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2.5. THE CHAIN RULE
understanding its algebraic structure and the basic functions that constitute it. It takes
substantial practice to get comfortable with navigating multiple rules in a single problem;
using proper notation and taking a few extra steps can be particularly helpful as well. We
demonstrate with an example and then provide further opportunity for practice in the
following activity.
Example 2.3. Find a formula for the derivative of h(t) = 3 t 2 +2t sec 4 (t).
Solution. We first observe that the most basic structure of h is that it is the product of two
functions: h(t) = a(t) · b(t) where a(t) = 3 t 2 +2t and b(t) = sec 4 (t). Therefore, we see that
we will need to use the product rule to differentiate h. When it comes time to differentiate
a and b in their roles in the product rule, we observe that since each is a composite
function, the chain rule will be needed. We therefore begin by working separately to
compute 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 g.
Doing so,
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, we now know that 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 overall 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 d
dt
[sec
4 (t)] + sec
4 (t)
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).
2.5. THE CHAIN RULE
understanding its algebraic structure and the basic functions that constitute it. It takes
substantial practice to get comfortable with navigating multiple rules in a single problem;
using proper notation and taking a few extra steps can be particularly helpful as well. We
demonstrate with an example and then provide further opportunity for practice in the
following activity.
Example 2.3. Find a formula for the derivative of h(t) = 3 t 2 +2t sec 4 (t).
Solution. We first observe that the most basic structure of h is that it is the product of two
functions: h(t) = a(t) · b(t) where a(t) = 3 t 2 +2t and b(t) = sec 4 (t). Therefore, we see that
we will need to use the product rule to differentiate h. When it comes time to differentiate
a and b in their roles in the product rule, we observe that since each is a composite
function, the chain rule will be needed. We therefore begin by working separately to
compute 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 g.
Doing so,
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, we now know that 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 overall 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 d
dt
[sec
4 (t)] + sec
4 (t)
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).