Three Trigonometric Derivatives from Familiar Rules
Secant, cosecant, and cotangent can be expressed using sine and cosine. That makes their derivatives accessible using the derivative formulas for sine and cosine, together with the quotient rule and the derivative of a reciprocal. The main care required is to keep track of negative signs and of the points where the original functions are undefined.
Throughout, \(x\) is measured in radians. The derivatives of sine and cosine used below are \(\frac{d}{dx}\sin x=\cos x\) and \(\frac{d}{dx}\cos x=-\sin x\). We will derive each of the three formulas rather than treating them as rules to memorize without explanation.
Secant: Differentiate the Reciprocal of Cosine
Because \(\sec x=1/\cos x\), the quotient rule gives its derivative wherever \(\cos x\ne 0\). For differentiable functions \(u\) and \(v\), the quotient rule states that \(\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}\), provided \(v\ne0\). Apply it with \(u(x)=1\) and \(v(x)=\cos x\):
The last line uses \(\sec x=1/\cos x\) and \(\tan x=\sin x/\cos x\). Both sides are defined when \(\cos x\ne0\). Since cosine is zero at \(x=\frac{\pi}{2}+k\pi\), where \(k\in\mathbb Z\), the derivative formula applies on the domain of secant.
Cosecant: The Negative Sign Comes from the Quotient Rule
Cosecant is the reciprocal of sine: \(\csc x=1/\sin x\). In the quotient-rule calculation, the derivative of the denominator is \(\cos x\), not \(-\cos x\). The negative sign in the final formula instead comes from the order \(u'v-uv'\) in the quotient rule.
This calculation requires \(\sin x\ne0\). Sine is zero at \(x=k\pi\), where \(k\in\mathbb Z\), so cosecant and its derivative formula are defined for all real \(x\) except those values.
Cotangent: Apply the Quotient Rule to Cosine over Sine
For cotangent, use \(\cot x=\cos x/\sin x\). Here the numerator has derivative \(-\sin x\), while the denominator has derivative \(\cos x\). Keeping both terms in the correct order gives:
The third line uses the Pythagorean identity \(\sin^2 x+\cos^2 x=1\), and the final line uses \(\csc x=1/\sin x\). The quotient and the resulting formula are defined when \(\sin x\ne0\), or \(x\ne k\pi\) for \(k\in\mathbb Z\).
Why the Formulas Work
The derivations above are instances of one general fact: the derivative of a reciprocal is controlled by the quotient rule. If \(g\) is differentiable and \(g(x)\ne0\), then apply the quotient rule to \(1/g(x)\), treating the numerator as the constant function \(1\):
For secant, take \(g(x)=\cos x\), so \(g'(x)=-\sin x\). The two negative signs cancel, giving \(\sin x/\cos^2x=\sec x\tan x\). For cosecant, take \(g(x)=\sin x\), so \(g'(x)=\cos x\), giving \(-\cos x/\sin^2x=-\csc x\cot x\). This explains both the formulas and why their domains exclude the zeros of the corresponding denominator.
Use \(1/\cos x\), \(1/\sin x\), or \(\cos x/\sin x\).
Use the sine and cosine derivative formulas and preserve the quotient-rule order.
Use reciprocal identities and, for cotangent, \(\sin^2x+\cos^2x=1\). Exclude every input that makes the original denominator zero.
Worked Examples
Worked Example: Evaluate the Derivative of Secant at \(\frac{\pi}{4}\)
Find \(\frac{d}{dx}\sec x\) at \(x=\frac{\pi}{4}\). First confirm that the input is in the domain: \(\cos(\pi/4)=\sqrt2/2\ne0\). Apply the formula and evaluate each factor:
Check directly with the quotient-rule form \(\sin x/\cos^2x\): at \(\pi/4\), this is \((\sqrt2/2)/((\sqrt2/2)^2)=(\sqrt2/2)/(1/2)=\sqrt2\). It matches the result.
Worked Example: Evaluate the Derivative of Cosecant at \(\frac{\pi}{6}\)
Find \(\frac{d}{dx}\csc x\) at \(x=\frac{\pi}{6}\). Since \(\sin(\pi/6)=1/2\ne0\), the function and derivative are defined there. Substitute exact trigonometric values:
Check with the quotient-rule expression \(-\cos x/\sin^2x\): at \(\pi/6\), it equals \(-(\sqrt3/2)/(1/2)^2=-(\sqrt3/2)/(1/4)=-2\sqrt3\), as required.
Worked Example: Evaluate the Derivative of Cotangent at \(\frac{\pi}{4}\)
Find \(\frac{d}{dx}\cot x\) at \(x=\frac{\pi}{4}\). The denominator in \(\cot x=\cos x/\sin x\) is nonzero because \(\sin(\pi/4)=\sqrt2/2\ne0\). Use the derivative formula:
Check using the unsimplified quotient-rule numerator: at \(\pi/4\), it is \(-\sin^2(\pi/4)-\cos^2(\pi/4)=-1/2-1/2=-1\). Dividing by \(\sin^2(\pi/4)=1/2\) gives \(-1/(1/2)=-2\), matching the formula.
Domains at a Glance
A derivative formula cannot make an original function defined at an input where the function itself has a zero denominator. The exclusions below therefore apply to both each function and its derivative formula.
| Function | Derivative | Inputs excluded |
|---|---|---|
| \(\sec x\) | \(\sec x\tan x\) | \(x=\frac{\pi}{2}+k\pi,\ k\in\mathbb Z\) |
| \(\csc x\) | \(-\csc x\cot x\) | \(x=k\pi,\ k\in\mathbb Z\) |
| \(\cot x\) | \(-\csc^2 x\) | \(x=k\pi,\ k\in\mathbb Z\) |
A Common Sign Error
A frequent mistake is to differentiate \(\csc x=1/\sin x\) as though the derivative were positive. For instance, an incorrect calculation might be written as
This misses the subtraction in the quotient rule. The correct numerator is \(0\cdot\sin x-1\cdot\cos x\), so
The error is visible at \(x=\pi/6\): the incorrect expression gives \(2\sqrt3\), while the correct derivative is \(-2\sqrt3\). Keeping the quotient-rule numerator in the written work helps prevent this sign reversal.
Putting the Three Results Together
All three derivatives follow from familiar derivative rules, but their conditions matter just as much as their algebra. Secant excludes zeros of cosine; cosecant and cotangent exclude zeros of sine. At an excluded input, the original function is undefined, so its derivative is not defined there either.
Check Your Understanding
For each question, pay attention to both the calculation and the domain.
- Use the quotient rule to derive \(\frac{d}{dx}\sec x=\sec x\tan x\).
- What sign appears in the derivative of \(\csc x\), and which quotient-rule terms produce it?
- At \(x=\frac{\pi}{3}\), evaluate the derivative of \(\cot x\) exactly.
- Which real inputs must be excluded from the domain of \(\sec x\) and its derivative?
- Explain why \(\cot x\) and its derivative are both undefined at \(x=k\pi\).
