# Differentiation - Trigonometry

Updated: Jun 22, 2021

__In this note, you will learn:__

· __Differentiation of sin(x), cos(x) and tan(x)__

· __Differentiation of sinn____f(x), cosn____f(x), tann____f(x) with the application of Chain Rule__

__Differentiation of sin(x), cos(x) and tan(x)__

__Differentiation of sin(x), cos(x) and tan(x)__

In the past few articles on differentiation, we have learnt about:

· __Definition of differentiation__

· __The five fundamental rules of differentiation__

· __Derivatives of higher degrees__

· __Differentiation with relations to equations of tangent and normal to a curve__

· __Differentiation with relations to connected rate of change__

Therefore, today we are going to be learning about differentiation with relations to trigonometric functions. Let’s begin!

As you can see from the diagram above, we will be learning about the differentiation of the sine, cosine and tangent functions. The derivatives are given by:

**d/dx [sin(x)] = cos(x)**

**d/dx [cos(x)] = -sin(x)**

**d/dx [tan(x)] = sec2(x)**

The **derivative of a sine function** will give you a **cosine** function as the output, the **derivative of a cosine function** will give you a **negative sine function** as the output and the **derivative of a tangent function** will give you a **secant* squared function** as the output.

*Note: A **secant** function is **the reciprocal of a cosine function**, while a **cosecant** function is **the reciprocal of a sine function**.

If you have difficulty memorizing or am afraid that you might mix up both the trigonometric terms, you can just try to remember that for **secant**, the third letter of the term is a ‘**c**’, which stands for **cosine**. Whereas for **cosecant**, the third letter of the term is a ‘**s**’, which stands for **sine**.

Let’s take a look at a few examples down below:

**Differentiate each of the following with respect to x, where x is in radians.**

**a)** **6 sin x** **b) 3x^2** **cos x** **c) y = (tan x)/ (3x + 5)**

Solution:

a) d/dx (6 sin x) = 6 cos x

b) d/dx (3x^2 cos x) = (3x^2) x d/dx (cos x) + (cos x) x d/dx (3x^2)

Note: What we are basically doing in the step above is to **separate the algebraic term (3x^2) from the trigonometric term (cos x)** and then apply the **Product Rule** which we have learnt previously. Hence, we will end up with:

= (3x^2) (-sin x) + (cos x) (6x)

= 6x cos x – 3x^2 sin x

c) y = (tan x)/ (3x + 5)

dy/dx = [(3x + 5) x d/dx (tan x) – (tan x) x d/dx (3x + 5)]/ (3x + 5)^2

Note: Since the equation involves a fraction, we are hence able to use the Quotient Rule for this question.

= [(3x + 5) sec^2 x – 3(tan x)]/ (3x + 5)^2

__Differentiation of sinn__ __f(x), cosn__ __f(x), tann__ __f(x) with the application of Chain Rule__

__Differentiation of sinn__

__f(x), cosn__

__f(x), tann__

__f(x) with the application of Chain Rule__

In the previous point above, the steps to find the derivatives of the trigonometric functions are relatively simple because the angle of the trigonometric function is just a simple variable, x. However, what if the angle is no longer x, but has become a **function of x**?

In such cases, we must then utilize the **Chain Rule** we have learnt previously to solve the question. Let us consider the function **y = sin f(x)**.

If we let **u = f(x)** and **y = sin u**, hence **dy/du = cos u** and **du/dx = f’(x)**.

If we were to apply the Chain Rule here, where dy/dx = (dy/du) x (du/dx),

Hence, we will end up with:

**d/dx [sin f(x)] = f’(x) cos f(x)**

Similarly, with the application of the same method above using Chain Rule, we can obtain the derivative for both cos f(x) and tan f(x) as well, which is as shown below:

**d/dx [cos f(x)] = -f’(x) sin f(x)**

**d/dx [tan f(x)] = f’(x) sec^2** **f(x)**

In a more detailed context, given that **f(x) = ax + b**, where **a** and **b** are constants, then we will obtain the following results by applying the formulae above:

**d/dx [sin (ax + b)] = a cos (ax + b)**

**d/dx [cos (ax + b)] = -a sin (ax + b)**

**d/dx [tan (ax + b)] = a sec^2** **(ax + b)**

Let’s take a look at the examples shown below of the above application:

Differentiate each of the following with respect to x, where x is in radians.

a) 3sin [4x + (π/2)]

b) 5x^2 tan 3x

c) y = (sin 3x)/ [cos (π – 3x)]

d) 6 sin^3 4x

Solution:

a) d/dx {3 sin [4x + (π/2)]} = 3 cos [4x + (π/2)] x 4

Note: For differentiation of trigonometric terms involving functions of x, we must first differentiate the trigonometric function (the ‘outside’ of the brackets) before we differentiate the function of x, which is the ‘inside’ of the brackets.

In this case, the ‘outside’ of the brackets will be the **cosine** function, and the ‘inside’ of the brackets will be **4x + (π/2)**.

= 12 cos [4x + (π/2)]

b) d/dx (5x^2 tan 3x) = (5x^2) x d/dx (tan 3x) + (tan 3x) x d/dx (5x^2)

= 5x^2 x [3 x (sec^2 3x)] + (tan 3x) (10x)

= 15x^2 sec^2 3x + 10x tan 3x

= 5x (3x sec^2 3x + 2 tan 3x)

c) y = (sin 3x)/ [cos (π – 3x)]

dy/dx = {[cos (π – 3x)] x d/dx (sin 3x) – (sin 3x) x d/dx [cos (π – 3x)]}/ [cos (π – 3x)]^2

= {[cos (π – 3x)] (3 cos 3x) – (sin 3x) [-3 sin (π – 3x)]}/ [cos (π – 3x)]^2

= [(-cos 3x) (3 cos 3x) – (sin 3x) (-3 sin 3x)]/ (-cos 3x)^2

= [(-3 cos^2 3x) – (-3 sin^2 3x)]/ cos^2 3x

= (-3 cos^2 3x + 3 sin^2 3x)/ cos^2 3x

d) d/dx (6 sin^3 4x) = 6 [3 x (sin^2 4x) x (cos 4x) x 4]

= 72 sin^2 4x cos 4x

Try it for yourself!

Differentiate the following equation with respect to x, where x is in radians.

**y = 3 cos (πx + 1)^2** **+ 6 tan^2[(x/2) + (π/3)]**

And that’s all for today, students! ** Math Lobby** hopes that after this article, you have a clear understanding on this topic of differentiation with relations to trigonometry, including the application of chain rule for trigonometric terms with functions of x!

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