Download 1. Derivatives for trigonometric functions We have almost completed

1. Derivatives for trigonometric functions
We have almost completed our repertoire for functions that we can differentiate.
All that is left is trigonometric functions (and their inverses). We present the
derivatives from sine and cosine below.
Theorem 1.1 (Derivatives for sine and cosine). Suppose that f (x) = sin(x) and
g(x) = cos(x). Then f ′ (x) = cos(x), and g ′ (x) = − sin(x).
If you have not seen this before, you should be very surprised (and relieved)
that the derivatives for sine and cosine are so simple. But if we think about it for
a bit, we shouldn’t be too surprised. Both sine and cosine are periodic functions
(i.e. they repeat over equal length intervals). So it makes sense that the slopes
of the corresponding tangent lines are periodic. With closer inspection of sin(x)
and cos(x), we see that the slopes remain relatively shallow. In fact, they oscillate
between −1 and 1. We won’t prove these derivatives. The proof requires that we
sin(x)
find lim
, which requires a bit too much geometry for us.
x→0
x
Also something nice to notice is that the derivatives of sine and cosine also
repeat. [sin(x)]′′′′ = [cos(x)]′′′ = [− sin(x)]′′ = [− cos(x)]′ = sin(x). So the fourth
derivative of either sine or cosine is the same function back. We will now turn to
sin(x)
tangent (the function). Recall that tan(x) = cos(x)
. Before we define it, we need
one more rule.
Theorem 1.2 (Quotient Rule). Suppose a function f (x) =
functions. Then f ′ (x) =
h(x) ⋅ g ′ (x) − g(x) ⋅ h′ (x)
.
[h(x)]2
g(x)
is a quotient of
h(x)
Unlike the product rule, the quotient rule is not commutative. That is, it matters
which function we have as g(x) and which is h(x). Notice that tan(x) is a quotient
of functions, so it is the perfect first example for the quotient rule.
Theorem 1.3 (Derivative of tan(x)). Suppose f (x) = tan(x). Then f ′ (x) =
1
sec2 (x) =
.
2
cos (x)
Proof. We will prove this using the quotient rule.
d
d
sin(x) − sin(x) ⋅ dx
cos(x)
d
d sin(x) cos(x) ⋅ dx
tan(x) =
=
2
dx
dx cos(x)
[cos (x)]
1
2
cos(x) ⋅ cos(x) − sin(x) ⋅ − sin(x) cos2 (x) + sin2 (x)
=
[cos2 (x)]
cos2 (x)
1
=
= sec2 (x).
2
cos (x)
=
Again, we will not prove the quotient rule. It is very much like the proof of
the product rule. Let’s see some examples of our new derivatives and the quotient
rule.
2. Some examples
Example 1. Find the derivative of f (x) = sin(ln(x)).
We have a composition of functions, so we need the chain rule. The outside
function is g(x) = sin(x). The inside function is h(x) = ln(x). g ′ (x) = cos(x) and
1 cos(ln(x))
h′ (x) = x1 . So by the chain rule, f ′ (x) = cos(ln(x)) ⋅ =
.
x
x
Example 2. Find the derivative of f (x) = cot(x) =
cos(x)
.
sin(x)
This is a great example to show that the quotient rule is not commutative. By
the quotient rule, we get
d
d
sin(x) ⋅ dx
cos(x) − cos(x) ⋅ dx
sin(x)
f (x) =
sin2 (x)
−1
− sin2 (x) − cos2 (x)
=
=
2
sin (x)
sin2 (x)
′
= − csc2 (x).
So we get the important identity
d
dx
cot(x) = − csc2 (x)
(x).
Example 3. Find the derivative of sec(x) =
1
cos(x) .
We can use either the quotient rule or the chain rule. Let’s use the quotient rule
for practice. We get
f ′ (x) =
cos(x) ⋅ 0 − 1 ⋅ − sin(x) sin(x)
=
= tan(x) ⋅ sec(x)
cos2 (x)
cos2 (x)
3
Example 4. Find the derivative of f (x) = csc(x) =
1
sin(x) .
We will use the chain rule for this one. Notice that we can write csc(x) as
(sin(x))−1 . So the ouside function is g(x) = x−1 , and the inside function is h(x) =
sin(x). Using the chain rule, we get
f ′ (x) = −(sin(x))−2 ⋅ cos(x) = −
cos(x)
= − cot(x) ⋅ csc(x).
sin2 (x)
Example 5. Find the derivative of f (x) = tan(x) ln(2x2 + 1).
This is a good problem to build our intuition as to how to approach the solution.
Looking at this function, do we need the product rule or the chain rule? We will
need the chain rule for ln(2x2 + 1), and we will need the product rule for the
entire function. So we need both rules. What should be do first? Looking at the
function, we see that the ln(2x2 + 1) is “inside” the product of the function. So we
will need to use the chain rule “inside” the product rule.
For the product rule, let g(x) = tan(x) and h(x) = ln(2x2 + 1). The g ′ (x) =
sec2 (x). But we need to find h′ (x); this is where we need the chain rule.
For the chain rule, let h1 (x) = ln(x) and h2 (x) = 2x2 + 1. Then the chain rule
1
4x
gives h′ (x) = 2
⋅ 4x = 2
. We are not done. The chain rule was just a
2x + 1
2x + 1
step in using the product rule.
We can now finish the product rule. By the product rule, we have
4x
+ ln(2x2 + 1) ⋅ sec2 (x).
f ′ (x) = tan(x) ⋅ 2
2x + 1