Introduction
-
- In addition to learning how to differentiate a variety of basic functions,
- we have also been developing our ability to use rules to differentiate certain algebraic combinations of them.
-
-
-
-
-
- State the rule(s) required to find the derivative of each of the following combinations of
- f(x) = \sin(x) and g(x) = x^2:
-
- s(x) = 3x^2 - 5\sin(x)
- ,
-
- p(x) = x^2 \sin(x), \text{and}
-
-
- q(x) = \frac{\sin(x)}{x^2}
- .
-
-
-
-
- Finding s' uses the sum and constant multiple rules,
- because s(x) = 3g(x) - 5f(x).
- Determining p' requires the product rule,
- because p(x) = g(x) \cdot f(x).
- To calculate q' we use the quotient rule,
- because q(x) =\frac{f(x)}{g(x)}.
-
-
-
-
-
- There is one more natural way to combine basic functions algebraically,
- and that is by composing them.
- For instance, let's consider the function
-
- C(x) = \sin(x^2)
- ,
- and observe that any input x passes through a
- chain of functions.
- In the process that defines the function C(x),
- x is first squared, and then the sine of the result is taken.
- Using an arrow diagram,
-
- x \longrightarrow x^2 \longrightarrow \sin(x^2)
- .
-
-
-
- In terms of the elementary functions f and g,
- we observe that x is the input for the function g,
- and the result is used as the input for f.
- We write
-
- C(x) = f(g(x)) = \sin(x^2)
-
- and say that C is the composition
- composition
- of f and g.
- We will refer to g,
- the function that is first applied to x,
- as the inner function, while f,
- the function that is applied to the result,
- is the outer function.
-
-
-
- Given a composite function
- C(x) = f(g(x)) that is built from differentiable functions f and g,
- how do we compute C'(x) in terms of f,
- g, f', and g'?
- In the same way that the rate of change of a product of two functions,
- p(x) = f(x) \cdot g(x),
- depends on the behavior of both f and g,
- it makes sense intuitively that the rate of change of a composite function
- C(x) = f(g(x)) will also depend on some combination of f and g and their derivatives.
- The rule that describes how to compute C' in terms of f and g and their derivatives is called the chain rule.
-
-
-
- But before we can learn what the chain rule says and why it works,
- we first need to be comfortable decomposing composite functions so that we can correctly identify the inner and outer functions,
- as we did in the example above with C(x) = \sin(x^2).
-
-
-
-
-
-
-
-
- The chain rule
-
- Often a composite function cannot be written in an alternate algebraic form.
- For instance, the function
- C(x) = \sin(x^2) cannot be expanded or otherwise rewritten,
- so it presents no alternate approaches to taking the derivative.
- But some composite functions can be expanded or simplified,
- and these provide a way to explore how the chain rule works.
-
-
-
-
-
- Let f(x) = -4x + 7 and g(x) = 3x - 5.
- Determine a formula for C(x) = f(g(x)) and compute C'(x).
- How is C' related to f and g and their derivatives?
-
-
-
-
- By the rules given for f and g,
-
- C(x) =\mathstrut \amp f(g(x))
- =\mathstrut \amp f(3x-5)
- =\mathstrut \amp -4(3x-5) + 7
- =\mathstrut \amp -12x + 20 + 7
- =\mathstrut \amp -12x + 27
- .
-
-
-
- Thus, C'(x) = -12.
- Noting that f'(x) = -4 and g'(x) = 3,
- we observe that C' appears to be the product of f' and g'.
-
-
-
-
-
- It may seem that
- Example
- is too elementary to illustrate how to differentiate a composite function.
- Linear functions are the simplest of all functions, and
- composing linear functions yields another linear function.
- While this example does not illustrate the full complexity of a composition of nonlinear functions,
- at the same time we remember that any differentiable function is
- locally linear,
- and thus any function with a derivative behaves like a line when viewed up close.
- The fact that the derivatives of the linear functions f and g are multiplied to find the derivative of their composition turns out to be a key insight.
-
-
-
- We now consider a composition involving a nonlinear function.
-
-
-
-
-
- Let C(x) = \sin(2x).
- Use the double angle identity to rewrite C as a product of basic functions,
- and use the product rule to find C'.
- Rewrite C' in the simplest form possible.
-
-
-
-
- Using the double angle identity for the sine function, we write
-
- C(x) = \sin(2x) = 2\sin(x)\cos(x)
- .
-
-
-
- Applying the product rule and simplifying, we find
-
- C'(x) = 2\sin(x)(-\sin(x)) + \cos(x)(2\cos(x)) = 2(\cos^2(x) - \sin^2(x))
- .
-
-
-
- Next, we recall that a double angle identity for the cosine tells us
-
- \cos(2x) = \cos^2(x) - \sin^2(x)
- .
-
-
-
- Substituting this result into our expression for C'(x),
- we now have that
-
- C'(x) = 2 \cos(2x)
- .
-
-
-
-
-
- In Example,
- if we let g(x) = 2x and f(x) = \sin(x),
- we observe that C(x) = f(g(x)).
- Now, g'(x) = 2 and f'(x) = \cos(x),
- so we can view the structure of C'(x) as
-
- C'(x) = 2\cos(2x) = g'(x) f'(g(x))
- .
-
+ Foundational Exercises
- In this example, as in the example involving linear functions,
- we see that the derivative of the composite function C(x) = f(g(x))
- is found by multiplying the derivatives of f and g,
- but with f' evaluated at g(x).
+ One of the most useful ways of constructing functions from other functions is composition
: doing one
+ function and then doing another function to the result. Rather than substituting a number in for the input
+ of a function, we substitute a whole second function in for the input of a function.
-
- It makes sense intuitively that these two quantities are involved in the rate of change of a composite function:
- if we ask how fast C is changing at a given x value,
- it clearly matters how fast g is changing at x,
- as well as how fast f is changing at the value of g(x).
- It turns out that this structure holds for all differentiable functions
- Like other differentiation rules,
- the Chain Rule can be proved formally using the limit definition of the derivative.
- as is stated in the Chain Rule.
+ We can use our knowledge of how to measure the rates of change of the constituent functions to calculate the
+ rate of change of the more complicated function. We treat the inside
or substituted
function
+ as if it were a single variable, take the derivative of the outer
function with respect to that variable,
+ and multiply by the derivative of the inside
function.
-
The Chain Rule
@@ -257,304 +63,124 @@
C'(x) = f'(g(x)) g'(x)
.
+
+ Another formulation of this rule which says the same thing using different notation is
+
+ \frac{dC}{dx}=\frac{df}{dg}\frac{dg}{dx}
+ .
+
-
- As with the product and quotient rules,
- it is often helpful to think verbally about what the chain rule says:
- If C is a composite function defined by an outer function f and an inner function g,
- then C' is given by the derivative of the outer function
- evaluated at the inner function,
- times the derivative of the inner function.
-
-
-
- It is helpful to identify clearly the inner function g and outer function f,
- compute their derivatives individually,
- and then put all of the pieces together by the chain rule.
-
-
-
-
-
- Determine the derivative of the function
-
- r(x) = (\tan(x))^2
- .
-
-
-
-
- The function r is composite,
- with inner function g(x) = \tan(x) and outer function f(x) = x^2.
- Organizing the key information involving f,
- g, and their derivatives, we have
-
-
-
-
- | f(x) = x^2 |
- |
- g(x) = \tan(x) |
-
-
- | f'(x) = 2x |
- |
- g'(x) = \sec^2(x) |
-
-
- | f'(g(x)) = 2\tan(x) |
- |
- |
-
-
-
-
- Applying the chain rule, we find that
-
- r'(x) = f'(g(x))g'(x) = 2\tan(x) \sec^2(x)
- .
-
-
-
-
-
- As a side note,
- we remark that r(x) is usually written as \tan^2(x).
- This is common notation for powers of trigonometric functions:
- \cos^4(x), \sin^5(x),
- and \sec^2(x) are all composite functions,
- with the outer function a power function and the inner function a trigonometric one.
-
-
-
-
-
-
- Using multiple rules simultaneously
-
- The chain rule now joins the sum,
- constant multiple, product,
- and quotient rules in our collection of techniques for finding the derivative of a function through understanding its algebraic structure and the basic functions that constitute it.
- It takes practice to get comfortable applying multiple rules to differentiate a single function, but
- using proper notation and taking a few extra steps will help.
-
-
-
-
-
- Find a formula for the derivative of h(t) = 3^{t^2 + 2t}\sec^4(t).
-
-
-
-
- We first observe that h is the product of two functions:
- h(t) = a(t) \cdot b(t), where
- a(t) = 3^{t^2 + 2t} and b(t) = \sec^4(t).
- We will need to use the product rule to differentiate h.
- And because a and b are composite functions,
- we will need
- the chain rule.
- We therefore begin by computing a'(t) and b'(t).
-
-
-
- Writing a(t) = f(g(t)) = 3^{t^2 + 2t},
- and finding the derivatives of f and g, we have
-
-
-
-
- | f(t) = 3^t |
- |
- g(t) = t^2 + 2t |
-
-
- | f'(t) = 3^t \ln(3) |
- |
- g'(t) = 2t+2 |
-
-
- | f'(g(t)) = 3^{t^2 + 2t}\ln(3) |
- |
- |
-
-
-
-
-
- Thus, by the chain rule,
- it follows that a'(t) = f'(g(t))g'(t) = 3^{t^2 + 2t}\ln(3) (2t+2).
-
-
-
- Turning next to b,
- we write b(t) = r(s(t)) = \sec^4(t) and find the derivatives of r and s.
-
-
-
-
- | r(t) = t^4 |
- |
- s(t) = \sec(t) |
-
-
- | r'(t) = 4t^3 |
- |
- s'(t) = \sec(t)\tan(t) |
-
-
- | r'(s(t)) = 4\sec^3(t) |
- |
- |
-
-
-
-
-
- By the chain rule,
-
- b'(t) = r'(s(t))s'(t) = 4\sec^3(t)\sec(t)\tan(t) = 4 \sec^4(t) \tan(t)
- .
-
-
-
- Now we are finally ready to compute the derivative of the function h.
- Recalling that h(t) = 3^{t^2 + 2t}\sec^4(t),
- by the product rule we have
-
- h'(t) = 3^{t^2 + 2t} \frac{d}{dt}[\sec^4(t)] + \sec^4(t) \frac{d}{dt}[3^{t^2 + 2t}]
- .
-
-
-
- From our work above with a and b,
- we know the derivatives of
- 3^{t^2 + 2t} and \sec^4(t), and therefore
-
- h'(t) = 3^{t^2 + 2t} 4\sec^4(t) \tan(t) + \sec^4(t) 3^{t^2 + 2t}\ln(3) (2t+2)
- .
-
-
-
-
-
-
-
- The chain rule now adds substantially to our ability to compute derivatives.
- Whether we are finding the equation of the tangent line to a curve,
- the instantaneous velocity of a moving particle,
- or the instantaneous rate of change of a certain quantity,
- if the function under consideration is a composition,
- the chain rule is often an essential tool.
-
-
-
+ Composition from a graph
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+
+ Composition from a graph
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+
+ Composition using algebra
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+ Composition using algebra
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+ Composition using algebra
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+ Composition using a table of values
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+ Composition using a table of values
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+ Composition using algebra
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+
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+ Identifying functions in a composition
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+
+ Identifying functions in a composition
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+
+ Composition using algebra
+
+
+ Composition using graphs
+
+
-
- The composite version of basic function rules
-
- As we gain more experience with differention,
- we will become more comfortable in simply writing down the derivative without taking multiple steps.
- This is particularly simple when the inner function is linear,
- since the derivative of a linear function is a constant.
-
-
-
-
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- For each of the following composite functions whose inside function is linear, find the overall function's derivative using the chain rule: f(x) = (5x+7)^{10}, g(x) = \tan(17x), and h(x) = e^{-3x}.
-
-
-
-
- For each of the three given functions, the derivative of the inner function is constant. By the chain rule, we see
-
- \frac{d}{dx} \left[ (5x+7)^{10} \right] = 10(5x+7)^9 \cdot 5
- ,
-
- \frac{d}{dx} \left[ \tan(17x) \right] = 17\sec^2(17x), \ \text{and}
-
-
- \frac{d}{dx} \left[ e^{-3x} \right] = -3e^{-3x}
- .
-
-
-
-
-
- More generally, we can think about how each basic function rule has a corresponding chain rule version. The next example demonstrates this for two familiar functions.
-
-
-
-
-
- Develop a chain rule version of the two basic derivative rules that state \frac{d}{dx}[\sin(x)] = \cos(x) and \frac{d}{dx}[a^x] = a^x \ln(a).
-
-
-
-
- To determine
-
- \frac{d}{dx}[\sin(u(x))]
- ,
- where u is a differentiable function of x,
- we use the chain rule with the sine function as the outer function.
- Applying the chain rule, we find that
-
- \frac{d}{dx}[\sin(u(x))] = \cos(u(x)) \cdot u'(x)
- .
- This rule is analogous to the basic derivative rule that \frac{d}{dx}[\sin(x)] = \cos(x).
-
-
-
- Similarly, since \frac{d}{dx}[a^x] = a^x \ln(a),
- it follows by the chain rule that
-
- \frac{d}{dx}[a^{u(x)}] = a^{u(x)} \ln(a) \cdot u'(x)
- .
- This rule is analogous to the basic derivative rule that \frac{d}{dx}[a^{x}] = a^{x} \ln(a).
-
-
-
-
-
- An excellent exercise for getting comfortable with the derivative rules is to complete Example for every basic function. That is,
- write down a list of all the basic functions whose derivatives you know, and
- list their corresponding derivatives.
- Then, corresponding to each basic rule, write a composite function with the inner function being an unknown function u(x) and the outer function being a basic function.
- Finally, write the chain rule for the composite function, such as \frac{d}{dx}[\sin(u(x))] = \cos(u(x)) \cdot u'(x).
-
-
-
-
-
- Summary
-
-
- -
-
- A composite function is one where the input variable x first passes through one function,
- and then the resulting output passes through another.
- For example, the function
- h(x) = 2^{\sin(x)} is composite since x \longrightarrow \sin(x) \longrightarrow 2^{\sin(x)}.
-
-
-
- -
-
- Given a composite function
- C(x) = f(g(x)) where f and g are differentiable functions,
- the chain rule tells us that
-
- C'(x) = f'(g(x)) g'(x)
- .
-
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+ Additional Practice Exercises
+
+ Chain rule
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+ Chain rule
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+ Chain rule
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+ Chain rule
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+ Chain rule
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+ Chain rule
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+ Chain rule
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+ Chain rule
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+ Using chain rule and information
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+ Using chain rule and information
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+ Chain rule from graph
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+ Chain rule from graph
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+ Chain rule from a table of values
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+ Chain rule from a table of values
+
+
+ Recall that the central difference is a way of approximating the value
+ of the derivative at a specified number by calculating the slope of the line through the two points
+ on either side of the specified number. That's what this problem is asking you to use along with the chain rule.
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+ Multiple rules
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+ Multiple rules with a table
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+ Multiple rules
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+ Multiple rules
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+ Multiple rules
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