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2.5. THE CHAIN RULE
Applying the product rule and simplifying,
C
′ (x) = 2 sin(x)(− sin(x)) + cos(x)(2 cos(x)) = 2(cos
2 (x) − sin
2 (x)).
Next, we recall that one of the double angle identities for the cosine function tells us that
cos(2x) = cos
2 (x) − sin
2 (x).
Substituting this result in our expression for C ′ (x), we now have that
C
′ (x) = 2 cos(2x).
So from Example 2.2, we see that if C(x) = sin(2x), then C ′ (x) = 2 cos(2x). Letting
g(x) = 2x and f (x) = sin(x), we observe that C(x) = f (g(x)). Moreover, with g ′ (x) = 2
and f ′ (x) = cos(x), it follows that we can view the structure of C ′ (x) as
C
′ (x) = 2 cos(2x) = g
′ (x) f
′ (g(x)).
In this example, we see that for the composite function C(x) = f (g(x)), the derivative C ′
is (as in the example involving linear functions) constituted by multiplying the derivatives
of f and g, but with the special condition that f ′ is evaluated at g(x), rather than at x.
It makes sense intuitively that these two quantities are involved in understanding the
rate of change of a composite function: if we are considering C(x) = f (g(x)) and asking
how fast C is changing at a given x value as x changes, 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 not only for the functions in Examples 2.1 and 2.2, but indeed for all
differentiable functions 7 as is stated in the Chain Rule.
Chain Rule: If g is differentiable at x and f is differentiable at g(x), then the
composite function C defined by C(x) = f (g(x)) is differentiable at x and
C
′ (x) = f
′ (g(x))g
′ (x).
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.”
At least initially in working particular examples requiring the chain rule, it can also
be helpful to clearly identify the inner function g and outer function f , compute their
7 Like other differentiation rules, the Chain Rule can be proved formally using the limit definition of the
derivative.
2.5. THE CHAIN RULE
Applying the product rule and simplifying,
C
′ (x) = 2 sin(x)(− sin(x)) + cos(x)(2 cos(x)) = 2(cos
2 (x) − sin
2 (x)).
Next, we recall that one of the double angle identities for the cosine function tells us that
cos(2x) = cos
2 (x) − sin
2 (x).
Substituting this result in our expression for C ′ (x), we now have that
C
′ (x) = 2 cos(2x).
So from Example 2.2, we see that if C(x) = sin(2x), then C ′ (x) = 2 cos(2x). Letting
g(x) = 2x and f (x) = sin(x), we observe that C(x) = f (g(x)). Moreover, with g ′ (x) = 2
and f ′ (x) = cos(x), it follows that we can view the structure of C ′ (x) as
C
′ (x) = 2 cos(2x) = g
′ (x) f
′ (g(x)).
In this example, we see that for the composite function C(x) = f (g(x)), the derivative C ′
is (as in the example involving linear functions) constituted by multiplying the derivatives
of f and g, but with the special condition that f ′ is evaluated at g(x), rather than at x.
It makes sense intuitively that these two quantities are involved in understanding the
rate of change of a composite function: if we are considering C(x) = f (g(x)) and asking
how fast C is changing at a given x value as x changes, 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 not only for the functions in Examples 2.1 and 2.2, but indeed for all
differentiable functions 7 as is stated in the Chain Rule.
Chain Rule: If g is differentiable at x and f is differentiable at g(x), then the
composite function C defined by C(x) = f (g(x)) is differentiable at x and
C
′ (x) = f
′ (g(x))g
′ (x).
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.”
At least initially in working particular examples requiring the chain rule, it can also
be helpful to clearly identify the inner function g and outer function f , compute their
7 Like other differentiation rules, the Chain Rule can be proved formally using the limit definition of the
derivative.