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Derivatives of Trigonometric Functions — Definition, Formula & Examples

Derivatives of trigonometric functions are the standard results that tell you the rate of change of each trig function with respect to its input. For example, the derivative of sin x is cos x, and the derivative of cos x is −sin x.

For each of the six standard trigonometric functions, the derivative is obtained from the limit definition of the derivative and can be expressed as a closed-form trigonometric or rational-trigonometric expression: ddxsin⁡x=cos⁡x\frac{d}{dx}\sin x = \cos x, ddxcos⁡x=−sin⁡x\frac{d}{dx}\cos x = -\sin x, ddxtan⁡x=sec⁡2x\frac{d}{dx}\tan x = \sec^2 x, ddxcot⁡x=−csc⁡2x\frac{d}{dx}\cot x = -\csc^2 x, ddxsec⁡x=sec⁡xtan⁡x\frac{d}{dx}\sec x = \sec x \tan x, and ddxcsc⁡x=−csc⁡xcot⁡x\frac{d}{dx}\csc x = -\csc x \cot x, each valid on the domain where the original function is defined.

Key Formula

ddxsin⁡x=cos⁡xddxcos⁡x=−sin⁡xddxtan⁡x=sec⁡2x\frac{d}{dx}\sin x = \cos x \qquad \frac{d}{dx}\cos x = -\sin x \qquad \frac{d}{dx}\tan x = \sec^2 x
Where:
  • xx = The variable (angle in radians) with respect to which you differentiate

How It Works

You apply these formulas whenever a trig function appears inside an expression you need to differentiate. If the argument is just xx, substitute the result directly. When the argument is more complex — say sin⁡(3x2)\sin(3x^2) — you also need the chain rule: differentiate the outer trig function first, then multiply by the derivative of the inner function. Memorizing the six basic derivatives lets you handle nearly every trig differentiation problem quickly. A helpful pattern: the derivatives of the "co-" functions (cos, cot, csc) all carry a negative sign.

Worked Example

Problem: Find the derivative of f(x) = 3sin(x) + x²cos(x).
Step 1: Differentiate the first term using the constant multiple rule and the derivative of sin x.
ddx[3sin⁡x]=3cos⁡x\frac{d}{dx}[3\sin x] = 3\cos x
Step 2: Differentiate the second term using the product rule: if u = x² and v = cos x, then (uv)' = u'v + uv'.
ddx[x2cos⁡x]=2xcos⁡x+x2(−sin⁡x)\frac{d}{dx}[x^2 \cos x] = 2x\cos x + x^2(-\sin x)
Step 3: Combine the results.
f′(x)=3cos⁡x+2xcos⁡x−x2sin⁡xf'(x) = 3\cos x + 2x\cos x - x^2\sin x
Answer: f′(x)=3cos⁡x+2xcos⁡x−x2sin⁡xf'(x) = 3\cos x + 2x\cos x - x^2\sin x

Another Example

Problem: Find the derivative of g(x) = tan(5x).
Step 1: Recognize that this requires the chain rule. The outer function is tan(u) and the inner function is u = 5x.
ddxtan⁡(u)=sec⁡2(u)⋅dudx\frac{d}{dx}\tan(u) = \sec^2(u) \cdot \frac{du}{dx}
Step 2: Compute the inner derivative and multiply.
g′(x)=sec⁡2(5x)⋅5=5sec⁡2(5x)g'(x) = \sec^2(5x) \cdot 5 = 5\sec^2(5x)
Answer: g′(x)=5sec⁡2(5x)g'(x) = 5\sec^2(5x)

Visualization

Why It Matters

These derivatives appear throughout AP Calculus AB and BC — in related-rates problems, optimization, and integration by parts. Physics and engineering courses rely on them constantly because sine and cosine model oscillations, waves, and circular motion. Mastering these six formulas is a prerequisite for nearly every advanced calculus technique.

Common Mistakes

Mistake: Forgetting the negative sign in the derivative of cos x and writing ddxcos⁡x=sin⁡x\frac{d}{dx}\cos x = \sin x.
Correction: The correct derivative is ddxcos⁡x=−sin⁡x\frac{d}{dx}\cos x = -\sin x. Remember: all three "co-" functions (cos, cot, csc) have a negative sign in their derivatives.
Mistake: Omitting the chain rule when the argument is not simply x, for example writing the derivative of sin⁡(4x)\sin(4x) as just cos⁡(4x)\cos(4x).
Correction: You must multiply by the derivative of the inner function: ddxsin⁡(4x)=cos⁡(4x)⋅4=4cos⁡(4x)\frac{d}{dx}\sin(4x) = \cos(4x) \cdot 4 = 4\cos(4x).

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