Derivatives of Tangent, Cotangent, Secant, and Cosecant

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Key Ideas Easy Medium Hard

\((\tan x)' = \sec^2 x,\quad (\sec x)' = \sec x\tan x\)

cot and csc each bring a minus sign

What to Remember

Four more derivatives

\(\dfrac{d}{dx}\tan x = \sec^2 x\)

\(\dfrac{d}{dx}\cot x = -\csc^2 x\)

\(\dfrac{d}{dx}\sec x = \sec x\tan x\)

\(\dfrac{d}{dx}\csc x = -\csc x\cot x\)

The two minus signs

Cotangent and cosecant both carry a minus sign. Secant does not.

Build tan from the quotient rule

\(\tan x = \dfrac{\sin x}{\cos x}\) produces \(\sec^2 x\) after the numerator simplifies to 1.

Special angles

\(\sec\!\bigl(\tfrac{\pi}{3}\bigr) = 2\), \(\tan\!\bigl(\tfrac{\pi}{3}\bigr) = \sqrt{3}\), and \(\csc\!\bigl(\tfrac{\pi}{6}\bigr) = 2\).

Level 1

Easy

Name the derivative. Do not evaluate yet.

Example

Find \(f'(x)\) for \(f(x) = \tan x\), starting from \(\tan x = \dfrac{\sin x}{\cos x}\).

\(\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}\)

\(u = \sin x,\quad u' = \cos x\)

\(v = \cos x,\quad v' = -\sin x\)

\(\dfrac{\cos x\cos x - \sin x(-\sin x)}{\cos^2 x} = \dfrac{\cos^2 x + \sin^2 x}{\cos^2 x}\)

\(\dfrac{1}{\cos^2 x} = \sec^2 x\)

Answer: \(\sec^2 x\)

Find \(f'(x)\) for \(f(x) = 4\tan x\).

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\(\dfrac{d}{dx}\tan x = \sec^2 x\)

\(4\sec^2 x\)

Answer: \(4\sec^2 x\)

Find \(f'(x)\) for \(f(x) = \cot x\).

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\(-\csc^2 x\)

Answer: \(-\csc^2 x\)

Find \(f'(x)\) for \(f(x) = \sec x\).

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\(\sec x\tan x\)

Answer: \(\sec x\tan x\)

Find \(f'(x)\) for \(f(x) = \csc x\).

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\(-\csc x\cot x\)

Answer: \(-\csc x\cot x\)

Find \(f'(x)\) for \(f(x) = 3\tan x - \sec x\).

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\(\dfrac{d}{dx}(3\tan x) = 3\sec^2 x\)

\(\dfrac{d}{dx}(-\sec x) = -\sec x\tan x\)

\(3\sec^2 x - \sec x\tan x\)

Answer: \(3\sec^2 x - \sec x\tan x\)

Level 2

Medium

Substitute a special angle after differentiating.

Example

For \(f(x) = \tan x\), find \(f'\!\bigl(\tfrac{\pi}{4}\bigr)\).

\(f'(x) = \sec^2 x\)

\(\sec\!\bigl(\tfrac{\pi}{4}\bigr) = \sqrt{2}\)

\(\bigl(\sqrt{2}\bigr)^2 = 2\)

Answer: \(2\)

For \(f(x) = \sec x\), find \(f'\!\bigl(\tfrac{\pi}{3}\bigr)\).

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\(f'(x) = \sec x\tan x\)

\(\sec\!\bigl(\tfrac{\pi}{3}\bigr) = 2\)

\(\tan\!\bigl(\tfrac{\pi}{3}\bigr) = \sqrt{3}\)

\(2\sqrt{3}\)

Answer: \(2\sqrt{3}\)

For \(f(x) = \cot x\), find \(f'\!\bigl(\tfrac{\pi}{6}\bigr)\).

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\(f'(x) = -\csc^2 x\)

\(\csc\!\bigl(\tfrac{\pi}{6}\bigr) = 2\)

\(-2^2 = -4\)

Answer: \(-4\)

For \(f(x) = \csc x\), find \(f'\!\bigl(\tfrac{\pi}{4}\bigr)\).

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\(f'(x) = -\csc x\cot x\)

\(\csc\!\bigl(\tfrac{\pi}{4}\bigr) = \sqrt{2}\)

\(\cot\!\bigl(\tfrac{\pi}{4}\bigr) = 1\)

\(-\sqrt{2}\cdot 1 = -\sqrt{2}\)

Answer: \(-\sqrt{2}\)

For \(f(x) = x + \tan x\), find \(f'\!\bigl(\tfrac{\pi}{4}\bigr)\).

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\(f'(x) = 1 + \sec^2 x\)

\(1 + \sec^2\!\bigl(\tfrac{\pi}{4}\bigr) = 1 + 2 = 3\)

Answer: \(3\)

Which expression is the derivative of \(\cot x + \sec x\)?

A. \(-\csc^2 x + \sec x\tan x\)

B. \(\csc^2 x + \sec x\tan x\)

C. \(-\csc^2 x - \sec x\tan x\)

D. \(-\sec^2 x + \sec x\tan x\)

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\(\dfrac{d}{dx}\cot x = -\csc^2 x\)

\(\dfrac{d}{dx}\sec x = \sec x\tan x\)

Answer: A

Level 3

Hard

A product that uses one of these derivatives.

Example

Find \(f'(x)\) for \(f(x) = x\sec x\).

\((uv)' = u'v + uv'\)

\(u = x,\quad u' = 1\)

\(v = \sec x,\quad v' = \sec x\tan x\)

\(f'(x) = \sec x + x\sec x\tan x\)

Answer: \(\sec x + x\sec x\tan x\)

Find \(f'(x)\) for \(f(x) = x\cot x\).

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\(u = x,\quad u' = 1\)

\(v = \cot x,\quad v' = -\csc^2 x\)

\(f'(x) = \cot x + x(-\csc^2 x)\)

\(\cot x - x\csc^2 x\)

Answer: \(\cot x - x\csc^2 x\)

For \(f(x) = e^x\tan x\), find \(f'(0)\).

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\(u = e^x,\quad u' = e^x\)

\(v = \tan x,\quad v' = \sec^2 x\)

\(f'(x) = e^x\tan x + e^x\sec^2 x\)

Substitute \(x = 0\).

\(e^0\tan 0 + e^0\sec^2 0 = 0 + 1 = 1\)

Answer: \(1\)

Find \(f'(x)\) for \(f(x) = \dfrac{\sec x}{x}\).

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\(u = \sec x,\quad u' = \sec x\tan x\)

\(v = x,\quad v' = 1\)

\(f'(x) = \dfrac{x\sec x\tan x - \sec x}{x^2}\)

\(\dfrac{\sec x\,(x\tan x - 1)}{x^2}\)

Answer: \(\dfrac{\sec x(x\tan x - 1)}{x^2}\)

For \(f(x) = \csc x\), find \(f'\!\bigl(\tfrac{\pi}{6}\bigr)\).

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\(f'(x) = -\csc x\cot x\)

\(\csc\!\bigl(\tfrac{\pi}{6}\bigr) = 2\)

\(\cot\!\bigl(\tfrac{\pi}{6}\bigr) = \sqrt{3}\)

\(-2\sqrt{3}\)

Answer: \(-2\sqrt{3}\)

Derive \(\dfrac{d}{dx}\sec x\) from \(\sec x = \dfrac{1}{\cos x}\).

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\(u = 1,\quad u' = 0\)

\(v = \cos x,\quad v' = -\sin x\)

\(\dfrac{0\cdot\cos x - 1\cdot(-\sin x)}{\cos^2 x} = \dfrac{\sin x}{\cos^2 x}\)

\(\dfrac{1}{\cos x}\cdot\dfrac{\sin x}{\cos x} = \sec x\tan x\)

Answer: \(\sec x\tan x\)

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