Differentiation of Trigonometric Functions

Neetesh Kumar | February 08, 2025 $\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space$ Share this Page on:

• 1. Definition of Trigonometric Derivatives
• 2. Explanation & Intuition
• 3. Solved Examples
• 4. Conclusion

The derivatives of basic trigonometric functions are:

$\dfrac{d}{dx} (\sin x) = \cos x$

$\dfrac{d}{dx} (\cos x) = -\sin x$

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

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

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

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

These rules help in differentiating trigonometric functions that frequently appear in physics, engineering, and calculus problems.

Example 1: Differentiating a Simple Trigonometric Function

Differentiate $f(x) = \sin x + \cos (e^x)$.

Solution:

Using the basic derivative rules:

$f'(x) = \dfrac{d}{dx} (\sin x) + \dfrac{d}{dx} (\cos (e^x))$

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

Thus, the final answer is: $\boxed{\cos x - e^x.\sin (e^x)}$.

Example 2: Differentiating a Product of a Polynomial and Trigonometric Function

Differentiate $f(x) = x^2 \tan x$.

Solution:

Using the product rule, let:

• $u(x) = x^2$, so $u'(x) = 2x$.
• $v(x) = \tan x$, so $v'(x) = \sec^2 (x)$.

Applying the product rule:

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

Thus, the final answer is: $\boxed{x^2. \sec^2 (x) + 2x. \tan x}$.

Example 3: Differentiating a Trigonometric Function with a Chain Rule

Differentiate $f(x) = \cos(3x^2)$.

Solution:

Using the chain rule, let:

• Outer function: $f(u) = \cos u$, so $f'(u) = -\sin u$.
• Inner function: $g(x) = 3x^2$, so $g'(x) = 6x$.

Applying the chain rule:

$f'(x) = -\sin(3x^2) \cdot 6x$.

Thus, the final answer is: $\boxed{-6x. \sin(3x^2)}$.

Example 4: Velocity in Harmonic Motion

The position of a particle in simple harmonic motion is given by: $s(t) = 5\cos(2t)$.

Find the velocity function $v(t)$.

Solution:

Since velocity is the derivative of position:

$v(t) = \dfrac{ds}{dt} = \dfrac{d}{dt} \big( 5\cos(2t) \big)$.

Using the chain rule:

• Outer function: $f(u) = \cos u$, so $f'(u) = -\sin u$.
• Inner function: $g(t) = 2t$, so $g'(t) = 2$.

Applying the chain rule:

$v(t) = 5 (-\sin(2t)) \cdot 2$.

$v(t) = -10 \sin(2t)$.

Thus, the velocity function is: $\boxed{v(t) = -10 \sin(2t)}$.