2.5. THE CHAIN RULE
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derivative. But other composite functions can be expanded or simplified, and these present
a way to begin to explore how the chain rule might have to work. To that end, we consider
two examples of composite functions that present alternate means of finding the derivative.
Example 2.1. 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?
Solution. By the rules given for f and g,
C(x) = f (g(x))
= f (3x − 5)
= −4(3x − 5) + 7
= −12x + 20 + 7
= −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 ′ .
From one perspective, Example 2.1 may be too elementary. Linear functions are the
simplest of all functions, and perhaps composing linear functions (which yields another
linear function) does not exemplify the true complexity that is involved with differentiating
a composition of more complicated functions. At the same time, we should remember
the perspective that any differentiable function is locally linear, so any function with a
derivative behaves like a line when viewed up close. From this point of view, the fact that
the derivatives of f and g are multiplied to find the derivative of their composition turns
out to be a key insight.
We now consider a second example involving a nonlinear function to gain further
understanding of how differentiating a composite function involves the basic functions that
combine to form it.
Example 2.2. 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.
Solution. By the double angle identity for the sine function,
C(x) = sin(2x) = 2 sin(x) cos(x).
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derivative. But other composite functions can be expanded or simplified, and these present
a way to begin to explore how the chain rule might have to work. To that end, we consider
two examples of composite functions that present alternate means of finding the derivative.
Example 2.1. 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?
Solution. By the rules given for f and g,
C(x) = f (g(x))
= f (3x − 5)
= −4(3x − 5) + 7
= −12x + 20 + 7
= −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 ′ .
From one perspective, Example 2.1 may be too elementary. Linear functions are the
simplest of all functions, and perhaps composing linear functions (which yields another
linear function) does not exemplify the true complexity that is involved with differentiating
a composition of more complicated functions. At the same time, we should remember
the perspective that any differentiable function is locally linear, so any function with a
derivative behaves like a line when viewed up close. From this point of view, the fact that
the derivatives of f and g are multiplied to find the derivative of their composition turns
out to be a key insight.
We now consider a second example involving a nonlinear function to gain further
understanding of how differentiating a composite function involves the basic functions that
combine to form it.
Example 2.2. 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.
Solution. By the double angle identity for the sine function,
C(x) = sin(2x) = 2 sin(x) cos(x).