# Differentiation Part 2

Updated: Jun 22, 2021

__In this note, you will learn:__

__1) Rules of differentiation (Part 2):__

- **Product Rule**

- **Quotient Rule**

__2) Higher derivatives of a function__

__Rules of differentiation (Part 2)__

__- Product Rule__

__- Product Rule__

The Product Rule suggests that when there is a product of two functions, the derivative of the whole expression is not as simple as it seems. Let’s say we are given two functions of x, **u** and **v**.

So, what is the derivative of the product of the two functions, **uv**? Many students will tend to give the answer: **d/dx (uv) = [d/dx (u)] x [d/dx (v)]**, which is not the case!

The derivative of the product of the two functions is given as such:

**d/dx (uv) = u (dv/dx) + v (du/dx)**

In layman terms, the derivative of the product of two functions is the **addition between the** **product of first function with the derivative of the second function** (**u (dv/dx)**), and **the product of the second function with the derivative of the first function** (**v (du/dx)**).

Let’s take a look at example of how the Product Rule works:

Find the derivative of the function **y = (x^4** **– 6) (x^2** **+ 5x + 4)** using the Product Rule.

Solution:

y = (x^4 – 6) (x^2 + 5x + 4)

Let **u = x^4** **– 6** and **v = x^2** **+ 5x + 4**, so **du/dx = 4x^3** and **dv/dx = 2x + 5**.

Therefore, dy/dx = u (dv/dx) + v (du/dx)

= (x^4 – 6) (2x + 5) + (x^2 +5x + 4) (4x^3)

= (2x^5 + 5x^4 – 12x – 30) + (4x^5 + 20x^4 + 16x^3)

= 6x^5 + 25x^4 + 16x^3 – 12x – 30

__- Quotient Rule__

__- Quotient Rule__

The Quotient Rule suggests that when a function is made up of a quotient of two or more functions, e.g. [x^6/ (x^4 – 3)], a special case of the Product Rule is applied here.

We can rewrite the main function [x6/ (x4 – 3)] as (x6) (x4 – 3)-1 and apply the product rule.

This would give us:

Let **u = x^6** and **v = (x^4** **– 3)^-1**

d/dx [ (x^6) (x^4 – 3)^-1] = (x^6) (-1) (x^4 – 3)^-2 (4x^3) + (x^4 – 3)^-1 (6x^5)

= x^5 (x^4 – 3)^-2 [-4x^4 + 6(x^4 – 3)]

= [x^5 (2x^4 – 18)]/ (x^4 – 3)^2

= [2x^5 (x^4 – 9)]/ (x^4 – 3)^2

To simplify things, we can derive another rule to differentiate functions like the one above. The result is the Quotient Rule.

If u and v are functions of x, then:

**d/dx (u/v) = [v(du/dx) – u(dv/dx)]/ v^2**

__Higher derivatives of a function__

__Higher derivatives of a function__

The function **dy/dx** is known to us as the **first derivative**, which there can be higher orders of derivatives like the **second (d^2y/dx^2)** and the **third (d^3y/dx^3)** derivatives.

How we obtain higher levels of derivatives will be through **the differentiation of the previous derivative**.

i.e. To get **d^2y/dx^2**, we will have to differentiate **dy/dx**. To get **d^3y/dx^3**, we will have to differentiate **d^2y/dx^2**.

An example of how to get the first, second and third derivative of a function:

E.g. **y = x^5** **+ 4x^3** **+ 2x^2** **– 9x + 5**

Hence,

**dy/dx = 5x^4** **+ 12x^2** **+ 4x – 9**

**d^2y/dx^2** **= 20x^3** **+ 24x + 4**

**d^3y/dx^3** **= 60x^2** **+ 24**

And that’s all for today, students! This sums up the second part of the article on Differentiation and ** Math Lobby** hopes that after this article, you have a clear understanding on the topic of differentiation, and is equipped with the necessary skills to deal with questions involving the application of the different rules of differentiation!

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