CC Derivatives of Other Trigonometric Functions

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Section 2.4 Derivatives of Other Trigonometric Functions

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Motivating Questions

What are the derivatives of the tangent, cotangent, secant, and cosecant functions?
How do the derivatives of $\tan (x),$ $\cot (x),$ $\sec (x),$ and $\csc (x)$ combine with other derivative rules we have developed to expand the library of functions we can quickly differentiate?

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Supplemental Videos.

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The main topics of this section are also presented in the following videos:

Deriving Derivative of Tangent
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unl.yuja.com/V/Video?v=7114276&node=34303254&a=205056485&autoplay=1
Deriving Derivative of Secant
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unl.yuja.com/V/Video?v=7114275&node=34303292&a=16613299&autoplay=1

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One of the powerful themes in trigonometry comes from a very simple idea: locating a point on the unit circle.

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Figure 2.50. The unit circle and the definition of the sine and cosine functions.

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Because each angle

θ

in standard position corresponds to one and only one point

(x, y)

on the unit circle, the

x

- and

y

-coordinates of this point are each functions of

θ .

In fact, this is the very definition of

\cos (θ)

and

\sin (θ)

i.e.

\cos (θ)

is the

x

-coordinate of the point on the unit circle corresponding to the angle

θ,

and

\sin (θ)

is the

y

-coordinate. From this simple definition, all of trigonometry is founded. For instance, the Pythagorean Identity,

\sin^{2} (θ) + \cos^{2} (θ) = 1,

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is a restatement of the Pythagorean Theorem, applied to the right triangle shown above in Figure 2.50.

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There are four other trigonometric functions, each defined in terms of the sine and/or cosine functions.

The tangent function is defined by $\tan (θ) = \frac{\sin (θ)}{\cos (θ)};$
the cotangent function is its reciprocal: $\cot (θ) = \frac{\cos (θ)}{\sin (θ)} .$
The secant function is the reciprocal of the cosine function, $\sec (θ) = \frac{1}{\cos (θ)};$
and the cosecant function is the reciprocal of the sine function, $\csc (θ) = \frac{1}{\sin (θ)} .$

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These six trigonometric functions together offer us a wide range of flexibility in problems involving right triangles.

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Because we know the derivatives of the sine and cosine functions, we can now develop shortcut differentiation rules for the tangent, cotangent, secant, and cosecant functions. In Example 2.51 we will work through the steps to find the derivative of

\tan (x) .

Example 2.51.

Consider the function

f (x) = \tan (x),

and remember that

\tan (x) = \frac{\sin (x)}{\cos (x)} .

What is the domain of $f ?$
Use the quotient rule to show that one expression for $f^{'} (x)$ is

$f^{'} (x) = \frac{\cos (x) \cos (x) + \sin (x) \sin (x)}{\cos^{2} (x)} .$
What is the Pythagorean Identity? How can this identity be used to find a simpler form for $f^{'} (x) ?$
Recall that $\sec (x) = \frac{1}{\cos (x)} .$ How can we express $f^{'} (x)$ in terms of the secant function?
For what values of $x$ is $f^{'} (x)$ defined? How does this set compare to the domain of $f ?$

Hint.

When is the denominator of $f$ equal to $0 ?$
Remember that

$\frac{d}{d x} [\frac{p (x)}{q (x)}] = \frac{p^{'} (x) q (x) - p (x) q^{'} (x)}{[q (x)]^{2}} .$
The Pythagorean Identity comes from the right triangle shown earlier in Figure 2.50.
Remember that $\cos^{2} (x) = [\cos (x)]^{2},$ and use the simpler form for $f^{'} (x)$ that you found in part (c).
When is the denominator of $f^{'}$ equal to $0 ?$

Answer.

$x \neq \frac{n π}{2}$ for any odd integer $n .$
$f^{'} (x) = \frac{(\cos (x)) \cos (x) - \sin (x) (- \sin (x))}{(\cos (x))^{2}} .$
$\sin^{2} (θ) + \cos^{2} (θ) = 1 .$
$f^{'} (x) = \sec^{2} (x) .$
$x \neq \frac{n π}{2}$ for any odd integer $n .$

Solution.

Since $\sin (x)$ is defined for every real value of $x$ and $\cos (x)$ is only equal to zero at odd integer multiples of $\frac{π}{2},$ it follows that the quotient $\tan (x) = \frac{\sin (x)}{\cos (x)}$ is defined whenever $x \neq \dots, - \frac{3 π}{2}, - \frac{π}{2}, \frac{π}{2}, \frac{3 π}{2}, \dots .$
Since $f (x) = \tan (x)$ is the quotient of $\sin (x)$ and $\cos (x),$ then to differentiate $f$ using the quotient rule we first need $\frac{d}{d x} [\sin (x)] = \cos (x)$ and $\frac{d}{d x} [\cos (x)] = - \sin (x) .$ Then

$\begin{aligned} f^{'} (x) = & \frac{d}{d x} [\frac{\sin (x)}{\cos (x)}] \\ = & \frac{\frac{d}{d x} [\sin (x)] \cos (x) - \sin (x) \frac{d}{d x} [\cos (x)]}{(\cos (x))^{2}} \\ = & \frac{(\cos (x)) \cos (x) - \sin (x) (- \sin (x))}{(\cos (x))^{2}} \\ = & \frac{\cos (x) \cos (x) + \sin (x) \sin (x)}{\cos^{2} (x)} \end{aligned}$
The Pythagorean Identity states that $\sin^{2} (θ) + \cos^{2} (θ) = 1$ for any real number $θ .$ We can rewrite the form of $f^{'}$ found in part (b) as

$f^{'} (x) = \frac{\cos^{2} (x) + \sin^{2} (x)}{\cos^{2} (x)},$

and use the Pythagorean Identity to reduce the numerator to $1 .$ Thus a simpler expression for $f^{'} (x)$ is

$f^{'} (x) = \frac{1}{\cos^{2} (x)} .$
Since $\sec (x) = \frac{1}{\cos (x)},$ then

$f^{'} (x) = \frac{1}{\cos^{2} (x)} = \sec^{2} (x) .$
Just like the original function $f,$ this derivative function $f^{'}$ can be expressed as a quotient, so its domain will be the values of $x$ at which the numerator is defined and the denominator is nonzero. Since $f^{'} (x) = \frac{1}{\cos^{2} (x)},$ the numerator is always defined and the denominator is zero exactly when $\cos (x) = 0,$ which occurs at odd integer multiples of $\frac{π}{2} .$ Thus the domain of $f^{'}$ consists of the values $x \neq \frac{n π}{2},$ where $n$ is an odd integer (positive or negative). This is the same as the domain of $f .$

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Subsection 2.4.1 Derivatives of the Cotangent, Secant, and Cosecant functions

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In Example 2.51 we found that the derivative of the tangent function can be expressed in several ways, with its simplest form written in terms of the secant function. Next, we develop the derivative of the cotangent function.

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Let

g (x) = \cot (x) .

To find

g^{'} (x),

we observe that

g (x) = \frac{\cos (x)}{\sin (x)}

and apply the quotient rule. Hence

\begin{aligned} g^{'} (x) = & \frac{(- \sin (x)) \sin (x) - \cos (x) \cos (x)}{\sin^{2} (x)} \\ = & - \frac{\sin^{2} (x) + \cos^{2} (x)}{\sin^{2} (x)} \end{aligned}

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Applying the Pythagorean Identity, we see that

g^{'} (x) = - \frac{1}{\sin^{2} (x)} .

Now recall that

\csc (x) = \frac{1}{\sin (x)};

it follows that we can express

g^{'}

by the rule

g^{'} (x) = - \csc^{2} (x) .

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Note that neither

g

nor

g^{'}

is defined when

\sin (x) = 0,

which occurs at every integer multiple of

π .

Hence we have the following rule.

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Derivative of the Cotangent Function.

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For all real numbers

x

such that

x \neq k π,

where

k = 0, \pm 1, \pm 2, \dots,

\frac{d}{d x} [\cot (x)] = - \csc^{2} (x) .

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Notice that the derivative of the cotangent function is very similar to the derivative of the tangent function we discovered in Example 2.51.

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Derivative of the Tangent Function.

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For all real numbers

x

such that

x \neq \frac{(2 k + 1) π}{2},

where

k = 0, \pm 1, \pm 2, \dots,

\frac{d}{d x} [\tan (x)] = \sec^{2} (x) .

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In the next two examples, we develop the rules for differentiating the secant and cosecant functions.

Example 2.52.

Let

h (x) = \sec (x)

and recall that

\sec (x) = \frac{1}{\cos (x)} .

What is the domain of $h ?$
Use the quotient rule to develop a formula for $h^{'} (x)$ that is expressed completely in terms of $\sin (x)$ and $\cos (x) .$
How can you use other relationships among trigonometric functions to write $h^{'} (x)$ only in terms of $\tan (x)$ and $\sec (x) ?$
What is the domain of $h^{'} ?$ How does this compare to the domain of $h ?$

Hint.

For what values of $x$ is $\cos (x) = 0 ?$
Don’t forget that $\frac{d}{d x} [1] = 0 .$
Consider rewriting $\frac{\sin (x)}{\cos^{2} (x)}$ as $\frac{1}{\cos (x)} \cdot \frac{\sin (x)}{\cos (x)} .$
Observe that $\cos (x)$ is still present in the denominator of $h^{'} (x) .$

Answer.

All real numbers $x$ such that $x \neq \frac{(2 k + 1) π}{2},$ where $k = 0, \pm 1, \pm 2, \dots .$
$h^{'} (x) = \frac{\sin (x)}{\cos^{2} (x)} .$
$h^{'} (x) = \sec (x) \tan (x) .$
$h$ and $h^{'}$ have the same domain: all real numbers $x$ such that $x \neq \frac{(2 k + 1) π}{2},$ where $k = 0, \pm 1, \pm 2, \dots .$

Solution.

$h (x) = \sec (x)$ is defined for all $x$ for which $\cos (x) \neq 0 .$ Hence the domain of $h$ is all real numbers $x$ such that $x \neq \frac{(2 k + 1) π}{2},$ where $k = 0, \pm 1, \pm 2, \dots .$
By the quotient rule,

$h^{'} (x) = \frac{0 - 1 (- \sin (x))}{\cos^{2} (x)} = \frac{\sin (x)}{\cos^{2} (x)} .$
Observe that $h^{'} (x) = \frac{\sin (x)}{\cos^{2} (x)} = \frac{1}{\cos (x)} \cdot \frac{\sin (x)}{\cos (x)},$ so

$h^{'} (x) = \sec (x) \tan (x) .$
The derivative $h^{'} (x)$ is, like $h (x),$ defined for all values of $x$ for which $\cos (x) \neq 0 .$ Therefore, $h$ and $h^{'}$ have the same domain: all real numbers $x$ such that $x \neq \frac{(2 k + 1) π}{2},$ where $k = 0, \pm 1, \pm 2, \dots .$

Example 2.53.

Let

p (x) = \csc (x)

and recall that

\csc (x) = \frac{1}{\sin (x)} .

What is the domain of $p ?$
Use the quotient rule to develop a formula for $p^{'} (x)$ that is expressed completely in terms of $\sin (x)$ and $\cos (x) .$
How can you use other relationships among trigonometric functions to write $p^{'} (x)$ only in terms of $\cot (x)$ and $\csc (x) ?$
What is the domain of $p^{'} ?$ How does this compare to the domain of $p ?$

Hint.

For what values of $x$ is $\sin (x) = 0 ?$
Don’t forget that $\frac{d}{d x} [1] = 0 .$
Consider rewriting $\frac{\cos (x)}{\sin^{2} (x)}$ as $\frac{1}{\sin (x)} \cdot \frac{\cos (x)}{\sin (x)} .$
Observe that $\sin (x)$ is still present in the denominator of $h^{'} (x) .$

Answer.

All real numbers $x$ such that $x \neq k π,$ where $k = 0, \pm 1, \pm 2, \dots .$
$h^{'} (x) = - \frac{\cos (x)}{\sin^{2} (x)} .$
$h^{'} (x) = - \csc (x) \cot (x) .$
$p$ and $p^{'}$ have the same domain: all real numbers $x$ such that $x \neq k π,$ where $k = 0, \pm 1, \pm 2, \dots .$

Solution.

$p (x) = \csc (x)$ is defined for all $x$ for which $\sin (x) \neq 0 .$ Hence the domain of $h$ is all real numbers $x$ such that $x \neq k π,$ where $k = 0, \pm 1, \pm 2, \dots .$
By the quotient rule,

$h^{'} (x) = \frac{0 - 1 \cdot (\cos (x))}{\sin^{2} (x)} = - \frac{\cos (x)}{\sin^{2} (x)} .$
Observe that $h^{'} (x) = - \frac{\cos (x)}{\sin^{2} (x)} = - \frac{1}{\sin (x)} \cdot \frac{\cos (x)}{\sin (x)},$ so

$h^{'} (x) = - \csc (x) \cot (x) .$
The derivative $p^{'} (x)$ is, like $p (x),$ defined for all values of $x$ for which $\sin (x) \neq 0 .$ Therefore, $p$ and $p^{'}$ have the same domain: all real numbers $x$ such that $x \neq k π,$ where $k = 0, \pm 1, \pm 2, \dots .$

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Using the quotient rule we have determined the derivatives of the tangent, cotangent, secant, and cosecant functions, expanding our overall library of functions we can differentiate. Observe that just as the derivative of any polynomial function is a polynomial, and the derivative of any exponential function is another exponential function, so it is that the derivative of any basic trigonometric function is another function that consists of basic trigonometric functions. This makes sense because all trigonometric functions are periodic, and hence their derivatives will be periodic, too.

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As always, the derivative retains all of its fundamental meaning, both as an instantaneous rate of change of the function under consideration and as the slope of the tangent line to the graph of that same function.

Example 2.54.

Answer each of the following questions. Where a derivative is requested, be sure to label the derivative function with its name using proper notation.

Let $f (x) = 5 \sec (x) - 2 \csc (x) .$ Find the slope of the tangent line to $f$ at the point where $x = \frac{π}{3} .$
Let $p (z) = z^{2} \sec (z) - z \cot (z) .$ Find the instantaneous rate of change of $p$ at the point where $z = \frac{π}{4} .$
Let $h (t) = \frac{\tan (t)}{t^{2} + 1} - 2 e^{t} \cos (t) .$ Find $h^{'} (t) .$
Let $g (r) = \frac{r \sec (r)}{5^{r}} .$ Find $g^{'} (r) .$
When a mass hangs from a spring and is set in motion, the object’s position oscillates in such a way that the size of the oscillations decreases. This is usually called a damped oscillation. Suppose that for a particular object, its displacement from equilibrium (where the object sits at rest) is modeled by the function

$s (t) = \frac{15 \sin (t)}{e^{t}} .$

Assume that $s$ is measured in inches and $t$ in seconds. Sketch a graph of this function for $t \geq 0$ to see how it represents the situation described. Then compute $\frac{d s}{d t},$ state the units of this function, and explain what it tells you about the object’s motion. Finally, compute and interpret $s^{'} (2) .$

Hint.

What rule(s) can help you determine $f^{'} (x) ?$
Note that $p$ is a sum of two functions. What rule is needed to differentiate each term in the sum?
Observe that $h$ is a sum of two functions; the first term in the sum is a quotient, while the second is a product.
What is the overall structure of $g ?$ What is the algebraic structure of the numerator of $g ?$
Keep in mind that the derivative of position tells us the instantaneous velocity.

Answer.

$10 \sqrt{3} + \frac{4}{3} .$
$p^{'} (\frac{π}{4}) = \frac{π}{2} \sqrt{2} + \frac{π^{2}}{16} \sqrt{2} - 1 + \frac{π}{2} .$
$h^{'} (t) = \frac{(t^{2} + 1) \sec^{2} (t) - 2 t \tan (t)}{(t^{2} + 1)^{2}} - 2 e^{t} \cos (t) + 2 e^{t} \sin (t) .$
$g^{'} (r) = \frac{\sec (r) + r \sec (r) \tan (r) - r \sec (r) \ln (5)}{5^{r}} .$
$s^{'} (2) = \frac{15 \cos (2) - 15 \sin (2)}{e^{2}} \approx - 2.69$ inches per second.

Solution.

Using the sum and constant multiple rules along with the formulas for the derivatives of $\sec (x)$ and $\csc (x),$ we find that

$f^{'} (x) = 5 \sec (x) \tan (x) + 2 \csc (x) \cot (x) .$

Therefore, the slope of the tangent line to $f$ at the point where $x = \frac{π}{3}$ is given by

$\begin{aligned} f^{'} (\frac{π}{3}) = & 5 \sec (\frac{π}{3}) \tan (\frac{π}{3}) + 2 \csc (\frac{π}{3}) \cot (\frac{π}{3}) \\ = & 5 \cdot 2 \cdot \sqrt{3} + 2 \cdot \frac{2}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}} = 10 \sqrt{3} + \frac{4}{3} . \end{aligned}$
By the sum rule and two applications of the product rule, we have

$\begin{aligned} p^{'} (z) = & \frac{d}{d z} [z^{2} \sec (z)] - \frac{d}{d z} [z \cot (z)] \\ = & [2 z \sec (z) + z^{2} \sec (z) \tan (z)] - [\cot (z) + z (- \csc^{2} (z))] \\ = & 2 z \sec (z) + z^{2} \sec (z) \tan (z) - \cot (z) + z \csc^{2} (z) . \end{aligned}$

Thus, the instantaneous rate of change of $p$ at the point where $z = \frac{π}{4}$ is

$\begin{aligned} p^{'} (\frac{π}{4}) = & 2 \cdot \frac{π}{4} \sec (\frac{π}{4}) + {(\frac{π}{4})}^{2} \sec (\frac{π}{4}) \tan (\frac{π}{4}) - \cot (\frac{π}{4}) + \frac{π}{4} \csc^{2} (\frac{π}{4}) \\ = & 2 \cdot \frac{π}{4} \cdot \sqrt{2} + \frac{π^{2}}{16} \cdot \sqrt{2} - 1 + \frac{π}{4} \cdot 2 \\ = & \frac{π}{2} \sqrt{2} + \frac{π^{2}}{16} \sqrt{2} - 1 + \frac{π}{2} . \end{aligned}$
Using the sum and constant multiple rules, followed by the quotient rule on the first term and the product rule on the second, we find that

$\begin{aligned} h^{'} (t) = & \frac{d}{d t} [\frac{\tan (t)}{t^{2} + 1}] - 2 \frac{d}{d t} [e^{t} \cos (t)] \\ = & \frac{\sec^{2} (t) (t^{2} + 1) - \tan (t) (2 t)}{(t^{2} + 1)^{2}} - 2 (e^{t} \cos (t) + e^{t} (- \sin (t))) \\ = & \frac{(t^{2} + 1) \sec^{2} (t) - 2 t \tan (t)}{(t^{2} + 1)^{2}} - 2 e^{t} \cos (t) + 2 e^{t} \sin (t) . \end{aligned}$
Note that $g$ is fundamentally a quotient, so we need to use the quotient rule. But the numerator of $g$ is a product, so the product rule will be required to compute the derivative of the top function. Executing the quotient rule and proceeding, we find that

$\begin{aligned} g^{'} (r) = & \frac{\frac{d}{d r} [r \sec (r)] 5^{r} - r \sec (r) \frac{d}{d r} [5^{r}]}{(5^{r})^{2}} \\ = & \frac{[\sec (r) + r \sec (r) \tan (r)] 5^{r} - r \sec (r) \cdot 5^{r} \ln (5)}{(5^{r})^{2}} \\ = & \frac{\sec (r) + r \sec (r) \tan (r) - r \sec (r) \ln (5)}{5^{r}} . \end{aligned}$
Figure 2.55. The graph of $y = s (t),$ where $s (t) = \frac{15 \sin (t)}{e^{t}}$ models the motion of a mass on a spring.

By the quotient rule,

$\frac{d s}{d t} = \frac{15 \cos (t) \cdot e^{t} - 15 \sin (t) \cdot e^{t}}{(e^{t})^{2}} = \frac{15 \cos (t) - 15 \sin (t)}{e^{t}} .$

The function $\frac{d s}{d t} = s^{'} (t)$ measures the instantaneous vertical velocity, in inches per second, of the mass that is attached to the spring. In particular, $s^{'} (2) = \frac{15 \cos (2) - 15 \sin (2)}{e^{2}} \approx - 2.69$ inches per second. This tells us that when precisely $2$ seconds have passed, the mass is moving downward at an instantaneous rate of 2.69 inches per second.

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Subsection 2.4.2 Summary

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The derivatives of the other four trigonometric functions are

$\frac{d}{d x} [\tan (x)] = \sec^{2} (x), \frac{d}{d x} [\cot (x)] = - \csc^{2} (x),$

$\frac{d}{d x} [\sec (x)] = \sec (x) \tan (x), and \frac{d}{d x} [\csc (x)] = - \csc (x) \cot (x) .$

Each derivative exists and is defined on the same domain as the original function. For example, both the tangent function and its derivative are defined for all real numbers $x$ such that $x \neq \frac{(2 k + 1) π}{2},$ where $k = 0, \pm 1, \pm 2, \dots .$
The four rules for the derivatives of the tangent, cotangent, secant, and cosecant can be used along with the rules for power functions, exponential functions, and the sine and cosine, as well as the sum, constant multiple, product, and quotient rules, to quickly differentiate a wide range of different functions.

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Exercises 2.4.3 Exercises

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1. A sum and product involving $\tan x$ .

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Find the derivative of

h (t) = t \tan t + \cos t

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h^{'} (t) =

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2. A quotient involving $\tan t$ .

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Let

f (x) = \frac{5 \tan (x)}{x} .

Find the following:

1.	$f^{'} (x)$	$=$
2.	$f^{'} (2)$	$=$

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3. A quotient of trigonometric functions.

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Let

f (x) = \frac{\tan (x) - 2}{\sec (x)} .

Find the following:

1.	$f^{'} (x)$	$=$
2.	$f^{'} (4)$	$=$

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4. A quotient that involves a product.

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Let

f (x) = \frac{2 x^{2} \tan (x)}{\sec (x)} .

Find the following:

1.	$f^{'} (x)$	$=$
2.	$f^{'} (1)$	$=$

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5. Finding a tangent line equation.

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Find the equation of the tangent line to the curve

y = 3 \tan x

at the point

(π / 4, 3) .

The equation of this tangent line can be written in the form

y = m x + b

where

m

is:

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and where

b

is:

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6. Oscillatory Motion.

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An object moving vertically has its height at time

t

(measured in feet, with time in seconds) given by the function

h (t) = 3 + \frac{2 \cos (t)}{{1.2}^{t}} .

What is the object’s instantaneous velocity when $t = 2 ?$
What is the object’s acceleration at the instant $t = 2 ?$
Describe in everyday language the behavior of the object at the instant $t = 2 .$

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7. A product of trigonometric functions.

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Let

f (x) = \sin (x) \cot (x) .

Use the product rule to find $f^{'} (x) .$
True or false: for all real numbers $x,$ $f (x) = \cos (x) .$
Explain why the function that you found in (a) is almost the opposite of the sine function, but not quite. Hint: convert all of the trigonometric functions in (a) to sines and cosines, and work to simplify. Think carefully about the domain of $f$ and the domain of $f^{'} .$

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8. Combining Differentiation Rules.

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Let

p (z)

be given by the rule

p (z) = \frac{z \tan (z)}{z^{2} \sec (z) + 1} + 3 e^{z} + 1 .

Determine $p^{'} (z) .$
Find an equation for the tangent line to $p$ at the point where $z = 0 .$
At $z = 0,$ is $p$ increasing, decreasing, or neither? Why?

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Section 2.4 Derivatives of Other Trigonometric Functions

Motivating Questions

Supplemental Videos.

Example 2.51.

Subsection 2.4.1 Derivatives of the Cotangent, Secant, and Cosecant functions

Derivative of the Cotangent Function.

Derivative of the Tangent Function.

Example 2.52.

Example 2.53.

Example 2.54.

Subsection 2.4.2 Summary

Exercises 2.4.3 Exercises

1. A sum and product involving tan⁡x.

2. A quotient involving tan⁡t.

3. A quotient of trigonometric functions.

4. A quotient that involves a product.

5. Finding a tangent line equation.

6. Oscillatory Motion.

7. A product of trigonometric functions.

8. Combining Differentiation Rules.

1. A sum and product involving $\tan x$ .

2. A quotient involving $\tan t$ .