Differentiating Basic Functions (Edexcel IGCSE Further Pure Maths): Revision Note

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 $\mathit{x}$ are differentiated according to the following formula:

  • If$\mathrm{f}\left(x\right)=x^n$ then$\mathrm{f}\text{'}\left(x\right)=n{x}^{n-1}$

    • 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 $\mathrm{f}\left(x\right)=a{x}^n$  then ${\mathrm{f}}^{\text{'}}\left(x\right)=an{x}^{n-1}$

• The alternative notation (to${\mathrm{f}}^{\text{'}}\left(x\right)$) is to use $\frac{\mathbf{d}\mathit{y}}{\mathbf{d}\mathit{x}}$

  • If $y=a{x}^n$ then $\frac{\mathrm{d}y}{\mathrm{d}x}=an{x}^{n-1}$

• Don't forget these two special cases:

  • If$\mathrm{f}\left(x\right)=ax$ then${\mathrm{f}}^{\text{'}}\left(x\right)=a$

    • e.g.  If $y=6x$ then $\frac{\mathrm{d}y}{\mathrm{d}x}=6$

  • If$\mathrm{f}\left(x\right)=a$ then${\mathrm{f}}^{\text{'}}\left(x\right)=0$

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

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

  • e.g.  $\mathrm{f}\left(x\right)=2\sqrt{x}$

    • rewrite as$\mathrm{f}\left(x\right)=2x^{\frac{1}{2}}$

    • then differentiate

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

  • e.g.  $\mathrm{f}\left(x\right)=\frac{4}{x}$

    • rewrite as$\mathrm{f}\left(x\right)=4x^{-1}$

    • 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. $\mathrm{f}\left(x\right)=5x^4-3x^{\frac{2}{3}}+4$

    • $\begin{array}{rcl}\mathrm{f}\text{'}\left(x\right)& =& 5\times 4x^{4-1}-3\times \frac{2}{3}{x}^{\frac{2}{3}-1}+0\end{array}$

    • $\begin{array}{rcl}\mathrm{f}\text{'}\left(x\right)& =& 20x^3-2x^{-\frac{1}{3}}\end{array}$
      

• Products and quotients cannot be differentiated in this way

  • These need to be expanded/simplifying first

    • e.g.  $\mathrm{f}\left(x\right)=(2x-3)(x^2-4)$

    • Expand to$\mathrm{f}\left(x\right)=2x^3-3x^2-8x+12$

    • 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

Differentiating Trig Functions

How do I differentiate sin and cos?

• The derivative of  $\mathit{y}\mathbf{=}\mathbf{sin}\mathit{x}$  is  $\frac{\mathbf{d}\mathit{y}}{\mathbf{d}\mathit{x}}\mathbf{=}\mathbf{cos}\mathit{x}$  

• The derivative of is  $\mathit{y}\mathbf{=}\mathbf{cos}\mathit{x}$  is  $\frac{\mathbf{d}\mathit{y}}{\mathbf{d}\mathit{x}}\mathbf{=}\mathbf{-}\mathbf{sin}\mathit{x}$

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

  • the derivative of  $\mathit{y}\mathbf{=}\mathbf{sin}\mathit{a}\mathit{x}$  is  $\frac{\mathbf{d}\mathit{y}}{\mathbf{d}\mathit{x}}\mathbf{=}\mathit{a}\mathbf{cos}\mathit{a}\mathit{x}$

  • the derivative of  $\mathit{y}\mathbf{=}\mathbf{cos}\mathit{a}\mathit{x}$  is  $\frac{\mathbf{d}\mathit{y}}{\mathbf{d}\mathit{x}}\mathbf{=}\mathbf{-}\mathit{a}\mathbf{sin}\mathit{a}\mathit{x}$

    • 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

Differentiating e^x

How do I differentiate exponentials?

• The derivative of  $\mathit{y}\mathbf{=}{\mathbf{e}}^{\mathit{x}}$  is  $\frac{\mathbf{d}\mathit{y}}{\mathbf{d}\mathit{x}}\mathbf{=}{\mathbf{e}}^{\mathit{x}}$

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

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

  • the derivative of  $\mathit{y}\mathbf{=}{\mathbf{e}}^{\mathit{a}\mathit{x}}$  is  $\frac{\mathbf{d}\mathit{y}}{\mathbf{d}\mathit{x}}\mathbf{=}\mathit{a}{\mathbf{e}}^{\mathit{a}\mathit{x}}$

    • This can be derived by using the chain rule

    • but it's easier (and quicker) just to remember it