frac{dA}{dr} cdot frac{dr}{dt}[/latex] to find the rate at which the area is expanding.
2. Use a. to find the rate at which the area is expanding at [latex]t=4[/latex] s.

44. [T] The formula for the volume of a sphere is [latex]S=frac{4}{3}pi r^3[/latex], where [latex]r[/latex] (in feet) is the radius of the sphere. Suppose a spherical snowball is melting in the sun.

1. Suppose [latex]r=frac{1}{(t+1)^2}-frac{1}{12}[/latex] where [latex]t[/latex] is time in minutes. Use the chain rule [latex]frac{dS}{dt}=frac{dS}{dr} cdot frac{dr}{dt}[/latex] to find the rate at which the snowball is melting.
2. Use a. to find the rate at which the volume is changing at [latex]t=1[/latex] min.

a. [latex]frac{dS}{dt}=-frac{8pi r^2}{(t+1)^3}[/latex]
b. The volume is decreasing at a rate of [latex]-frac{pi}{36} , text{ft}^3/text{min}[/latex].

45. [T] The daily temperature in degrees Fahrenheit of Phoenix in the summer can be modeled by the function [latex]T(x)=94-10 cos [frac{pi}{12}(x-2)][/latex], where [latex]x[/latex] is hours after midnight. Find the rate at which the temperature is changing at 4 p.m.

46. [T] The depth (in feet) of water at a dock changes with the rise and fall of tides. The depth is modeled by the function [latex]D(t)=5 sin (frac{pi}{6} t-frac{7pi}{6})+8[/latex], where [latex]t[/latex] is the number of hours after midnight. Find the rate at which the depth is changing at 6 a.m.

Glossary

chain rule

the chain rule defines the derivative of a composite function as the derivative of the outer function evaluated at the inner function times the derivative of the inner function

Consider the product of two simple functions, say$ds f(x)=(x^2+1)(x^3-3x)$. An obvious guess for the derivative of $f$ isthe product of the derivatives of the constituent functions:$ds (2x)(3x^2-3)=6x^3-6x$. Is this correct? We can easily check, byrewriting $f$ and doing the calculation in a way that is known towork. First, $ds f(x)=x^5-3x^3+x^3-3x=x^5-2x^3-3x$, and then$ds f'(x)=5x^4-6x^2-3$. Not even close! What went 'wrong'? Well,nothing really, except the guess was wrong.

So the derivative of $f(x)g(x)$ is NOT as simple as$f'(x)g'(x)$. Surely there is some rule for such a situation? Thereis, and it is instructive to 'discover' it by trying to do thegeneral calculation even without knowing the answer in advance.$$eqalign{{dover dx}(&f(x)g(x)) = lim_{Delta x to0} {f(x+Delta x)g(x+Delta x) - f(x)g(x)over Delta x}cr&=lim_{Delta x to0} {f(x+Delta x)g(x+Delta x)-f(x+Delta x)g(x) + f(x+Delta x)g(x)- f(x)g(x)over Delta x}cr &=lim_{Delta x to0} {f(x+Delta x)g(x+Delta x)-f(x+Delta x)g(x)over Delta x} + lim_{Delta x to0} {f(x+Delta x)g(x)- f(x)g(x)over Delta x}cr &=lim_{Delta x to0} f(x+Delta x){ g(x+Delta x)-g(x)over Delta x} + lim_{Delta x to0} {f(x+Delta x)- f(x)over Delta x}g(x)cr &=f(x)g'(x) + f'(x)g(x)cr}$$A couple of items here need discussion. First, we used a standardtrick, 'add and subtract the same thing', to transform what we hadinto a more useful form. After some rewriting, we realize that we havetwo limits that produce $f'(x)$ and $g'(x)$. Of course, $f'(x)$ and$g'(x)$ must actually exist for this to make sense.We also replaced$ds lim_{Delta xto0}f(x+Delta x)$ with $f(x)$—why is this justified?

What we really need to know here is that $ds lim_{Delta xto 0}f(x+Delta x)=f(x)$, or in the language ofsection 2.5, that $f$ is continuousat $x$. We already know that $f'(x)$ exists (or the whole approach,writing the derivative of $fg$ in terms of $f'$ and $g'$, doesn't makesense). This turns out to imply that $f$ is continuous as well. Here'swhy:$$eqalign{lim_{Delta xto 0} f(x+Delta x) &= lim_{Delta xto 0} (f(x+Deltax) -f(x) + f(x))cr&= lim_{Delta xto 0} {f(x+Delta x) -f(x)over Delta x}Delta x +lim_{Delta xto 0} f(x)cr&=f'(x)cdot 0 + f(x) = f(x)cr}$$

To summarize: the product rule says that$${dover dx}(f(x)g(x)) = f(x)g'(x) + f'(x)g(x).$$

Returning to the example we started with, let $ds f(x)=(x^2+1)(x^3-3x)$.Then $ds f'(x)=(x^2+1)(3x^2-3)+(2x)(x^3-3x)=3x^4-3x^2+3x^2-3+2x^4-6x^2=5x^4-6x^2-3$, as before. In this case it is probably simpler tomultiply $f(x)$ out first, then compute the derivative; here's anexample for which we really need the product rule.

Example 3.3.1 Compute the derivative of $ds f(x)=x^2sqrt{625-x^2}$. We havealready computed $ds {dover dx}sqrt{625-x^2}={-xoversqrt{625-x^2}}$. Now$$f'(x)=x^2{-xoversqrt{625-x^2}}+2xsqrt{625-x^2}={-x^3+2x(625-x^2)over sqrt{625-x^2}}={-3x^3+1250xover sqrt{625-x^2}}.$$

Exercises 3.3

In 1–4, find the derivatives of the functions using the product rule.

3.3 Chain Ruleap Calculus 14th Edition

Ex 3.3.1$ds x^3(x^3-5x+10)$(answer)

Ex 3.3.2$ds (x^2+5x-3)(x^5-6x^3+3x^2-7x+1)$(answer)

Ex 3.3.3$ds sqrt{x}sqrt{625-x^2}$(answer)

Ex 3.3.4$displaystyle {sqrt{625-x^2}over x^{20}}$(answer)

Ex 3.3.5Use the product rule to compute the derivative of $ds f(x)=(2x-3)^2$. Sketch the function. Find an equation of the tangent line to the curve at $x=2$. Sketch the tangent line at $x=2$.(answer)

Ex 3.3.6Suppose that $f$, $g$, and $h$ are differentiable functions.Show that $(fgh)'(x) = f'(x) g(x)h(x) + f(x)g'(x) h(x) + f(x) g(x)h'(x)$.

Ex 3.3.7State and prove a rule to compute $(fghi)'(x)$, similar to the rule in the previous problem.

Remark 3.3.2 {Product notation}Suppose $ds f_1 , f_2 , ldots f_n$ are functions.The product of all these functions can be written$$ prod _{k=1 } ^n f_k.$$This is similar to the use of $ds sum$ to denote a sum.For example,$$prod _{k=1 } ^5 f_k =f_1 f_2 f_3 f_4 f_5$$and$$prod _ {k=1 } ^n k = 1cdot 2 cdot ldots cdot n = n!.$$We sometimes use somewhat more complicated conditions; for example$$prod _{k=1 , kneq j } ^n f_k$$denotes the product of $ds f_1$ through $ds f_n$ except for $ds f_j$.For example, $$prod _{k=1 , kneq 4} ^5 x^k = xcdot x^2 cdot x^3 cdot x^5 =x^{11}.$$

3.3 Chain Ruleap Calculus Calculator

Ex 3.3.8The generalized product rule says that if $ds f_1 , f_2 ,ldots ,f_n$ are differentiable functions at $x$ then$${dover dx}prod _{k=1 } ^n f_k(x) = sum _{j=1 } ^n left(f'_j (x) prod _{k=1 , kneq j} ^n f_k (x)right).$$Verify that this is the same as your answer to the previous problemwhen $n=4$,and write out what this says when $n=5$.