Derivatives of Trigonometric Functions

Why Trigonometric Derivatives Use Radians

Trigonometric functions use the same sum, product, quotient, and chain rules as algebraic functions. What changes is the set of basic derivatives we start from.

The formulas below assume that angles are measured in radians. If an angle is measured in degrees, the chain rule introduces the additional factor $\pi/180$.

Basic Trigonometric Derivatives

The sine and cosine formulas follow from the limit definition of the derivative. The remaining four can then be derived from trigonometric identities together with the product or quotient rule.

These are the basic derivatives of the six trigonometric functions:

1. Derivative of Sine:

   $$f(x) = \sin x \implies f'(x) = \cos x$$

2. Derivative of Cosine:

   $$f(x) = \cos x \implies f'(x) = -\sin x$$

3. Derivative of Tangent:

   $$f(x) = \tan x \implies f'(x) = \sec^2 x$$

4. Derivative of Cotangent:

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

5. Derivative of Secant:

   $$\begin{aligned} f(x) &= \sec x \\ &\implies f'(x) \\ &= \sec x \tan x \end{aligned}$$

6. Derivative of Cosecant:

   $$\begin{aligned} f(x) &= \csc x \\ &\implies f'(x) \\ &= -\csc x \cot x \end{aligned}$$

Applying the Rules to Trigonometric Functions

The examples below combine these formulas with familiar derivative rules. Identify the outer rule first, then differentiate the angle and multiply, because the derivative of $\sin u$ is $\cos u$ multiplied by the derivative of $u$.

Combination of Algebra and Trigonometry

Find the derivative of $y = 2 \sin x + 5x$.

Solution:

We can differentiate this function term by term using the sum rule.

Using the Product Rule

Find the derivative of $y = 2 \sin x \tan x$.

Solution:

Use the product rule $y' = u'v + uv'$.

Let $u = 2 \sin x$ and $v = \tan x$.

Then $u' = 2 \cos x$ and $v' = \sec^2 x$.

Using the Quotient Rule

On its domain $\sin x \ne 0$, find the derivative of $y = \frac{1 + \cos x}{\sin x}$.

Solution:

Use the quotient rule $y' = \frac{u'v - uv'}{v^2}$.

Let $u = 1 + \cos x$ and $v = \sin x$.

Then $u' = -\sin x$ and $v' = \cos x$.

Compute the numerator first:

Use the Pythagorean identity:

Factor the expression:

The simplified formula is valid only on the original domain, so values with $\sin x=0$ remain excluded even if the final expression appears defined there.

Exercises

Each problem asks for the first derivative of a trigonometric expression, and the second one is a product. Choose the rule that matches the structure.

1. Find the first derivative of $f(x) = 4x^3 - 5 \cos x$.
2. Find the first derivative of $f(x) = \sin x \cos x$.

Answer Key

1. Solution:

   Use the subtraction rule to differentiate each term separately.

   Step 1: Differentiate the first term

   The derivative of $4x^3$ using the power rule is $3 \cdot 4x^{3-1} = 12x^2$.

   Step 2: Differentiate the second term

   The derivative of $-5 \cos x$ is $-5(-\sin x) = 5 \sin x$.

   Step 3: Combine the results

   $$f'(x) = 12x^2 + 5 \sin x$$

   So, the derivative of the function is $12x^2 + 5 \sin x$.

2. Solution:

   Use the product rule, $f'(x) = u'v + uv'$.

   Step 1: Determine $u$, $v$, $u'$, and $v'$

   Let $u = \sin x$ and $v = \cos x$.

   Then, $u' = \cos x$ and $v' = -\sin x$.

   Step 2: Apply the Product Rule

   $$\begin{aligned} f'(x) &= (\cos x)(\cos x) \\ &\quad{}+ (\sin x)(-\sin x) \end{aligned}$$

   $$f'(x) = \cos^2 x - \sin^2 x$$

   Step 3: Use a Trigonometric Identity (Optional)

   The result can also be simplified using the double angle identity, $\cos(2x) = \cos^2 x - \sin^2 x$.

   $$f'(x) = \cos(2x)$$

   So, the derivative of the function is $\cos(2x)$.