Finding a derivative of a function is an important concept of calculus. The derivative of a function is defined as follows:

"A derivative is an instantaneous rate of change of a function at a given point"

The process of finding derivatives is known as differentiation. The two types of functions that are generally differentiated are explicit and implicit functions. In this article, we will study how to differentiate explicit trigonometric functions. We use differentiation rules to find the derivatives of explicit functions. Some of the common rules of differentiation are constant rule, sum rule, difference rule, power rule, product rule, quotient rule (used for fractions), and chain rule.

Before proceeding to the relevant formulae and examples, let us find out what are the trigonometric functions first.

What are Trigonometric Functions?

Trigonometry is an important branch of mathematics in which we study the relationship between sides and angles of the triangle. Trigonometric functions tell us how the angles of a triangle are related to its sides. The other names for trig functions are circular functions, goniometric, and angle functions. However, the word "trigonometric" is widely used to describe these kinds of functions. The most common trigonometric functions are sine, cosine, tangent, cotangent, co-secant, and secant. The inverse functions of trig functions are known as inverse trigonometric functions. Inverse trig functions are also known as arcus functions, cyclometric functions, or anti trigonometric functions.

Before proceeding to examples let us find out the derivatives of common trigonometric functions. It is always the best idea to memorize them as they are used to differentiate the functions that involve trigonometric functions.

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Derivative of the Sine Function

f(x) = sin u

f '(x) = u' \cdot cos u

Derivative of the Cosine Function

f(x) = cos u

f ' (x) = - u' \cdot sin u

Derivative of the Tangent Function

f(x) = tan u

f ' (x) = \frac {u'} {cos ^2 u} = u' \cdot sec ^2 u = u' \cdot (1 + tan ^2 u)

Derivative of the Cotangent Function

f(x) = cot u

f '(x) = -\frac {u'} {sin ^2 u } = -u' \cdot csc ^2 u

It is also equal to  f ' (x) = -u' \cdot (1 + cot ^2 u)

Derivative of the Secant Function

f(x) = sec u

f '(x) = \frac {u' \cdot sin u} {cos ^2 u} = u' \cdot sec u \cdot tan u

Derivative of the Cosecant Function

f(x) = csc u

f ' (x) = -\frac {u' \cdot cos u} {sin ^2 u} = - u' \cdot csc u \cdot cot u

Examples of Derivatives of Trigonometric Functions

Example 1

Compute the derivative of f(x) = sin 4x

Solution

Let u = 4x

We know that if f(x) = sin u, then f ' (x) = u' \cdot cos u.It means that we will multiply the derivative of the inner function u with the derivative of the outer function sin.

In this example, u = 4x. Therefore, u' = 4. We will substitute it in the above formula to get the final answer:

f ' (x) = 4 \cdot cos u

Since u = 4x, hence f ' (x) = 4 cos 4x.

Example 2

Differentiate f(x) = sin x ^4

Solution

Suppose x ^4 = u.

f(x) = sin u

We know that if f(x) = sin u, then f ' (x) = u' cos u.

Here, u = x ^4. We will apply power rule to find u '.

u' = 4 x ^ 3

f ' (x) = u'cos u

f ' (x) = 4x ^3 cos x ^4

 

Example 3

Differentiate y = sin ^ 3 \sqrt {x}

Solution

This is a complex trig function. We will use chain rule again and again to find the derivative of this function.

y ' = 3 sin ^2 (\sqrt {x}) \cdot \frac {d} {d (x)} sin (\sqrt {x})

Remember that sin \sqrt {x} = \sqrt {x}' \cdot cos \sqrt {x}.

= 3 sin ^2 (\sqrt {x}) cos \sqrt {x} \frac {d} {d(x)} \sqrt {x}

= 3 sin ^2 (\sqrt {x}) cos \sqrt {x} \cdot \frac {1} {2} x ^ {\frac{1} {2} - 1}

= 3 sin ^2 (\sqrt {x}) cos \sqrt {x} \cdot \frac {1} {2} x ^ {-\frac {1} {2}}

= \frac {3 sin ^2 (\sqrt {x}) cos \sqrt {x}} {2 \sqrt {x}}

 

Example 4

Differentiate f(x) = \frac {cos x} {5}

Solution

This function can be written as f(x) = \frac {1} {5} cos x

We know that the derivative of a cosine function is - u' sin u. In the above function, u = \frac {1} {5} x, therefore u' = \frac{1} {5}

y' = -\frac {1} {5} sin x

Example 5

Differentiate f(x) = cos (3x ^2 + x - 1)

Solution

Suppose u = 3x ^2 + x - 1.

We know that if f(x) = cos u, then f ' (x) = u' sin u. It means that the derivative of the inner function will be multiplied with the derivative of the outer function cosine.

If u = 3x ^2 + x - 1, then u' = 6x + 1.

f ' (x) = - (6x + 1) sin (3x ^2 + x - 1)

Example 6

Differentiate f(x) = tan \sqrt {x}

Solution

Suppose \sqrt {x} = u.

We know that if f(x) = tan u, then f ' (x) = \frac {u'} {cos ^2 u}

Here, u = \sqrt {x}. Hence, u' = \frac {1} {2 \sqrt {x}}.

f '(x) = \frac {1} {2 \sqrt {x}} \cdot \frac {1} {cos ^2 \sqrt{x}}

Example 7

Differentiate f(x) = cot 4 x ^2

Solution

Suppose 4x ^2 = u.

u' = 8x

We know that if f(x) = cot u, then f ' (x) =- \frac{u'} {sin ^2 u}.

f ' (x) = -\frac {8x} {sin ^2 4x ^2}

Example 8

Differentiate f(x) = sec 5x

Solution

Suppose u = 5x

u' = 5

We know that if f(x) = sec u, then f ' (x) = \frac {u' sin u} {cos ^2 u}. Now, we will substitute u = 5x and u' = 5 to get the derivative of the function.

f ' (x) = \frac {5 \cdot sin 5x} {cos ^2 5x}

Example 9

Differentiate f(x) = csc (\frac {x} {2} )

Solution

Suppose \frac {x} {2} = u.

u' = \frac {1} {2}

We know that if f(x) = csc u, then f '(x) = - \frac {cos u} {sin ^2 u}

The substitution of u = \frac {x} {2} and u' = \frac {1} {2} will get us the final answer of this question.

f ' (x) = - \frac {cos (\frac{x}{2})} { 2 \cdot sin ^2 \frac {x} {2}}

 

Example 10

Differentiate f(x) = sin ^4 x - cos ^4 x.

Solution

This function involves two trigonometric functions, sine, and cosine. To differentiate this function, we will use the derivative difference rule and derivative power rule.

= \frac {d} {dx} sin ^4 x - \frac {d} {dx} cos ^4 x

= 4 sin ^3 x \frac {d} {dx} sin x - 4 cos ^3 x \frac {d} {dx} cos x

= 4 sin ^3 x cos x - 4 cos ^3 x (- sin x)

= 4 sin ^3 x cos x + 4 cos ^3 x sin x

We will simplify and rewrite it as:

= 4 cos x sin x (cos ^2 x sin ^2 x)

 

Example 11

Differentiate f(x) = cos x + \frac {1} {4} tan ^4 x

Solution

We will use the derivative sum rule to differentiate the above function.

= cos x' + \frac {1}{4} tan ^4 x '

= - sin x + \frac {1}{4} \cdot 4 tan ^3 x \cdot (tan x) '

After simplification, we will get the following derivative of the function.

= - sin x + tan ^3 x \cdot \frac {1} {cos ^2 x}

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Rafia Shabbir