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

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

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

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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