Derivative of Trigonometric Function

The Rules of Trigonometric Functions

Finding the derivative of a trigonometric function is not much different from an algebraic function. We still use the familiar derivative properties, such as the product and quotient rules.

However, before that, we need to know the basic derivatives of the main trigonometric functions like sine and cosine. These basic rules are the foundation for solving more complex derivatives.

Basic Trigonometric Derivatives

Just like the derivative of a power function, the derivative for a trigonometric function also has its basic pattern. The derivatives for sine and cosine, for example, can be proven directly from the limit definition of a derivative.

Here are the basic derivatives of the six trigonometric functions that we need to memorize:

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:

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

6. Derivative of Cosecant:

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

Applying the Rules to Trigonometric Functions

Now let's see how to apply these rules in a few examples.

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

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 $v' = \cos x$.

Exercises

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

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

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

   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 $\cos(2x)$.