Differentiating Basic Functions (Edexcel IGCSE Further Maths)

Differentiating Powers of x

Spec point — 9A

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$f'(x)=nx^{n-1}$

    • 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 $f(x)=ax^{n}$ then $f^{'}(x)=anx^{n-1}$

• The alternative notation (to$f^{'}(x)$) is to use $\frac{dy}{dx}$

  • If $y=ax^{n}$ then $\frac{dy}{dx}=anx^{n-1}$

• Don't forget these two special cases:

  • If$f(x)=ax$ then$f^{'}(x)=a$

    • e.g. If $y=6x$ then $\frac{dy}{dx}=6$

  • If$f(x)=a$ then$f^{'}(x)=0$

    • e.g. If $y=5$ (or if $y$ equals any constant) then $\frac{dy}{dx}=0$

• Functions involving roots will need to be rewritten as fractional powers

  • e.g. $f(x)=2\sqrt{x}$

    • rewrite as$f(x)=2x^{\frac{1}{2}}$
    • then differentiate

• Functions involving fractions with $x$ in the denominator will need to be rewritten as negative powers

  • e.g. $f(x)=\frac{4}{x}$

    • rewrite as$f(x)=4x^{-1}$
    • 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. $f(x)=5x^{4}-3x^{\frac{2}{3}}+4$
    • $\begin{gathered} \mathrm{f} ' ( x ) & = 5 \times 4 x^{4 - 1} - 3 \times \frac{2}{3} x^{ \frac{2}{3} - 1} + 0 \end{gathered}$
    • $\begin{gathered} \mathrm{f} ' ( x ) & = 20 x^{3} - 2 x^{- 1 \frac{1}{3}} \end{gathered}$

• Products and quotients cannot be differentiated in this way

  • These need to be expanded/simplifying first

    • e.g. $f(x)=(2x-3)(x^{2}-4)$
    • Expand to$f(x)=2x^{3}-3x^{2}-8x+12$
    • Then differentiate term by term
    • You can't just multiply the derivatives of $2x-3$ and $x^{2}-4$ together!

  • These can also be differentiated using the product rule or quotient rule

Exam Hint

• 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

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

Find the derivative of $\mathrm{f} ( x )$.

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

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

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

Now differentiate as powers of $x$

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

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

Differentiating Trig Functions

Spec point — 9A

How do I differentiate sin and cos?

• The derivative of $y=\mathrm{sin}x$ is $\frac{dy}{dx}=\mathrm{cos}x$

• The derivative of is $y=\mathrm{cos}x$ is $\frac{dy}{dx}=-\mathrm{sin}x$

• If $x$ is multiplied by a constant $a$ then

  • the derivative of $y=\mathrm{sin}ax$ is $\frac{dy}{dx}=a\mathrm{cos}ax$

  • the derivative of $y=\mathrm{cos}ax$ is $\frac{dy}{dx}=-a\mathrm{sin}ax$

    • 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 Hint

• As soon as you see a question involving differentiation and trigonometry

  • put your calculator into radians mode

Worked Example (a) Given the function $\mathrm{f} ( x ) = \cos 5 x$, find $f^{'}(x)$.

Use $y=\mathrm{cos}ax\Rightarrow\frac{dy}{dx}=-a\mathrm{sin}ax$ with $a=5$

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

(b) A curve has the equation $y=3\mathrm{sin}\frac{x}{2}$. Find the gradient of the curve at the point where $x=\frac{π}{3}$, giving your answer as an exact value.

Start by finding $\frac{dy}{dx}$

Use $y=\mathrm{sin}ax\Rightarrow\frac{dy}{dx}=a\mathrm{cos}ax$ with $a=\frac{1}{2}$

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

Substitute $x=\frac{π}{3}$ into $\frac{dy}{dx}$ to find the gradient

$\text{gradient} = \frac{3}{2} \text{cos} ( \frac{ ( \frac{\pi}{3} ) }{2} ) = \frac{3}{2} \text{cos} \frac{\pi}{6} = \frac{3}{2} ( \frac{\sqrt{3}}{2} ) = \frac{3 \sqrt{3}}{4}$

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

Differentiating e^x

Spec point — 9A

How do I differentiate exponentials?

• The derivative of $y=e^{x}$ is $\frac{dy}{dx}=e^{x}$

  • Note that $e^{x}$ is its own derivative!

• If $x$ is multiplied by a constant $a$ then

  • the derivative of $y=e^{ax}$ is $\frac{dy}{dx}=ae^{ax}$

    • This can be derived by using the chain rule
    • but it's easier (and quicker) just to remember it

Exam Hint

• Remember this is not a 'powers of $x$' differentiation

  • the derivative of $e^{kx}$ is $ke^{kx}$, NOT $kxe^{kx-1}$

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

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

Differentiate using $y=e^{ax}\Rightarrow\frac{dy}{dx}=ae^{ax}$

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

Substitute $x=\frac{1}{2}$ into $\frac{dy}{dx}$ to find the gradient

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

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

Use a calculator to find the decimal version

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

$-1.34(3s.f.)$