Derivatives of Trig Functions

The derivatives of trig functions are: the derivative of sin x is cos x, of cos x is −sin x, and of tan x is sec²x. The other three are \frac{d}{dx}\sec x = \sec x\tan x, \frac{d}{dx}\csc x = -\csc x\cot x and \frac{d}{dx}\cot x = -\csc^2 x. These formulas work only when x is in radians.

Before this lesson: Chain Rule, Quotient Rule, Unit Circle and Squeeze Theorem

Why the derivative of sine is cosine

Look at the graph of y = sin x. At x = 0 it climbs with slope 1. At x = \frac{\pi}{2} it is at its peak, so its slope is 0. At x = π it falls with slope −1. Those slopes, 1, 0 and −1, are exactly the values of cos x at the same points.

The proof uses the limit \lim_{h \to 0} \frac{\sin h}{h} = 1, which comes from the squeeze theorem. That limit is true only when angles are measured in radians, so calculus with trig functions always uses radians.

Trig derivative formulas

A memory trick: the three “co” functions (cosine, cosecant, cotangent) have derivatives with a minus sign. Each one is the matching formula with every function swapped for its co-function and a minus sign added.

With the chain rule, an inside function u adds a factor of u′:

\frac{d}{dx}\left[\sin u\right] = \cos u \cdot u', \qquad \frac{d}{dx}\left[\cos u\right] = -\sin u \cdot u', \qquad \frac{d}{dx}\left[\tan u\right] = \sec^2 u \cdot u'

Derivatives of trig functions examples

Example 1: A sum of trig functions

Differentiate y = 3\sin x - 2\cos x.

y' = 3\cos x - 2(-\sin x)

y' = 3\cos x + 2\sin x

Answer: y' = 3\cos x + 2\sin x.

Example 2: Product rule with sine

Differentiate y = x^2\sin x.

y' = (2x)(\sin x) + (x^2)(\cos x)

y' = 2x\sin x + x^2\cos x

Answer: y' = 2x\sin x + x^2\cos x.

Example 3: Chain rule with sine

Differentiate y = \sin(3x^2).

u = 3x^2, \; u' = 6x

y' = \cos(3x^2) \cdot 6x

y' = 6x\cos(3x^2)

Answer: y' = 6x\cos(3x^2). A common slip is writing \cos(6x). The inside stays 3x².

Example 4: Tangent line to y = tan x

Find the tangent line to y = \tan x at x = \frac{\pi}{4}.

y = \tan\frac{\pi}{4} = 1

y' = \sec^2 x

\sec\frac{\pi}{4} = \frac{1}{\cos\frac{\pi}{4}} = \sqrt{2}

y'\left(\frac{\pi}{4}\right) = (\sqrt{2})^2 = 2

y - 1 = 2\left(x - \frac{\pi}{4}\right)

y = 2x - \frac{\pi}{2} + 1

Answer: y = 2x - \frac{\pi}{2} + 1, or about y = 2x − 0.571.

Example 5: Word problem with a spring

A weight on a spring moves up and down. Its position is s(t) = 4\sin(2t) cm after t seconds. Find its velocity and acceleration at t = \frac{\pi}{6}.

v(t) = 4\cos(2t) \cdot 2 = 8\cos(2t)

a(t) = 8(-\sin(2t)) \cdot 2 = -16\sin(2t)

v\left(\frac{\pi}{6}\right) = 8\cos\frac{\pi}{3} = 8 \cdot \frac{1}{2} = 4

a\left(\frac{\pi}{6}\right) = -16\sin\frac{\pi}{3} = -16 \cdot \frac{\sqrt{3}}{2}

a\left(\frac{\pi}{6}\right) = -8\sqrt{3} \approx -13.86

Answer: at t = \frac{\pi}{6} s, the weight moves up at 4 cm/s, and its acceleration is about −13.86 cm/s². See position, velocity and acceleration for what these signs mean.

Common mistakes

Related methods

Getting the other four from sine and cosine

Once you know the derivatives of sin x and cos x, the quotient rule gives the rest. For tan x:

\frac{d}{dx}\left[\frac{\sin x}{\cos x}\right] = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x

The step \cos^2 x + \sin^2 x = 1 is the Pythagorean identity (see fundamental trig identities). The same method works for sec x, csc x and cot x.

The cycle of derivatives

Differentiating sin x over and over gives a cycle of four: sin x, cos x, −sin x, −cos x, then back to sin x. So the 4th derivative of sin x is sin x, and the 10th derivative is the same as the 2nd, −sin x (see higher-order derivatives).

Practice problems

1. Differentiate y = 5\cos x + x^3.
2. Differentiate y = \sin x\cos x.
3. Differentiate y = x\tan x.
4. Differentiate y = \sin(4x).
5. Differentiate y = \cos(x^2 + 1).
6. Differentiate y = \sin^2 x.
7. Differentiate y = \sec(3x).
8. Differentiate y = \frac{\sin x}{x}.
9. Find every x in [0, 2\pi] where f(x) = 2\sin x + x has a horizontal tangent.
10. Find the tangent line to y = \cos x at x = \frac{\pi}{3}.

y' = -5\sin x + 3x^2

y' = \cos x \cdot \cos x + \sin x \cdot (-\sin x)

y' = \cos^2 x - \sin^2 x

y' = 1 \cdot \tan x + x\sec^2 x

y' = \cos(4x) \cdot 4

y' = -\sin(x^2 + 1) \cdot 2x

y = (\sin x)^2

y' = 2\sin x \cdot \cos x

y' = \sec(3x)\tan(3x) \cdot 3

y' = \frac{\cos x \cdot x - \sin x \cdot 1}{x^2}

f'(x) = 2\cos x + 1

2\cos x + 1 = 0

\cos x = -\frac{1}{2}

x = \frac{2\pi}{3} \text{ or } x = \frac{4\pi}{3}

y = \cos\frac{\pi}{3} = \frac{1}{2}

y' = -\sin x

y'\left(\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}

y - \frac{1}{2} = -\frac{\sqrt{3}}{2}\left(x - \frac{\pi}{3}\right)

Derivatives of trig functions FAQ

What is the derivative of sin x?

The derivative of sin x is cos x, when x is in radians. With an inside function, \frac{d}{dx}\sin u = \cos u \cdot u'.

Why is the derivative of cos x negative?

Right after x = 0, the cosine graph falls from its peak of 1, so its slope is negative. The formula -\sin x matches: it is negative for x between 0 and π.

Why do trig derivatives need radians?

The key limit \frac{\sin h}{h} \to 1 is true only in radians. In degrees the derivative of sin x would be \frac{\pi}{180}\cos x, which is messy.

How do you remember the six trig derivatives?

Learn sin → cos, tan → sec² and sec → sec tan. For the co-functions, swap each function for its co-function and add a minus sign.

Summary

• (\sin x)' = \cos x and (\cos x)' = -\sin x.
• (\tan x)' = \sec^2 x, (\sec x)' = \sec x\tan x, (\csc x)' = -\csc x\cot x, (\cot x)' = -\csc^2 x.
• The co-functions have derivatives with a minus sign.
• Use the chain rule for inside functions: (\sin u)' = \cos u \cdot u'.
• Always work in radians.

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