Differentiating Basic Functions (Edexcel IGCSE Further Maths)

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Author

Roger

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Maths

Differentiating Powers of x

What is differentiation?

Differentiation is the process of finding the derivative (gradient function) of a function

How do I differentiate powers of x?

Powers of $x$ are differentiated according to the following formula:
- If $f (x) = x^{n}$ then $f' (x) = n x^{n - 1}$
  - Bring the power down in front as a multiplier
  - Then subtract 1 from the power
- This formula is not on the exam formula sheet, so you need to remember it
If the power of $x$ term is multiplied by a constant $a$
- then the derivative is also multiplied by that constant
  - If $f (x) = a x^{n}$ then $f^{'} (x) = a n x^{n - 1}$
The alternative notation (to $f^{'} (x)$ ) is to use $\frac{d y}{d x}$
- If $y = a x^{n}$ then $\frac{d y}{d x} = a n x^{n - 1}$
Don't forget these two special cases:
- If $f (x) = a x$ then $f^{'} (x) = a$
  - e.g. If $y = 6 x$ then $\frac{d y}{d x} = 6$
- If $f (x) = a$ then $f^{'} (x) = 0$
  - e.g. If $y = 5$ (or if $y$ equals any constant) then $\frac{d y}{d x} = 0$
Functions involving roots will need to be rewritten as fractional powers
- e.g. $f (x) = 2 \sqrt{x}$
  - rewrite as $f (x) = 2 x^{\frac{1}{2}}$
  - then differentiate
Functions involving fractions with in the denominator will need to be rewritten as negative powers
- e.g. $f (x) = \frac{4}{x}$
  - rewrite as $f (x) = 4 x^{- 1}$
  - then differentiate

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

The formulae can be used to differentiate sums or differences of powers of $x$
- Just differentiate term by term
  - e.g. $f (x) = 5 x^{4} - 3 x^{\frac{2}{3}} + 4$
  - $\begin{array}{rcl} f' (x) & = & 5 \times 4 x^{4 - 1} - 3 \times \frac{2}{3} x^{\frac{2}{3} - 1} + 0 \end{array}$
  - $\begin{array}{rcl} f' (x) & = & 20 x^{3} - 2 x^{- \frac{1}{3}} \end{array}$
Products and quotients cannot be differentiated in this way
- These need to be expanded/simplifying first
  - e.g. $f (x) = (2 x - 3) (x^{2} - 4)$
  - Expand to $f (x) = 2 x^{3} - 3 x^{2} - 8 x + 12$
  - Then differentiate term by term
  - You can't just multiply the derivatives of $2 x - 3$ and $x^{2} - 4$ together!
- These can also be differentiated using the product rule or quotient rule

Exam Tip

Be careful with negative and fractional powers
- It's easy to make a mistake when subtracting 1 from these

Worked example

The function $f (x)$ is given by

$f (x) = 2 x^{3} + \frac{4}{\sqrt{x}}$ , where $x > 0$

Find the derivative of $f (x)$ .

Start by rewriting the $\frac{4}{\sqrt{x}}$ term as a power of $x$

By laws of indices, $\frac{1}{\sqrt{x}} = x^{- \frac{1}{2}}$

$f (x) = 2 x^{3} + 4 x^{- \frac{1}{2}}$

Now differentiate as powers of $x$

$\begin{array}{rcl} f^{'} (x) & = & 2 (3 x^{3 - 1}) + 4 (- \frac{1}{2} x^{- \frac{1}{2} - 1}) \\ = & 2 (3 x^{2}) + 4 (- \frac{1}{2} x^{- \frac{3}{2}}) \\ = & 6 x^{2} - 2 x^{- \frac{3}{2}} \end{array}$

$f^{'} (x) = 6 x^{2} - 2 x^{- \frac{3}{2}}$

Differentiating Trig Functions

How do I differentiate sin and cos?

The derivative of $y = \sin x$ is $\frac{d y}{d x} = \cos x$
The derivative of is $y = \cos x$ is $\frac{d y}{d x} = - \sin x$
If $x$ is multiplied by a constant $a$ then
- the derivative of $y = \sin a x$ is $\frac{d y}{d x} = a \cos a x$
- the derivative of $y = \cos a x$ is $\frac{d y}{d x} = - a \sin a x$
  - These can be derived by using the chain rule
  - but it's easier (and quicker) just to remember them
None of these formulae are on the exam formula sheet, so you need to remember them
For calculus with trigonometric functions angles must be measured in radians
- Make sure you know how to change the angle mode on your calculator

Exam Tip

As soon as you see a question involving differentiation and trigonometry
- put your calculator into radians mode

Worked example

(a)

Given the function

f (x) = \cos 5 x

, find

f^{'} (x)

Use

y = \cos a x \Rightarrow \frac{d y}{d x} = - a \sin a x

with

a = 5

f^{'} (x) = - 5 \sin 5 x

(b)

A curve has the equation

y = 3 \sin \frac{x}{2}

.

Find the gradient of the curve at the point where

x = \frac{π}{3}

, giving your answer as an exact value.

Start by finding

\frac{d y}{d x}

Use

y = \sin a x \Rightarrow \frac{d y}{d x} = a \cos a x

with

a = \frac{1}{2}

\frac{d y}{d x} = 3 (\frac{1}{2} \cos \frac{x}{2}) = \frac{3}{2} \cos \frac{x}{2}

Substitute

x = \frac{π}{3}

into

\frac{d y}{d x}

to find the gradient

gradient = \frac{3}{2} \cos (\frac{(\frac{π}{3})}{2}) = \frac{3}{2} \cos \frac{π}{6} = \frac{3}{2} (\frac{\sqrt{3}}{2}) = \frac{3 \sqrt{3}}{4}

\frac{3 \sqrt{3}}{4}

Differentiating e^x

How do I differentiate exponentials?

The derivative of $y = e^{x}$ is $\frac{d y}{d x} = e^{x}$
- Note that $e^{x}$ is its own derivative!
If $x$ is multiplied by a constant $a$ then
- the derivative of $y = e^{a x}$ is $\frac{d y}{d x} = a e^{a x}$
  - This can be derived by using the chain rule
  - but it's easier (and quicker) just to remember it

Exam Tip

Remember this is not a 'powers of ' differentiation
- the derivative of $e^{k x}$ is $k e^{k x}$ , NOT $k x e^{k x - 1}$

Worked example

A curve has the equation $y = 2 e^{- 3 x}$ .

Find the gradient of the curve at the point where $x = \frac{1}{2}$ , giving your answer correct to 3 significant figures.

Differentiate using $y = e^{a x} \Rightarrow \frac{d y}{d x} = a e^{a x}$

$\frac{d y}{d x} = 2 (- 3 e^{- 3 x}) = - 6 e^{- 3 x}$

Substitute $x = \frac{1}{2}$ into $\frac{d y}{d x}$ to find the gradient

$gradient = - 6 e^{- 3 (\frac{1}{2})} = - 6 e^{- \frac{3}{2}}$

Note that $- 6 e^{- \frac{3}{2}} = - \frac{6}{e \sqrt{e}}$ is the exact value answer

Use a calculator to find the decimal version

$- 6 e^{- \frac{3}{2}} = - 1.338780 . . .$

$- 1.34 (3 s . f .)$

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Differentiating Basic Functions (Edexcel IGCSE Further Maths)

Revision Note

What is differentiation?

How do I differentiate powers of x?

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

How do I differentiate sin and cos?

How do I differentiate exponentials?

You've read 0 of your 0 free revision notes

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Author: Roger