Nathan Wakefield, Christine Kelley, Marla Williams, Michelle Haver, Lawrence Seminario-Romero, Robert Huben, Aurora Marks, Stephanie Prahl, Based upon Active Calculus by Matthew Boelkins

Section2.4Derivatives of Other Trigonometric Functions

Motivating Questions

What are the derivatives of the tangent, cotangent, secant, and cosecant functions?

How do the derivatives of \(\tan(x)\text{,}\) \(\cot(x)\text{,}\) \(\sec(x)\text{,}\) and \(\csc(x)\) combine with other derivative rules we have developed to expand the library of functions we can quickly differentiate?

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

Because each angle \(\theta\) 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 \(\theta\text{.}\) In fact, this is the very definition of \(\cos(\theta)\) and \(\sin(\theta)\) i.e. \(\cos(\theta)\) is the \(x\)-coordinate of the point on the unit circle corresponding to the angle \(\theta\text{,}\) and \(\sin(\theta)\) is the \(y\)-coordinate. From this simple definition, all of trigonometry is founded. For instance, the Pythagorean Identity,

is a restatement of the Pythagorean Theorem, applied to the right triangle shown above in Figure2.50.

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

The tangent function is defined by \(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\text{;}\)

the cotangent function is its reciprocal: \(\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}\text{.}\)

The secant function is the reciprocal of the cosine function, \(\sec(\theta) = \frac{1}{\cos(\theta)}\text{;}\)

and the cosecant function is the reciprocal of the sine function, \(\csc(\theta) = \frac{1}{\sin(\theta)}\text{.}\)

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

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 Example2.51 we will work through the steps to find the derivative of \(\tan(x)\text{.}\)

Example2.51

Consider the function \(f(x) = \tan(x)\text{,}\) and remember that \(\tan(x) = \frac{\sin(x)}{\cos(x)}\text{.}\)

What is the domain of \(f\text{?}\)

Use the quotient rule to show that one expression for \(f'(x)\) is

Since \(\sin(x)\) is defined for every real value of \(x\) and \(\cos(x)\) is only equal to zero at odd integer multiples of \(\frac{\pi}2\text{,}\) it follows that the quotient \(\tan(x)=\frac{\sin(x)}{\cos(x)}\) is defined whenever \(x\neq \ldots,-\frac{3\pi}2,-\frac{\pi}2,\frac{\pi}2,\frac{3\pi}2,\ldots\text{.}\)

Since \(f(x)=\tan(x)\) is the quotient of \(\sin(x)\) and \(\cos(x)\text{,}\) then to differentiate \(f\) using the quotient rule we first need \(\frac{d}{dx}[\sin(x)]=\cos(x)\) and \(\frac{d}{dx}[\cos(x)]=-\sin(x)\text{.}\) Then

The Pythagorean Identity states that \(\sin^2(\theta)+\cos^2(\theta)=1\) for any real number \(\theta\text{.}\) We can rewrite the form of \(f'\) found in part (b) as

Just like the original function \(f\text{,}\) 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 \(f'(x)=\frac{1}{\cos^2(x)}\text{,}\) the numerator is always defined and the denominator is zero exactly when \(\cos(x)=0\text{,}\) which occurs at odd integer multiples of \(\frac{\pi}2\text{.}\) Thus the domain of \(f'\) consists of the values \(x\neq\frac{n\pi}{2}\text{,}\) where \(n\) is an odd integer (positive or negative). This is the same as the domain of \(f\text{.}\)

SubsectionDerivatives of the Cotangent, Secant, and Cosecant functions

In Example2.51 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.

Let \(g(x) = \cot(x)\text{.}\) To find \(g'(x)\text{,}\) we observe that \(g(x) = \frac{\cos(x)}{\sin(x)}\) and apply the quotient rule. Hence

Applying the Pythagorean Identity, we see that \(g'(x) = -\frac{1}{\sin^2(x)}\text{.}\) Now recall that \(\csc(x) = \frac{1}{\sin(x)}\text{;}\) it follows that we can express \(g'\) by the rule

Note that neither \(g\) nor \(g'\) is defined when \(\sin(x) = 0\text{,}\) which occurs at every integer multiple of \(\pi\text{.}\) Hence we have the following rule.

Derivative of the Cotangent Function

For all real numbers \(x\) such that \(x \ne k\pi\text{,}\) where \(k = 0, \pm 1, \pm 2, \ldots\text{,}\)

All real numbers \(x\) such that \(x \ne \frac{(2k+1)\pi}{2}\text{,}\) where \(k =0, \pm 1, \pm 2, \ldots\text{.}\)

\(h'(x) = \frac{\sin(x)}{\cos^2(x)}\text{.}\)

\(h'(x) = \sec(x) \tan(x)\text{.}\)

\(h\) and \(h'\) have the same domain: all real numbers \(x\) such that \(x \ne \frac{(2k+1)\pi}{2}\text{,}\) where \(k =0, \pm 1, \pm 2, \ldots\text{.}\)

\(h(x) = \sec(x)\) is defined for all \(x\) for which \(\cos(x) \ne 0\text{.}\) Hence the domain of \(h\) is all real numbers \(x\) such that \(x \ne \frac{(2k+1)\pi}{2}\text{,}\) where \(k =0, \pm 1, \pm 2, \ldots\text{.}\)

The derivative \(h'(x)\) is, like \(h(x)\text{,}\) defined for all values of \(x\) for which \(\cos(x) \ne 0\text{.}\) Therefore, \(h\) and \(h'\) have the same domain: all real numbers \(x\) such that \(x \ne \frac{(2k+1)\pi}{2}\text{,}\) where \(k = 0, \pm 1, \pm 2, \ldots\text{.}\)

Example2.53

Let \(p(x) = \csc(x)\) and recall that \(\csc(x) = \frac{1}{\sin(x)}\text{.}\)

What is the domain of \(p\text{?}\)

Use the quotient rule to develop a formula for \(p'(x)\) that is expressed completely in terms of \(\sin(x)\) and \(\cos(x)\text{.}\)

How can you use other relationships among trigonometric functions to write \(p'(x)\) only in terms of \(\cot(x)\) and \(\csc(x)\text{?}\)

What is the domain of \(p'\text{?}\) How does this compare to the domain of \(p\text{?}\)

\(p(x) = \csc(x)\) is defined for all \(x\) for which \(\sin(x) \ne 0\text{.}\) Hence the domain of \(h\) is all real numbers \(x\) such that \(x\neq k\pi\text{,}\) where \(k = 0, \pm 1, \pm 2, \ldots\text{.}\)

The derivative \(p'(x)\) is, like \(p(x)\text{,}\) defined for all values of \(x\) for which \(\sin(x) \ne 0\text{.}\) Therefore, \(p\) and \(p'\) have the same domain: all real numbers \(x\) such that \(x \ne k\pi\text{,}\) where \(k = 0, \pm 1, \pm 2, \ldots\text{.}\)

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.

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.

Example2.54

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 \(f(x) = 5 \sec(x) - 2\csc(x)\text{.}\) Find the slope of the tangent line to \(f\) at the point where \(x =\frac{\pi}{3}\text{.}\)

Let \(p(z) = z^2\sec(z) - z\cot(z)\text{.}\) Find the instantaneous rate of change of \(p\) at the point where \(z = \frac{\pi}{4}\text{.}\)

Let \(g(r) = \displaystyle \frac{r \sec(r) }{5^r}\text{.}\) Find \(g'(r)\text{.}\)

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

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

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

The function \(\frac{ds}{dt} = s'(t)\) 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.

SubsectionSummary

The derivatives of the other four trigonometric functions are

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 \(x \ne \frac{(2k+1)\pi}{2}\text{,}\) where \(k =0, \pm 1, \pm 2, \ldots\text{.}\)

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.

An object moving vertically has its height at time \(t\) (measured in feet, with time in seconds) given by the function \(h(t) = 3 + \frac{2\cos(t)}{1.2^t}\text{.}\)

What is the object's instantaneous velocity when \(t =2\text{?}\)

What is the object's acceleration at the instant \(t = 2\text{?}\)

Describe in everyday language the behavior of the object at the instant \(t = 2\text{.}\)

True or false: for all real numbers \(x\text{,}\) \(f(x) = \cos(x)\text{.}\)

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 \(f'\text{.}\)