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Trigonometric and Exponential Derivatives · Tutorial 150 of 999

Where Trigonometric Derivative Formulas are Defined

Use the zeros of sine and cosine to identify the exact domains of four trigonometric functions and their derivative formulas.

Intermediate 9 min read

What You'll Learn

  • Locate every real zero of sine and cosine using integer multiples of pi.
  • State the domains of tangent, secant, cosecant, and cotangent.
  • Determine where each corresponding derivative formula is defined.
  • Explain why a derivative cannot be defined outside its original function’s domain.
  • Avoid confusing the excluded values for functions with different denominators.

Why Domains Matter for Derivatives

A derivative describes how a function changes near an input. For that description to make sense, the function must first be defined at the input. This is especially important for trigonometric functions written as quotients: their denominators determine which inputs are allowed.

The sine and cosine functions are defined for every real number, but they each equal zero at particular angles. Since tangent, secant, cosecant, and cotangent involve division by sine or cosine, those zero locations determine where the functions—and their derivatives—are defined. Throughout, angles are measured in radians and \(x\) is real.

Key idea: A quotient is defined only where its denominator is nonzero. The derivative formulas for tangent, secant, cosecant, and cotangent also have denominators that vanish at exactly the points excluded from the corresponding function’s domain.

Where Sine and Cosine Are Zero

The unit circle provides a direct way to locate the zeros. At angle \(x\), the point on the unit circle has coordinates \((\cos x,\sin x)\). The sine is zero where the point lies on the horizontal axis, which occurs at integer multiples of \(\pi\). The cosine is zero where the point lies on the vertical axis, which occurs at an odd multiple of \(\frac{\pi}{2}\).

$$ \sin x=0 \quad\Longleftrightarrow\quad x=n\pi,\qquad n\in\mathbb{Z}, $$
$$ \cos x=0 \quad\Longleftrightarrow\quad x=\frac{\pi}{2}+n\pi,\qquad n\in\mathbb{Z}. $$

The integer \(n\) accounts for every full turn in either direction. For example, the cosine zeros include \(-\frac{3\pi}{2},-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2}\), and continue indefinitely. Listing only the positive zeros would miss valid excluded inputs.

Domains of the Four Functions

The definitions as quotients immediately identify the restrictions. Tangent and secant have cosine in the denominator, while cosecant and cotangent have sine in the denominator.

Definition: The domains of the four functions are $$ \begin{aligned} \tan x=\frac{\sin x}{\cos x},\quad \sec x=\frac{1}{\cos x} &\quad\text{for } \cos x\ne 0,\\ \csc x=\frac{1}{\sin x},\quad \cot x=\frac{\cos x}{\sin x} &\quad\text{for } \sin x\ne 0. \end{aligned} $$

Using the zero locations gives the domains in set notation:

$$ \begin{aligned} \operatorname{Dom}(\tan)&=\operatorname{Dom}(\sec) =\mathbb{R}\setminus\left\{\frac{\pi}{2}+n\pi:n\in\mathbb{Z}\right\},\\ \operatorname{Dom}(\csc)&=\operatorname{Dom}(\cot) =\mathbb{R}\setminus\left\{n\pi:n\in\mathbb{Z}\right\}. \end{aligned} $$

The set notation means that each function is defined at every real number except the listed values. For instance, \(\tan x\) is defined at \(x=0\), because \(\cos 0=1\), but \(\csc x\) is not, because \(\sin 0=0\).

Worked Example: Find the Domains on a Closed Interval

Find the portions of \([-\pi,2\pi]\) on which each of the four functions is defined. Include the endpoints in your check.

For tangent and secant, exclude the zeros of cosine:

$$ \frac{\pi}{2}+n\pi\in[-\pi,2\pi]. $$

The values in this interval are \(-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2}\). The next values, \(-\frac{3\pi}{2}\) and \(\frac{5\pi}{2}\), lie outside the interval. Thus tangent and secant are defined on

$$ [-\pi,2\pi]\setminus\left\{-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2}\right\}. $$

For cosecant and cotangent, exclude the zeros of sine:

$$ n\pi\in[-\pi,2\pi] \quad\Longrightarrow\quad n=-1,0,1,2. $$

Therefore cosecant and cotangent are defined on

$$ [-\pi,2\pi]\setminus\{-\pi,0,\pi,2\pi\}. $$

The endpoints were included in the starting interval, but \(-\pi\) and \(2\pi\) must be removed for cosecant and cotangent because their sine values are zero. Neither endpoint is a cosine zero, so both are allowed for tangent and secant. This checks the interval endpoints as well as its interior.

Derivative Formulas and Their Domains

The derivative formulas follow from quotient differentiation, which requires the original quotient’s denominator to be nonzero. For example, differentiating tangent as \(\frac{\sin x}{\cos x}\) gives

$$ \frac{d}{dx}\left(\frac{\sin x}{\cos x}\right) = \frac{\cos x\cos x-\sin x(-\sin x)}{\cos^2 x} = \frac{\cos^2 x+\sin^2 x}{\cos^2 x} = \sec^2 x, \qquad \cos x\ne0. $$

The same quotient rule applied to the other definitions gives the remaining formulas. The condition beside each one is essential: it specifies where the original function exists and the differentiation calculation is valid.

$$ \begin{aligned} \frac{d}{dx}(\tan x)&=\sec^2 x, &\cos x&\ne0,\\ \frac{d}{dx}(\sec x)&=\sec x\tan x, &\cos x&\ne0,\\ \frac{d}{dx}(\csc x)&=-\csc x\cot x, &\sin x&\ne0,\\ \frac{d}{dx}(\cot x)&=-\csc^2 x, &\sin x&\ne0. \end{aligned} $$

These conditions can also be read directly from the right sides. The tangent derivative \(\sec^2x\) and secant derivative \(\sec x\tan x\) involve division by cosine. The cosecant derivative \(-\csc x\cot x\) and cotangent derivative \(-\csc^2x\) involve division by sine. Squaring a denominator does not make a zero denominator valid.

Theorem: Each of these four derivative functions has the same domain as its original function: $$ \begin{aligned} \operatorname{Dom}\left(\frac{d}{dx}\tan x\right) &=\operatorname{Dom}(\tan x),& \operatorname{Dom}\left(\frac{d}{dx}\sec x\right) &=\operatorname{Dom}(\sec x),\\ \operatorname{Dom}\left(\frac{d}{dx}\csc x\right) &=\operatorname{Dom}(\csc x),& \operatorname{Dom}\left(\frac{d}{dx}\cot x\right) &=\operatorname{Dom}(\cot x). \end{aligned} $$

Why the Derivative Domains Match

There are two points to establish. First, each function is differentiable wherever its denominator is nonzero. Sine and cosine are differentiable on \(\mathbb{R}\), and the quotient rule applies at any input where the denominator is nonzero. The formulas above therefore show that all four derivatives exist at every input in the corresponding function domain.

Second, the derivative of a function cannot be defined at an input where the function itself is not defined. In the usual definition, the derivative at \(a\) is a limit of difference quotients involving \(f(a)\):

$$ f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}. $$

If \(f(a)\) does not exist, this difference quotient is not defined as written, so \(f'(a)\) cannot exist for that function at \(a\). Consequently, the derivative domain is neither smaller nor larger than the function domain here: quotient differentiation gives a derivative at every allowed point, and no derivative can be assigned at an excluded point of the original function.

1
Identify the denominator.
Tangent and secant use cosine; cosecant and cotangent use sine.
2
Locate its zeros.
Cosine vanishes at \(\frac{\pi}{2}+n\pi\); sine vanishes at \(n\pi\), for \(n\in\mathbb{Z}\).
3
Exclude those inputs.
Use the same exclusions for the corresponding derivative formula.

Worked Example: Check a Point in Each Domain

Evaluate all four derivative formulas at \(x=\frac{\pi}{4}\), and confirm that this input is allowed.

First, \(\sin\frac{\pi}{4}=\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}\). Neither value is zero, so all four functions and their derivatives are defined there.

For tangent, \(\sec\frac{\pi}{4}=\frac{1}{\cos(\pi/4)}=\sqrt{2}\). Therefore

$$ \left.\frac{d}{dx}\tan x\right|_{x=\pi/4} =\sec^2\frac{\pi}{4} =(\sqrt{2})^2=2. $$

For secant, \(\tan\frac{\pi}{4}=\frac{\sin(\pi/4)}{\cos(\pi/4)}=1\), so

$$ \left.\frac{d}{dx}\sec x\right|_{x=\pi/4} =\sec\frac{\pi}{4}\tan\frac{\pi}{4} =\sqrt{2}\cdot1=\sqrt{2}. $$

For cosecant, \(\csc\frac{\pi}{4}=\sqrt{2}\) and \(\cot\frac{\pi}{4}=1\). Thus

$$ \left.\frac{d}{dx}\csc x\right|_{x=\pi/4} =-\csc\frac{\pi}{4}\cot\frac{\pi}{4} =-\sqrt{2}\cdot1=-\sqrt{2}. $$

Finally,

$$ \left.\frac{d}{dx}\cot x\right|_{x=\pi/4} =-\csc^2\frac{\pi}{4} =-(\sqrt{2})^2=-2. $$

As a check, the evaluated inputs have nonzero sine and cosine, and every value follows by substitution into its stated derivative formula. No excluded denominator is used in any calculation.

Worked Example: Detect an Excluded Input

Determine which of the four functions and derivatives are defined at \(x=\frac{\pi}{2}\).

At this angle, \(\sin\frac{\pi}{2}=1\) and \(\cos\frac{\pi}{2}=0\). Tangent and secant both have cosine in the denominator, so neither function is defined. Their derivatives are not defined there either.

Cosecant and cotangent have sine in the denominator, which is nonzero at this input. In fact,

$$ \csc\frac{\pi}{2}=1, \qquad \cot\frac{\pi}{2}=0. $$

Their derivative values are

$$ -\csc\frac{\pi}{2}\cot\frac{\pi}{2} =-(1)(0)=0, \qquad -\csc^2\frac{\pi}{2}=-(1)^2=-1. $$

Thus, at \(x=\frac{\pi}{2}\), cosecant and cotangent and their derivatives are defined, while tangent and secant and their derivatives are not. The results agree with the cosine-zero exclusion \(\frac{\pi}{2}+n\pi\).

A Common Domain Mistake

A frequent error is to exclude the zeros of the numerator instead of the zeros of the denominator. For example, someone might reason that \(\tan x=\frac{\sin x}{\cos x}\) is undefined whenever \(\sin x=0\), then conclude that \(x=n\pi\) must be excluded from the tangent domain.

$$ \text{Incorrect: }\sin(n\pi)=0 \quad\Longrightarrow\quad \tan(n\pi)\text{ is undefined.} $$

That conclusion is false. At \(x=n\pi\), the tangent numerator is zero, but its denominator is \(\cos(n\pi)=(-1)^n\), which is nonzero. In fact, \(\tan(n\pi)=0\). The correct exclusion comes from setting the denominator equal to zero:

$$ \cos x=0 \quad\Longleftrightarrow\quad x=\frac{\pi}{2}+n\pi. $$

The distinction matters for both the function and its derivative. The reliable procedure is to identify the denominator first, locate its zeros, and only then state the domain.

Summary

The zeros of sine and cosine divide the four trigonometric functions into two pairs. Tangent and secant are excluded where cosine is zero; cosecant and cotangent are excluded where sine is zero. Since the derivative formulas require those same nonzero denominators, each derivative has the same domain as its original function.

Check Your Understanding

Use the denominator and its zero locations to answer each question.

  1. List the excluded inputs for \(\tan x\) and \(\sec x\) using \(n\in\mathbb{Z}\).
  2. List the excluded inputs for \(\csc x\) and \(\cot x\) using \(n\in\mathbb{Z}\).
  3. Are \(\tan x\) and \(\csc x\) defined at \(x=\pi\)? Explain using the sine and cosine values.
  4. Are \(\sec x\) and \(\cot x\) defined at \(x=\frac{\pi}{2}\)? Explain using their denominators.
  5. Why can none of these derivatives be defined at a point excluded from its original function’s domain?

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