Derivatives of the Reciprocal Trigonometric Functions

You've differentiated sine, cosine, and tangent. Their three reciprocals — \(\sec x\), \(\csc x\), and \(\cot x\) — have derivatives of their own, and you don't have to take them on faith: the quotient rule manufactures all three.

By the end you'll be able to differentiate any expression built from \(\sec x\), \(\csc x\), and \(\cot x\), and derive each formula yourself with one pass of the quotient rule.

Predict: where does \(\sec x\) have a horizontal tangent (slope 0)? Its slope is \(\sec x \tan x\), and \(\sec x\) is never zero — so drag the point and check: the only flat spots are where \(\tan x = 0\), that is \(x = 0\) and \(x = \pm\pi\).

The curve is the chosen reciprocal function; the dashed vertical lines are its asymptotes (where the function it's 1-over hits zero). Drag the point (or use the slider / arrow keys) and watch the tangent's live slope match the derivative formula — the same number, computed two ways. Hover the curve for exact values.

Function

d/dx [sec x] = sec x · tan x

x = 0.60 · sec x = 1.212 · tangent slope = sec x · tan x = 0.829

Curve, draggable tangent point, and asymptotes — the tangent's slope always equals the formula's value
function tangent (slope = formula) vertical asymptote

The derivatives of the reciprocal trigonometric functions, each obtainable via the quotient rule, are \(\tfrac{d}{dx}[\cot x] = -\csc^2 x\), \(\tfrac{d}{dx}[\sec x] = \sec x \tan x\), and \(\tfrac{d}{dx}[\csc x] = -\csc x \cot x\).

Formal

None of these are new axioms — each is the quotient rule applied to a ratio you already know how to differentiate: \(\sec x = \tfrac{1}{\cos x}\), \(\csc x = \tfrac{1}{\sin x}\), \(\cot x = \tfrac{\cos x}{\sin x}\). Take \(\sec x\): \[\frac{d}{dx}\left[\frac{1}{\cos x}\right] = \frac{0\cdot\cos x - 1\cdot(-\sin x)}{\cos^2 x} = \frac{\sin x}{\cos^2 x} = \frac{1}{\cos x}\cdot\frac{\sin x}{\cos x} = \sec x \tan x.\] That last step — splitting the fraction into \(\sec x\) times \(\tan x\) — is why the answer comes out in reciprocal-function notation. The other two derivations run the same way, and a sign pattern falls out: every "co-" function (cosine, cotangent, cosecant) picks up a negative in its derivative, while sine, tangent, and secant stay positive. One caution: like all trig derivatives, these hold only when \(x\) is measured in radians.

Here are all three, with the tangent-family rules from the previous lesson for contrast — keep \(\tfrac{d}{dx}[\sec x] = \sec x \tan x\) firmly separate from \(\tfrac{d}{dx}[\tan x] = \sec^2 x\); mixing those two is the classic error:

  • \(\tfrac{d}{dx}[\sec x] = \sec x \tan x\)
  • \(\tfrac{d}{dx}[\csc x] = -\csc x \cot x\)
  • \(\tfrac{d}{dx}[\cot x] = -\csc^2 x\)
Worked example

Differentiate \(y = \sec x + 2\cot x\). Term by term: \(\tfrac{d}{dx}[\sec x] = \sec x \tan x\), and \(\tfrac{d}{dx}[2\cot x] = 2\cdot(-\csc^2 x) = -2\csc^2 x\). So \[y' = \sec x \tan x - 2\csc^2 x.\] Two sign checks: sec (no "co-") stays positive; cot (a "co-" function) went negative.

Your turn

Differentiate \(y = 4\sec x - \csc x\). First term: \(\tfrac{d}{dx}[4\sec x] = 4\sec x \tan x\). Now finish it: \(\tfrac{d}{dx}[-\csc x] = \) ____, so \(y' = 4\sec x \tan x \) ____.

Reveal the answer

\(\tfrac{d}{dx}[\csc x] = -\csc x \cot x\), so the leading minus flips it: \(\tfrac{d}{dx}[-\csc x] = +\csc x \cot x\). Altogether \(y' = 4\sec x \tan x + \csc x \cot x\). If you got a minus on the second term, you either dropped the csc rule's built-in negative or forgot the minus already in front of \(\csc x\) — two negatives cancel.

More info — deriving cot x, and where the Pythagorean identity sneaks in

The \(\cot x\) derivation shows the quotient rule and the Pythagorean identity working together. With top \(= \cos x\) and bottom \(= \sin x\): \[\frac{d}{dx}\left[\frac{\cos x}{\sin x}\right] = \frac{(-\sin x)(\sin x) - (\cos x)(\cos x)}{\sin^2 x} = \frac{-(\sin^2 x + \cos^2 x)}{\sin^2 x} = \frac{-1}{\sin^2 x} = -\csc^2 x.\] The identity collapses the messy numerator to \(-1\) — the same move that turned the quotient-rule computation of \(\tan x\) into \(\sec^2 x\) in the last lesson. The \(\csc x\) derivation is the mirror image of the \(\sec x\) one shown above. For every step spelled out, see the Paul's Online Notes link under Dive deeper.

Check your understanding

Question 1 of 4

Differentiate \(y = 3\csc x - \sec x\).

Question 2 of 4

Which of the six trig derivatives carry a negative sign — and what's the pattern that makes this easy to remember?

Question 3 of 4

At which \(x\)-values does \(y = \sec x\) have a horizontal tangent (slope 0)?

Question 4 of 4

Derive \(\tfrac{d}{dx}[\csc x]\) by writing \(\csc x = \tfrac{1}{\sin x}\) and applying the quotient rule. What do you get?

Recap

  • \(\tfrac{d}{dx}[\sec x] = \sec x \tan x\), \(\tfrac{d}{dx}[\csc x] = -\csc x \cot x\), \(\tfrac{d}{dx}[\cot x] = -\csc^2 x\).
  • Sign pattern: every "co-" function (cos, cot, csc) gets a negative derivative; sin, tan, sec stay positive.
  • All three come from one pass of the quotient rule on \(1/\cos x\), \(1/\sin x\), and \(\cos x/\sin x\) — no new axioms.
  • Don't confuse \(\tfrac{d}{dx}[\sec x] = \sec x \tan x\) with \(\tfrac{d}{dx}[\tan x] = \sec^2 x\), and remember: radians only.

Dive deeper

Sources

  • Derivatives of the Reciprocal Trigonometric Functions