__In this note, you will learn:__

__1) What is differentiation?__

__2) Rules of differentiation (Part 1):__

**- Scalar Multiple Rule**

**- Addition/Subtraction Rule**

**- Chain Rule**

__1) What is differentiation?__

__1) What is differentiation?__

Differentiation, in mathematical terms, is the process of finding the derivative, or the rate of change, of a function with respect to one of its variables.

Therefore, the **derivative of y, with respect to x**, is written as such: **dy/dx**, pronounced “dee y over dee x”. Another typical way of presenting the derivative of a function, f(x), is written as such: f’(x), pronounced “f prime x”.

The general rule of differentiation to find the derivative of a function is to **shift the power of the variable you are trying to differentiate and multiply it with constant of the term, followed by decreasing the power of the variable by 1**,

or:

*d/dx (x^n) = nx^(n-1), where n is a real number*

Let’s take a look at how it’s done!

Questions to practice:

Find the derivative of each of the following with respect to x.

a) 3/x

b) y = x^(1/3)

c) 5/x^4

d) f(x) = 8

__2) Rules of differentiation__

__2) Rules of differentiation__

### · __Scalar Multiple Rule__

__Scalar Multiple Rule__

The Scalar Multiple Rule suggests that considering we have a function k f(x), where k is a constant and f(x) is a function, k will hence be called a **scalar multiple**.

This means that k, the scalar multiple, can be moved to the front of the term and be omitted from the differentiation of the term. This is as shown below:

**d/dx [k f(x)] = (k) {d/dx [f(x)]}**

### · __Addition/Subtraction Rule__

__Addition/Subtraction Rule__

The Addition/Subtraction Rule suggests that if f(x) and g(x) are functions, then the differentiation when there is an additional or subtraction of the two functions can be split into differentiation of the individual functions. This is as shown below:

**d/dx [f(x) +/- g(x)] = d/dx [f(x)] +/- d/dx [g(x)]**

Let’s take a look at an example of how this rule applies:

Given that **f(x) = (x^3 – 2x^2 + 5)/ x^2**, find **f’(x)**.

Solution:

f(x) = (x^3 – 2x^2 + 5)/ x^2

= x – 2 + 5x^-2

Therefore,

f’(x) = 1 – 0 + 5^(-2x^-3)

= 1 – (10/x^3)

### · __Chain Rule__

__Chain Rule__

The Chain Rule suggests that for complex composite functions e.g. (2x^2 + 5)^9, it will be unwise to try and expand and differentiate each term separately.

Hence, what we can do is to **let the complex portion of the composite function be a variable**. i.e. Let **u = 2x^2 + 5**, which then **u is a function of x**. In function notation, it will be **y = f(u)**, where **u = g(x)**.

Hence, **y = f[g(x)]**.

Therefore, **y** is called a **composite function of f and g**. In this manner, we are essentially breaking up the function into smaller parts so that we can easily differentiate the pieces and then combining it back together afterwards. The breakdown is as shown below:

**dy/dx = (dy/du) x (du/dx)**

How this works:

- We are basically differentiating **y with respect to u**, and **u with respect to x** first before combining the terms back together

- ‘du’ in the denominator of the ‘dy/du’ term will cancel out with the ‘du’ term in the nominator of the ‘du/dx’ term, which does not change the original derivative we are trying to find!

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

Find the derivative of the function **y = (2x^2 + 5)^9** using the Chain Rule.

Solution:

Let **u = 2x^2 + 5**, i.e. **y = u^9**, so **dy/du = 9u^8** and **du/dx = 4x**.

Therefore, dy/dx = (dy/du) x (du/dx)

= (9u^8) x (4x)

= 9(2x^2 + 5)^8 x (4x)

= 36x (2x^2 + 5)^8

And that’s all for today, students! Do note that this is only the first part of the article on Differentiation, there will be a part two continuation of the article coming soon, so stay tuned to find out!

** 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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