Techniques of Differentiation (Edexcel IGCSE Further Pure Maths): Revision Note

Product Rule

What is the product rule?

• The product rule states that if  $y=\mathrm{f}\left(x\right)\mathrm{g}\left(x\right)$  is the product of two functions $\mathrm{f}\left(x\right)$ and $\mathrm{g}\left(x\right)$ then

  •  $\frac{\mathrm{d}y}{\mathrm{d}x}=\mathrm{f}\text{'}\left(x\right)\mathrm{g}\left(x\right)+\mathrm{f}\left(x\right)\mathrm{g}\text{'}\left(x\right)$

• This is not given on the exam formula sheet, so you need to remember it

• This is sometimes written as  $y=uv$  where $u$ and $v$ are both functions of $x$

  • Then $y\text{'}=u\text{'}v+uv\text{'}$

    • where  $y\text{'}=\frac{\mathrm{d}y}{\mathrm{d}x}$,  $u\text{'}=\frac{\mathrm{d}u}{\mathrm{d}x}$  and  $v\text{'}=\frac{\mathrm{d}v}{\mathrm{d}x}$

• For your final answer make sure you match the notation used in the question

How do I know when to use the product rule?

• The product rule is used to differentiate a product of two functions

  • This can easily be confused with a 'function of a function' (see the Chain Rule note)

    • $\sin \left(\cos x\right)$ is a function of a function, “sin of cos of $x$”

    •  $\sin x\cos x$ is a product of two functions, “sin x times cos $x$”

How do I use the product rule?

• To differentiate  $y=\mathrm{f}\left(x\right)\mathrm{g}\left(x\right)$

  • you'll need to make it clear what $\mathrm{f}\left(x\right),\mathrm{g}\left(x\right),{\mathrm{f}}^{\text{'}}\left(x\right)$ and ${\mathrm{g}}^{\text{'}}\left(x\right)$ are

    • Arranging them in a square can help

• STEP 1
  Identify the two functions, $\mathrm{f}\left(x\right)$ and $\mathrm{g}\left(x\right)$

  • Then differentiate each one with respect to$x$ to find ${\mathrm{f}}^{\text{'}}\left(x\right)$ and ${\mathrm{g}}^{\text{'}}\left(x\right)$

• STEP 2
  Obtain $\frac{\mathrm{d}y}{\mathrm{d}x}$ by applying the product rule formula

  • If  $y=\mathrm{f}\left(x\right)\mathrm{g}\left(x\right)$,  then  $\frac{\mathrm{d}y}{\mathrm{d}x}={\mathrm{f}}^{\text{'}}\left(x\right)\mathrm{g}\left(x\right)+\mathrm{f}\left(x\right){\mathrm{g}}^{\text{'}}\left(x\right)$

  • Simplify the answer if

    • it is straightforward to do so

    • or if the question requires a particular form

• In trickier problems chain rule may have to be used when finding $u\text{'}$ and $v\text{'}$

Quotient Rule

What is the quotient rule?

• The quotient rule states if  $y=\frac{\mathrm{f}\left(x\right)}{\mathrm{g}\left(x\right)}$  is the quotient of two functions $\mathrm{f}\left(x\right)$ and $\mathrm{g}\left(x\right)$ then

  • $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{{\mathrm{f}}^{\text{'}}\left(x\right)\mathrm{g}\left(x\right)-\mathrm{f}\left(x\right){\mathrm{g}}^{\text{'}}\left(x\right)}{{\left[\mathrm{g}\left(x\right)\right]}^2}$

    • This is given on the exam formula sheet, so you don't need to remember it

• This is sometimes written as  $y=\frac{u}{v}$  where $u$ and $v$ are both functions of $x$

  • Then  $y\text{'}=\frac{u\text{'}v-uv\text{'}}{v^2}$

    • where  $y\text{'}=\frac{\mathrm{d}y}{\mathrm{d}x}$,  $u\text{'}=\frac{\mathrm{d}u}{\mathrm{d}x}$  and  $v\text{'}=\frac{\mathrm{d}v}{\mathrm{d}x}$

• For your final answer make sure you match the notation used in the question

How do I know when to use the quotient rule?

• The quotient rule is used to differentiate a quotient of two functions

  • If the numerator is a constant, negative powers can be used

    • e.g  $\frac{k}{\mathrm{g}\left(x\right)}=k{\left(\mathrm{g}\left(x\right)\right)}^{-1}$

    • The chain rule can be used here

    • Note that ${\left(\mathrm{g}\left(x\right)\right)}^{-1}$ is different from the inverse function ${\mathrm{g}}^{-1}\left(x\right)$

  • If the denominator is a constant, treat it as a factor of the expression

    • $\frac{\mathrm{f}\left(x\right)}{k}=\frac{1}{k}\mathrm{f}\left(x\right)$

  • The quotient rule will still work for both those cases

    • But it might not be the quickest method

How do I use the quotient rule?

• To differentiate  $y=\frac{\mathrm{f}\left(x\right)}{\mathrm{g}\left(x\right)}$

  • you'll need to make it clear what $\mathrm{f}\left(x\right),\mathrm{g}\left(x\right),{\mathrm{f}}^{\text{'}}\left(x\right)$ and ${\mathrm{g}}^{\text{'}}\left(x\right)$ are

    • Arranging them in a square can help

• STEP 1
  Identify the two functions, $\mathrm{f}\left(x\right)$ and $\mathrm{g}\left(x\right)$

  • Then differentiate each one with respect to$x$ to find ${\mathrm{f}}^{\text{'}}\left(x\right)$ and ${\mathrm{g}}^{\text{'}}\left(x\right)$

• STEP 2
  Obtain $\frac{\mathrm{d}y}{\mathrm{d}x}$ by applying the quotient rule formula

  • If  $y=\frac{\mathrm{f}\left(x\right)}{\mathrm{g}\left(x\right)}$,  then  $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{{\mathrm{f}}^{\text{'}}\left(x\right)\mathrm{g}\left(x\right)-\mathrm{f}\left(x\right){\mathrm{g}}^{\text{'}}\left(x\right)}{{\left[\mathrm{g}\left(x\right)\right]}^2}$

  • Simplify the answer if

    • it is straightforward to do so

    • or if the question requires a particular form

• In trickier problems chain rule may have to be used when finding $u\text{'}$ and $v\text{'}$

Chain Rule

What is the chain rule?

• The chain rule is used to differentiate a composite function 

  • A function of a function

• The chain rule is given by the formula

  • $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}u}\times \frac{\mathrm{d}u}{\mathrm{d}x}$

    • where $y$ is a function of $u$

    • and $u$ is a function of $x$

    • (Of course this ultimately makes $y$ a function of $x$ as well!)

• It can also be written in function notation as

  • If  $y=\mathrm{f}\left(\mathrm{g}\right(x\left)\right)$,  then  $\frac{\mathrm{d}y}{\mathrm{d}x}=\mathrm{f}\text{'}\left(\mathrm{g}\right(x\left)\right)\mathrm{g}\text{'}\left(x\right)$

• The chain rule makes it possible to differentiate a function of a function

  • For example  $y={\left(5x^4-2x\right)}^7$

    • $y=u^7$

    • $u=5x^4-2x$

  • Or  $y=\cos \left(3x^2-1\right)$

    • $y=\cos u$

    • $u=3x^2-1$

How can I use the chain rule to differentiate the power of a function?

•  The chain rule can be used to differentiate a 'power of a function'

  • This can save you a lot of work

    • e.g. finding the derivative of $y={\left(x^2-5x+7\right)}^7$

    • The other option would require trying to expand ${\left(x^2-5x+7\right)}^7$ first!

  • You may need to use laws of indices
    

    • e.g. finding the derivative of  $y=\frac{1}{\sqrt{2x-3}}$

    • Rewrite first as $y={\left(2x-3\right)}^{-\frac{1}{2}}$

• A power of a function can be differentiated using this special case of the chain rule:

  • If $y={\left[\mathrm{f}\left(x\right)\right]}^n$

    • i.e. if $y$ is the function $\mathrm{f}\left(x\right)$ raised to the power $n$

  • then  $\frac{\mathrm{d}y}{\mathrm{d}x}=n{\mathrm{f}}^{\text{'}}\left(x\right){\left[\mathrm{f}\left(x\right)\right]}^{n-1}$

    • i.e. $n$ times the derivative of $\mathrm{f}\left(x\right)$, times $\mathrm{f}\left(x\right)$ to the power of $n-1$

    • compare $\frac{\mathrm{d}}{\mathrm{d}x}\left(x^n\right)=n{x}^{n-1}$

  • $n$ can be any power (including fractional and negative powers)

  • This formula is not on the exam formula sheet, so you need to remember it

• The power of a function may also be differentiated using the general chain rule method in the next section

  • But remembering the 'special case' formula is a lot quicker

How can I use the general chain rule to differentiate a function?

• STEP 1
  Identify $y$ and $u$

  • e.g.  $y=\cos \left(3x^2-1\right)$

    • $y=\cos u$

    • $u=3x^2-1$

• STEP 2
  Find  $\frac{\mathrm{d}y}{\mathrm{d}u}$ and $\frac{\mathrm{d}u}{\mathrm{d}x}$

  • $\frac{\mathrm{d}y}{\mathrm{d}u}=-\sin u$

  • $\frac{\mathrm{d}u}{\mathrm{d}x}=6x$

• STEP 3
  Substitute into  $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}u}\times \frac{\mathrm{d}u}{\mathrm{d}x}$

  • $\frac{\mathrm{d}y}{\mathrm{d}x}=\left(-\sin u\right)\times \left(6x\right)$

• STEP 4
  Substitute the expression for $u$ back in

  • This gets $\frac{\mathrm{d}y}{\mathrm{d}x}$ entirely in terms of $x$

    • $\frac{\mathrm{d}y}{\mathrm{d}x}=\left(-\sin \left(3x^2-1\right)\right)\times \left(6x\right)=-6x\sin \left(3x^2-1\right)$