maths.free › Calculus › 2. Computing Derivatives › Derivatives of other trigonometric functions

Derivatives of other trigonometric functions

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

Introduction

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 \(\theta\). In fact, this is the very definition of \(\cos(\theta)\) and \(\sin(\theta)\): \(\cos(\theta)\) is the \(x\)-coordinate of the point on the unit circle corresponding to the angle \(\theta\), and \(\sin(\theta)\) is the \(y\)-coordinate. From this simple definition, all of trigonometry is founded. For instance, the Fundamental Trigonometric Identity, \[\begin{aligned}\end{aligned}\], is a restatement of the Pythagorean Theorem, applied to the right triangle shown in Figure.

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

\(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\)
\(\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}\)
\(\sec(\theta) = \frac{1}{\cos(\theta)}\)
\(\csc(\theta) = \frac{1}{\sin(\theta)}\)

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 function, we can now develop shortcut differentiation rules for the tangent, cotangent, secant, and cosecant functions. In this section's preview activity, we work through the steps to find the derivative of \(y = \tan(x)\).

Exploration

Derivatives of the cotangent, secant, and cosecant functions

In Preview Activity, we found that the derivative of the tangent function can be expressed in several ways, but most simply in terms of the secant function. Next, we develop the derivative of the cotangent function.

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 \[\begin{aligned}g'(x) =\mathstrut \amp \frac{\sin(x)(-\sin(x)) - \cos(x) \cos(x)}{\sin^2(x)} \\ =\mathstrut \amp -\frac{\sin^2(x) + \cos^2(x)}{\sin^2(x)}\end{aligned}\]

By the Fundamental Trigonometric Identity, we see that \(g'(x) = -\frac{1}{\sin^2(x)}\), and recalling that \(\csc(x) = \frac{1}{\sin(x)}\), it follows that we can express \(g'\) by the rule \[\begin{aligned}\end{aligned}\].

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

For all real numbers \(x\) such that \(x \ne k\pi\), where \(k = 0, \pm 1, \pm 2, \ldots\), \[\begin{aligned}\end{aligned}\].

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

For all real numbers \(x\) such that \(x \ne \frac{(2k+1)\pi}{2}\), where \(k = \pm 1, \pm 2, \ldots\), \[\begin{aligned}\end{aligned}\].

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

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.

The derivative retains all of its fundamental meaning as an instantaneous rate of change and as the slope of the tangent line to the function under consideration.

Summary

The derivatives of the other four trigonometric functions are \[\begin{aligned}\end{aligned}\], \[\begin{aligned}\end{aligned}\]. 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 \(x \ne \frac{k\pi}{2}\), where \(k = \pm 1, \pm 2, \ldots\).
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.

Practice (9)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Let \(h(x) = \sec(x)\) and recall that \(\sec(x) = \frac{1}{\cos(x)}\). Apply the quotient rule to find a formula for \(h'(x)\) in terms of \(\sin(x)\) and \(\cos(x)\). Don't simplify your result.
Let \(p(x) = \csc(x)\) and recall that \(\csc(x) = \frac{1}{\sin(x)}\). Apply the quotient rule to find a formula for \(p'(x)\) in terms of \(\sin(x)\) and \(\cos(x)\). Don't simplify your result.
What do your results from the preceding questions have in common with one another? what aspects of your results are different?
An object moving vertically has its height at time \(t\) (measured in feet, with time in seconds) given by the function \(h(t) = 3 + \frac{2\cos(t)}{1.2^t}\).
1. What is the object's instantaneous velocity when \(t =2\)?
2. What is the object's acceleration at the instant \(t = 2\)?
3. Describe in everyday language the behavior of the object at the instant \(t = 2\).
Vis svaret
1. The object's instantaneous velocity at \(t=2\) is given by \(h'(2)\). Since \(h(t) = 3 + \frac{2\cos(t)}{1.2^t}\), by the quotient rule we can determine that \[\begin{aligned}h'(t) &= \frac{1.2^t \cdot (-2\sin(t)) - 2\cos(t) \cdot 1.2^t \ln(1.2)}{(1.2^t)^2} \\ &= \frac{1.2^t (-2\sin(t) - 2\cos(t) \ln(1.2))}{(1.2^t)^2} \\ &= \frac{-2\sin(t) - 2\cos(t) \ln(1.2)}{1.2^t}\end{aligned}\]. Thus, \(h'(2) = \frac{-2\sin(2) - 2\cos(2) \ln(1.2)}{1.2^2} \approx -1.1575\) feet per second is the instanateous (vertical) velocity of the object at \(t=2\).
2. To find the object's acceleration at \(t=2\), we compute \(h''(2)\). First, by the quotient rule, \[\begin{aligned}h''(t) &= \frac{1.2^t(-2\cos(t) + 2\sin(t) \ln(1.2)) - (-2\sin(t) - 2\cos(t) \ln(1.2))1.2^t \ln(1.2)}{(1.2^t)^2} \\ &= \frac{-2\cos(t) + 2\sin(t) \ln(1.2)) + 2\sin(t)\ln(1.2) + 2\cos(t) \ln^2(1.2)}{1.2^t} \\ &= \frac{\cos(t)(-2 + 2\ln^2(1.2)) + 4\ln(1.2)\sin(t))}{1.2^t}\end{aligned}\]. Therefore, the acceleration at \(t = 2\) is \(h''(2) = \frac{\cos(2)(-2 + 2\ln^2(1.2)) + 4\ln(1.2)\sin(2))}{1.2^2} \approx 1.0193\) feet per second per second.
3. At \(t=2\), the object is falling and slowing down. We know the object is falling because its velocity is negative, and slowing down because its negative velocity is increasing (due to \(h''(2)\) being positive) and thus getting less negative.
Let \(f(x) = \sin(x) \cot(x)\).
1. Use the product rule to find \(f'(x)\).
2. True or false: for all real numbers \(x\), \(f(x) = \cos(x)\).
3. 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 \(f'\).)
Vis svaret
1. Since \(f(x) = \sin(x) \cot(x)\), by the product rule we know that \(f'(x) = \sin(x) \cdot (-\csc^2(x)) + \cot(x) \cdot \cos(x)\).
2. False. Since \(\cot(x) = \frac{\cos(x)}{\sin(x)}\), we can write \(f(x) = \sin(x) \cdot \frac{\cos(x)}{\sin(x)}\). From here, it is tempting to cancel the \(\sin(x)\) term in the numerator and denominator. However, note that because of the \(\sin(x)\) in the denominator, \(f\) is not defined for \(x = \frac{\pi}{2} + k\pi\) for any integer \(k\). In other words, the domain of \(f\) is all real numbers except odd multiples of \(\frac{\pi}{2}\). Thus, we can say that \(f(x) = \cos(x)\) for all real numbers except odd multiples of \(\frac{\pi}{2}\). But it is not the case that \(f(x) = \cos(x)\) for every value of \(x\).
3. Taking our result from (a) and converting \(\csc(x)\) and \(\cot(x)\) to expressions involving the sine and cosine functions, we have \[\begin{aligned}f'(x) &= \sin(x) \cdot (-\csc^2(x)) + \cot(x) \cdot \cos(x) \\ &= \sin(x) \cdot \left(-\frac{1}{\sin^2(x)}\right) + \frac{\cos(x)}{\sin(x)} \cdot \cos(x)\end{aligned}\] Provided that \(x \ne \frac{\pi}{2} + k\pi\) for some integer value of \(k\), it follows that \[\begin{aligned}\end{aligned}\]. By the Fundamental Trigonometric Identity, recall that \(\cos^2(x) + \sin^2(x) = 1\), and thus \(\cos^2(x)-1 = -\sin^2(x)\). Hence, for \(x \ne \frac{\pi}{2} + k\pi\) for some integer value of \(k\), we have \(f'(x) = \frac{-\sin^2(x)}{\sin(x)} = -\sin(x)\). Again, this equality holds only for \(x \ne \frac{\pi}{2} + k\pi\) for some integer value of \(k\) because those values of \(x\) are not in the domain of the original function \(f\).
Let \(p(z)\) be given by the rule \[\begin{aligned}\end{aligned}\].
1. Determine \(p'(z)\).
2. Find an equation for the tangent line to \(p\) at the point where \(z = 0\).
3. At \(z = 0\), is \(p\) increasing, decreasing, or neither? Why?
Vis svaret
1. We use the quotient, product, sum, and constant multiple rules to obtain \[\begin{aligned}p'(z) &= \frac{d}{dz}\left(\frac{z\tan(z)}{z^2\sec(z) + 1} + 3 e^z + 1\right) \\ &= \frac{d}{dz}\left(\frac{z\tan(z)}{z^2\sec(z) + 1}\right) + \frac{d}{dz} \left(3 e^z \right) + \frac{d}{dz}(1) \\ &= \frac{\left(z^2\sec(z) +1 \right)\frac{d}{dz}\left(z\tan(z)\right) - z\tan(z) \frac{d}{dz}\left(z^2\sec(z) + 1\right)}{\left(z^2\sec(z) + 1\right)^2} +3e^z \\ &= \frac{\left(z^2\sec(z) +1 \right)\left(z\sec^2(z)+\tan(z)\right) - z\tan(z) \left(z^2\sec(z)\tan(z)+2z\sec(z)\right)}{\left(z^2\sec(z) + 1\right)^2} +3e^z\end{aligned}\]
2. We have \(p(0) = 0+3+1 = 4\) and \(p'(0) = 0+3=3\), so an equation for the tangent line to \(p\) at the point where \(z=0\) is \(y - 4 = 3(x-0)\).
3. We found that \(p'(0) = 3\) is positive, so \(p\) is increasing at \(t=0\).
An object moving vertically has its height at time \(t\) (measured in feet, with time in seconds) given by the function \(h(t) = 3 + \frac{2\cos(t)}{1.2^t}\).
1. What is the object's instantaneous velocity when \(t =2\)?
2. What is the object's acceleration at the instant \(t = 2\)?
3. Describe in everyday language the behavior of the object at the instant \(t = 2\).
Vis svaret
1. The object's instantaneous velocity at \(t=2\) is given by \(h'(2)\). Since \(h(t) = 3 + \frac{2\cos(t)}{1.2^t}\), by the quotient rule we can determine that \[\begin{aligned}h'(t) &= \frac{1.2^t \cdot (-2\sin(t)) - 2\cos(t) \cdot 1.2^t \ln(1.2)}{(1.2^t)^2} \\ &= \frac{1.2^t (-2\sin(t) - 2\cos(t) \ln(1.2))}{(1.2^t)^2} \\ &= \frac{-2\sin(t) - 2\cos(t) \ln(1.2)}{1.2^t}\end{aligned}\]. Thus, \(h'(2) = \frac{-2\sin(2) - 2\cos(2) \ln(1.2)}{1.2^2} \approx -1.1575\) feet per second is the instanateous (vertical) velocity of the object at \(t=2\).
2. To find the object's acceleration at \(t=2\), we compute \(h''(2)\). First, by the quotient rule, \[\begin{aligned}h''(t) &= \frac{1.2^t(-2\cos(t) + 2\sin(t) \ln(1.2)) - (-2\sin(t) - 2\cos(t) \ln(1.2))1.2^t \ln(1.2)}{(1.2^t)^2} \\ &= \frac{-2\cos(t) + 2\sin(t) \ln(1.2)) + 2\sin(t)\ln(1.2) + 2\cos(t) \ln^2(1.2)}{1.2^t} \\ &= \frac{\cos(t)(-2 + 2\ln^2(1.2)) + 4\ln(1.2)\sin(t))}{1.2^t}\end{aligned}\]. Therefore, the acceleration at \(t = 2\) is \(h''(2) = \frac{\cos(2)(-2 + 2\ln^2(1.2)) + 4\ln(1.2)\sin(2))}{1.2^2} \approx 1.0193\) feet per second per second.
3. At \(t=2\), the object is falling and slowing down. We know the object is falling because its velocity is negative, and slowing down because its negative velocity is increasing (due to \(h''(2)\) being positive) and thus getting less negative.
Let \(f(x) = \sin(x) \cot(x)\).
1. Use the product rule to find \(f'(x)\).
2. True or false: for all real numbers \(x\), \(f(x) = \cos(x)\).
3. 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 \(f'\).)
Vis svaret
1. Since \(f(x) = \sin(x) \cot(x)\), by the product rule we know that \(f'(x) = \sin(x) \cdot (-\csc^2(x)) + \cot(x) \cdot \cos(x)\).
2. False. Since \(\cot(x) = \frac{\cos(x)}{\sin(x)}\), we can write \(f(x) = \sin(x) \cdot \frac{\cos(x)}{\sin(x)}\). From here, it is tempting to cancel the \(\sin(x)\) term in the numerator and denominator. However, note that because of the \(\sin(x)\) in the denominator, \(f\) is not defined for \(x = \frac{\pi}{2} + k\pi\) for any integer \(k\). In other words, the domain of \(f\) is all real numbers except odd multiples of \(\frac{\pi}{2}\). Thus, we can say that \(f(x) = \cos(x)\) for all real numbers except odd multiples of \(\frac{\pi}{2}\). But it is not the case that \(f(x) = \cos(x)\) for every value of \(x\).
3. Taking our result from (a) and converting \(\csc(x)\) and \(\cot(x)\) to expressions involving the sine and cosine functions, we have \[\begin{aligned}f'(x) &= \sin(x) \cdot (-\csc^2(x)) + \cot(x) \cdot \cos(x) \\ &= \sin(x) \cdot \left(-\frac{1}{\sin^2(x)}\right) + \frac{\cos(x)}{\sin(x)} \cdot \cos(x)\end{aligned}\] Provided that \(x \ne \frac{\pi}{2} + k\pi\) for some integer value of \(k\), it follows that \[\begin{aligned}\end{aligned}\]. By the Fundamental Trigonometric Identity, recall that \(\cos^2(x) + \sin^2(x) = 1\), and thus \(\cos^2(x)-1 = -\sin^2(x)\). Hence, for \(x \ne \frac{\pi}{2} + k\pi\) for some integer value of \(k\), we have \(f'(x) = \frac{-\sin^2(x)}{\sin(x)} = -\sin(x)\). Again, this equality holds only for \(x \ne \frac{\pi}{2} + k\pi\) for some integer value of \(k\) because those values of \(x\) are not in the domain of the original function \(f\).
Let \(p(z)\) be given by the rule \[\begin{aligned}\end{aligned}\].
1. Determine \(p'(z)\).
2. Find an equation for the tangent line to \(p\) at the point where \(z = 0\).
3. At \(z = 0\), is \(p\) increasing, decreasing, or neither? Why?
Vis svaret
1. We use the quotient, product, sum, and constant multiple rules to obtain \[\begin{aligned}p'(z) &= \frac{d}{dz}\left(\frac{z\tan(z)}{z^2\sec(z) + 1} + 3 e^z + 1\right) \\ &= \frac{d}{dz}\left(\frac{z\tan(z)}{z^2\sec(z) + 1}\right) + \frac{d}{dz} \left(3 e^z \right) + \frac{d}{dz}(1) \\ &= \frac{\left(z^2\sec(z) +1 \right)\frac{d}{dz}\left(z\tan(z)\right) - z\tan(z) \frac{d}{dz}\left(z^2\sec(z) + 1\right)}{\left(z^2\sec(z) + 1\right)^2} +3e^z \\ &= \frac{\left(z^2\sec(z) +1 \right)\left(z\sec^2(z)+\tan(z)\right) - z\tan(z) \left(z^2\sec(z)\tan(z)+2z\sec(z)\right)}{\left(z^2\sec(z) + 1\right)^2} +3e^z\end{aligned}\]
2. We have \(p(0) = 0+3+1 = 4\) and \(p'(0) = 0+3=3\), so an equation for the tangent line to \(p\) at the point where \(z=0\) is \(y - 4 = 3(x-0)\).
3. We found that \(p'(0) = 3\) is positive, so \(p\) is increasing at \(t=0\).

Symbols used here

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\theta

theta
The usual name for an angle.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

Euler's number
2.71828…, the base whose exponential is its own derivative.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: Derivatives of other trigonometric functions

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?

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

Prøv din egen

Parts of this page are adapted from Boelkins, Active Calculus (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.

Mer i Calculus

Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests