Introduction to Differentiation · Revision Notes · Further Maths

Definition of Gradientvideo

Definition of Gradient

Definition of gradient

What is the gradient of a curve?

At a given point the gradient of a curve is defined as the gradient of the tangent to the curve at that point
A tangent to a curve is a line that just touches the curve at one point but doesn't cut the curve at that point

Def Grad Illustr 1, A Level & AS Maths: Pure revision notes

A tangent may cut the curve somewhere else on the curve

Def Grad Illustr 2, A Level & AS Maths: Pure revision notes

It is only possible to draw one tangent to a curve at any given point
Note that unlike the gradient of a straight line, the gradient of a curve is constantly changing

Definition of Derivatives

Definition of derivatives

What is a derivative?

Calculus is about rates of change
- the way a car’s position on a road changes is its speed (velocity)
- the way the car’s speed changes is its acceleration
The gradient (rate of change) of a (non-linear) function varies with $x$
The derivative of a function is a function that relates the gradient to the value of $x$
- For example, the derivative of $y = x^{2}$ is $2 x$
  - This means that when $x = 1$ , the gradient of $y = x^{2}$ is $2 (1) = 2$
  - And when $x = 5$ , the gradient of $y = x^{2}$ is $2 (5) = 10$
The derivative is also called the gradient function

Worked example

The derivative of $y = x^{3} - 2 x^{2} - x + 3$ is $3 x^{2} - 4 x - 1$ .

Use the derivative to find the gradient of $y = x^{3} - 2 x^{2} - x + 3$ at the point $A (2, 1)$ .

Substitute $x = 2$ into the derivative, $3 x^{2} - 4 x - 1$

$3 {(2)}^{2} - 4 (2) - 1 = 12 - 8 - 1$

gradient = 3

Note that the answer is the same as in the method above

Differentiating Powers of xvideo

Differentiating Powers of x

Differentiating powers of x

What is differentiation?

Differentiation is the process of finding an expression for the derivative (gradient function) from the equation of a curve
- The equation of the curve is written $y = . . .$ and the gradient function is written $\frac{d y}{d x} = . . .$

How do I differentiate powers of x?

Powers of $x$ are differentiated according to the following formula:
- If $y = a x^{n}$ then $\frac{d y}{d x} = a n x^{n - 1}$
  - e.g. If $y = 4 x^{3}$ then $\frac{d y}{d x} = 4 \times 3 \times x^{3 - 1} = 12 x^{2}$
  - you "bring down the power" then "subtract one from the power"
Don't forget these two special cases:
- If $y = a x$ then $\frac{d y}{d x} = a$
  - e.g. If $y = 6 x$ then $\frac{d y}{d x} = 6$
- If $y = a$ then $\frac{d y}{d x} = 0$
  - e.g. If $y = 5$ then $\frac{d y}{d x} = 0$
- These allow you to differentiate linear terms in $x$ and constants
Functions involving fractions with denominators in terms of $x$ will need to be rewritten as negative powers of $x$ first
- e.g. If $y = \frac{4}{x}$ then rewrite as $y = 4 x^{- 1}$ and differentiate

How do I differentiate sums and differences of powers of x?

The formulae for differentiating powers of $x$ work for a sum or difference of powers of $x$
- e.g. If $y = 5 x^{4} + 3 x^{- 2} + 4$ then $\frac{d y}{d x} = 5 \times 4 x^{4 - 1} + 3 \times (- 2) x^{- 2 - 1} + 0$ $\frac{d y}{d x} = 20 x^{3} - 6 x^{- 3}$
- This is sometimes referred to differentiating 'term-by-term'
Products and quotients (divisions) cannot be differentiated in this way so they need expanding/simplifying first
- e.g. If $y = (2 x - 3) (x^{2} - 4)$ then expand to $y = 2 x^{3} - 3 x^{2} - 8 x + 12$ which is a sum/difference of powers of $x$ and can then be differentiated

What can I do with derivatives (gradient functions)?

The derivative can be used to find the gradient of a function at any point
- The gradient of a function at a point is equal to the gradient of the tangent to the curve at that point
- A question may refer to the gradient of the tangent

Worked example

Find the derivative of

(a) $\begin{matrix} \end{matrix}$ $y = 5 x^{3} + 2 x + \frac{3}{x^{2}} + 8$

Rewrite the $\frac{3}{x^{2}}$ term

$y = 5 x^{3} + 2 x + 3 x^{- 2} + 8$

Apply the rule for differentiating powers ( $y = a x^{n}, \frac{d y}{d x} = a n x^{n - 1}$ ) and apply the special cases for the terms $2 x$ and 8 ( $y = a x, \frac{d y}{d x} = a$ and $y = a, \frac{d y}{d x} = 0$ )

$\frac{d y}{d x} = 15 x^{2} + 2 - 6 x^{- 3}$

Unless a question specifies there is not usually a need to rewrite/simplify the answer

$\frac{d y}{d x} = 15 x^{2} + 2 - \frac{6}{x^{3}}$

(b) $y = {(2 x + 3)}^{2}$

This is a product of two (equal) brackets so cannot be differentiated directly Expand the brackets to get an expression in powers of $x$ Take time to get the expansion correct, writing stages out in full if necessary

$\begin{array}{rcl} y & = & (2 x + 3) (2 x + 3) \\ y & = & 4 x^{2} + 6 x + 6 x + 9 \\ y & = & 4 x^{2} + 12 x + 9 \end{array}$

Differentiate 'term-by-term', looking out for those special cases

$\frac{d y}{d x} = 8 x + 12$

There is a factor of 4 but there is no demand to factorise the final answer in the question

$\frac{d y}{d x} = 8 x + 12$

(c) $\begin{matrix} \end{matrix}$ $y = \frac{8 x^{6} - x^{3}}{2 x^{4}}$

This is a quotient so cannot be differentiated directly Spot the single denominator which means we can split the fraction by the two terms on the numerator

$y = \frac{8 x^{6}}{2 x^{4}} - \frac{x^{3}}{2 x^{4}}$

Simplify using the laws of indices

$y = 4 x^{6 - 4} - \frac{1}{2} x^{3 - 4} y = 4 x^{2} - \frac{1}{2} x^{- 1}$

Each term is now a power of $x$ , so differentiate 'term-by-term'

$\frac{d y}{d x} = 8 x + \frac{1}{2} x^{- 2}$

There is demand to simplify or write the answer in a particular form