**Calculus is the study of how the output of a function changes
as its input changes.**

# Gentle Introduction

Take the function f(x) = x + 2. Plotting y = f(x) looks something like this:

| / | / | / |/ / | ------------

Every time you go along the x-axis by 1 unit, you also go up the y-axis by 1 unit. In other words, when the input of f(x) is increased by 1, its output is also increased by 1.

If the function was instead f(x) = 0.5x, we'd get this.

| | -/ | -/ | -/ | -/ | -/ -/----------

As the gradient of the line is 0.5, every time you go along the x-axis by 1 unit, you go up the y-axis by 0.5 of a unit. In other words, when the input of f(x) is increased by 1, its output is increased by 0.5.

(Note that this will hold true regardless of the y-intercept, so we can translate the graph up and down to our heart's content)

We can therefore mathematically describe the gradient of a linear function as the change of the function's output divided by the change of the function's input.

So,

f(x + h) - f(x) gradient of f(x) = ---------------, (x + h) - (x)

where h is "the change".

For example, if f(x) = 3x - 2:

f(x + h) - f(x) (3(x + h) - 2) - (3x - 2) 3x + 3h - 2 - 3x + 2 3h --------------- = ------------------------- = -------------------- = -- = 3. h h h h

I.e. the gradient of f(x) = 3. We'll write this as f'(x) = 3, although there are alternate notations.

Practise questions: [make sure use the formula - you'll see why in a bit!]

- 1. What is the gradient of f(x) = 4x + 2?
- 2. In the above formula, why can't h = 0?
- 3. Find g'(x) where g(x) = -3x + 7.
- 4. (a) If h'(x) = 10, what possible functions could h(x) be?
- (b) What do you think h''(x) is? What about h'''(x)?

f'(x) = 4

Because you can't divide by 0.

g'(x) = -3

(a) h(x) = 10x + c

(b) h''(x) = h'''(x) = 0

As discussed in question 2, h cannot be 0. However, we can consider what will happen as h approaches 0: the same thing for any other value for h - we'll still be able to calculate our gradient.

This is useful, because it allows us to find the gradient of more complex functions.

Consider f(x) = x ^ 2.

| | | | | / | / | -/ | -/ -/----------

As you can see, it gets steeper as x increases. This means that the gradient is not constant! The gradient changes as x changes.

We can't just see what happens to the output of the function when we increase the input by 1, because it'll depend on: a) Where we start b) And the size of our 'change'

So how can we work out what the gradient is, at, let's say, x = 1? The simplest thing to do is to draw a line tangent to the curve at this x = 1, and then find the gradient of that line, but that's not very mathmetical, is it?

If only there was a way to algebraically consider for any starting value of x, what would happen to the output of the function if x was 'nudged' by an infinitely small amount...

Oh wait, there is.

lim f(x + h) - f(x) f'(x) = h->0 --------------- (x + h) - (x) lim ((x + h) ^ 2) - (x ^ 2) = h->0 ----------------------- h lim (x ^ 2 + 2xh + h ^ 2) - (x ^ 2) = h->0 ------------------------------- h lim 2xh + h ^ 2 lim h(2x + h) lim = h->0 ----------- = h->0 --------- = h->0 2x+h h h = 2x

This means at any point on the x^2 graph, the tangent line at the point will have a gradient of 2x.

This means that the gradient of f(x), f'(x) = 2x.

And that the gradient when x = 1 is f'(1) = **2**. (In other words, the tangent drawn at x = 1 would be of the form y = **2**x + c).

Now, let's take a step back.
The process of going from f(x) to f'(x) is called *differentiating*, or *finding the derivative [of f(x)]*.

You'll often also see the following notation to indicate the derivative:

y = x^2 dy -- = 2x dx

The d- prefix means "a small change of", so it's easy to work out what this means.

It's also worth noting that this would also imply that:

dy = 2x dx

You'll understand why this is useful if you study implicit differentiation.

Going from f'(x) to f(x) is called *integrating*, or *finding the antiderivative*.
It's also notated like this:

/- | | 2x dx = x^2 | -/

Well, almost. Remember question 4 (a)? There are a range of functions that all have the same derivative, i.e. any translation on the y-axis. This means we actually have to write:

/- | | 2x dx = x^2 + c | -/

Forgetting the "+ c" is quite the common mistake, but if you do forget it, you'll end up in quite a mess with more advanced integration concepts. "c" is called the constant of integration.

By the way, integrating is usually more difficult than finding the derivative of a function. In fact, there are bunch of that just don't have an antiderivative.

Trying to do this formula

lim f(x + h) - f(x) f'(x) = h->0 --------------- (x + h) - (x)

in reverse just isn't very easy. Instead, there are a bunch of tricks and methods to use to find the anti-derivative of a function. Knowing which one to apply isn't always obvious, and'll take a lot of practise.

One last thing. It's important to understand what a "function of <something>" means. In calculus this phrase is used to denote any expression that depends on the value of a given variable. It doesn't have to be explicitly written as a function.

For example, if you were told `y = 2x^2 + 4x + 6`

, there are several things that could be called "functions of x":

- y
- x^2
- 2x^2
- 4x
- 2x^2 + 4x

However, things like the "6" and the factor of 2 would both be considered "constants".

Practise questions:

- 1. f(x) = x^3. What is f'(5)?
- 2. g'(x) = 3x^2. What is g(x)?
- 3. y = 10x. What is dx/dy?

75

x^3 + c

1/10

# Finding the Derivative

Notation: f(x), g(x), h(x) etc. are functions of x, and a, b, c, etc. are constants.

## a

y = a dy -- = 0 dx

## f(x)

y = f(x) dy -- = f'(x) dx

## f(x) + a

y = f(x) + a dy -- = f'(x) dx

## f(x) + g(x)

y = f(x) + g(x) dy -- = f'(x) + g'(x) dx

## f(x) * a

y = f(x) * a dy -- = f'(x) * a dx

## f(x) * g(x)

y = f(x) * g(x) dy -- = f(x)*g'(x) + g(x)*f'(x) dx

## f(a * x)

y = f(a * x) dy -- = f'(x) * a dx

## f(g(x))

y = f(g(x)) dy -- = f'(g(x)) * g'(x) dx

## sin(x)

y = sin(x) dy -- = cos(x) dx

## cos(x)

y = cos(x) dy -- = -1 * sin(x) dx

## x^a

y = x^a where a != 0 dy -- = a * x^(a-1) dx

## ln(x)

y = ln(x) dy 1 -- = - dx x

## a^x

y = a^x dy -- = ln(a) * a^x dx

### e^x

y = e^x dy -- = e^x dx