Techniques of Differentiation (Edexcel IGCSE Further Maths)

Product Rule

Spec point — 9B

What is the product rule?

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

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

• 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 $y'=u'v+uv'$

    • where $y'=\frac{dy}{dx}$, $u'=\frac{du}{dx}$ and $v'=\frac{dv}{dx}$

• 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)

    • $\mathrm{sin}(\mathrm{cos}x)$ is a function of a function, “sin of cos of $x$”
    • $\mathrm{sin}x\mathrm{cos}x$ is a product of two functions, “sin x times cos $x$”

How do I use the product rule?

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

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

    • Arranging them in a square can help

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

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

• STEP 2 Obtain $\frac{dy}{dx}$ by applying the product rule formula

  • If $y = \mathrm{f} ( x ) \mathrm{g} ( x )$, then $\frac{dy}{dx}=f^{'}(x)g(x)+f(x)g^{'}(x)$

  • 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 $u'$ and $v'$

Exam Hint

• Using $u,v,u'$ and $v'$ can save time writing

  • lay them out in a 2x2 'square' to help keep which is which straight

• For trickier functions chain rule may be required along with product rule

  • i.e. either $u$ and/or $v$ could be a 'function of a function'
  • So chain rule needed to find $u'$ and $v'$

Worked Example Find the derivative of $y=5x^{2}\mathrm{cos}3x$.

We'll use $y=uv$ form

Identify the functions $u$ and $v$

$u=5x^{2}v=\mathrm{cos}3x$

Differentiate those to find $u'$ and $v'$

$u'=10xv'=-3\mathrm{sin}3x$

Put the pieces together using $y'=u'v+uv'$

$\frac{ \mathrm{d} y}{ \mathrm{d} x} = ( 10 x ) ( \cos 3 x ) + ( 5 x^{2} ) ( - 3 \sin 3 x )$

Expand the brackets

$\frac{dy}{dx}=10x\mathrm{cos}3x-15x^{2}\mathrm{sin}3x$

Quotient Rule

Spec point — 9B

What is the quotient rule?

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

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

    • 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 $y'=\frac{u'v-uv'}{v^{2}}$

    • where $y'=\frac{dy}{dx}$, $u'=\frac{du}{dx}$ and $v'=\frac{dv}{dx}$

• 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 $\frac{k}{ \mathrm{g} ( x )} = k ( \mathrm{g} ( x ) )^{- 1}$
    • The chain rule can be used here
    • Note that $( \mathrm{g} ( x ) )^{- 1}$ is different from the inverse function $\mathrm{g}^{- 1} ( x )$

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

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

  • 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 $y = \frac{ \mathrm{f} ( x )}{ \mathrm{g} ( x )}$

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

    • Arranging them in a square can help

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

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

• STEP 2 Obtain $\frac{dy}{dx}$ by applying the quotient rule formula

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

  • 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 $u'$ and $v'$

Exam Hint

• For trickier functions chain rule may be required along with product rule

  • i.e. either $u$ and/or $v$ could be a 'function of a function'
  • So chain rule needed to find $u'$ and $v'$

• Look out for functions of the form $y=f(x)(g(x))^{-1}$

  • These can be differentiated using a combination of chain rule and product rule

    • It would be good practice to try this sometime!

  • But it's probably easier to use laws of indices to rewrite as $y = \frac{ \mathrm{f} ( x )}{ \mathrm{g} ( x )}$

    • and then use the quotient rule

Worked Example Given the function $f(x)=\frac{\mathrm{cos}2x}{3x+2}$, find $\mathrm{f}^{'} ( x )$.

We'll use $y = \frac{ \mathrm{f} ( x )}{ \mathrm{g} ( x )}$ form

Identify the functions $\mathrm{f} ( x )$ and $\mathrm{g} ( x )$

$\mathrm{f} ( x ) = \cos 2 x \mathrm{g} ( x ) = 3 x + 2$

Differentiate those to find $\mathrm{f} ' ( x )$ and $\mathrm{g}^{'} ( x )$

$\mathrm{f}^{'} ( x ) = - 2 \sin 2 x \mathrm{g}^{'} ( x ) = 3$

Put the pieces together using $\frac{ \mathrm{d} }{ \mathrm{d} x} ( \frac{ \mathrm{f} ( x )}{ \mathrm{g} ( x )} ) = \frac{ \mathrm{f}^{'} ( x ) \mathrm{g} ( x ) - \mathrm{f} ( x ) \mathrm{g}^{'} ( x )}{[ \mathrm{g} ( x ) ]^{2}}$ from the exam formula sheet

$\mathrm{f} ' ( x ) = \frac{( - 2 \sin 2 x ) ( 3 x + 2 ) - ( \cos 2 x ) ( 3 )}{( 3 x + 2 )^{2}}$

Simplify the numerator (The denominator is simplest left as it is)

$f^{'}(x)=\frac{-2(3x+2)\mathrm{sin}2x-3\mathrm{cos}2x}{(3x+2)^{2}}$

Chain Rule

Spec point — 9B

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

  • $\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}$

    • 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 $y=f(g(x))$, then $\frac{dy}{dx}=f'(g(x))g'(x)$

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

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

    • $y=u^{7}$
    • $u=5x^{4}-2x$

  • Or $y = \cos ( 3 x^{2} - 1 )$

    • $y=\mathrm{cos}u$
    • $u=3x^{2}-1$

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

• You may need to use laws of indices

  • e.g. finding the derivative of $y=\frac{1}{\sqrt{2x-3}}$
  • Rewrite first as $y = ( 2 x - 3 )^{- 1 \frac{1}{2}}$

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

  • If $y = [ \mathrm{f} ( x ) ]^{n}$

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

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

    • i.e. $n$ times the derivative of $\mathrm{f} ( x )$, times $\mathrm{f} ( x )$ to the power of $n-1$
    • compare $\frac{ \mathrm{d} }{ \mathrm{d} x} ( x^{n ) = n x^{n} - 1}$

  • $n$ 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. $y = \cos ( 3 x^{2} - 1 )$

    • $y=\mathrm{cos}u$
    • $u=3x^{2}-1$

• STEP 2 Find $\frac{dy}{du}$ and $\frac{du}{dx}$

  • $\frac{dy}{du}=-\mathrm{sin}u$
  • $\frac{du}{dx}=6x$

• STEP 3 Substitute into $\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}$

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

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

  • This gets $\frac{dy}{dx}$ entirely in terms of $x$

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

Exam Hint

• When asked to differentiate a 'power of a function', the chain rule is usually your best option

  • Look out for 'hidden powers'

    • e.g. square roots (fractional powers)
    • or functions in a denominator (negative powers)

• To integrate a general 'function of a function' the chain rule is your only option!

Worked Example (a) Find the derivative of$y=(x^{2}-5x+7)^{7}$.

This is $y = [ \mathrm{f} ( x ) ]^{n}$ with $n=7$ and $\mathrm{f} ( x ) = x^{2} - 5 x + 7$

Start by finding the derivative of $\mathrm{f} ( x )$

$\frac{ \mathrm{d} }{ \mathrm{d} x} ( x^{2} - 5 x + 7 ) = 2 x - 5$

Now use $\frac{ \mathrm{d} y}{ \mathrm{d} x} = n \mathrm{f}^{' ( x ) [ \mathrm{f} ( x ) ]^{n} - 1}$

$\frac{ \mathrm{d} y}{ \mathrm{d} x} = 7 ( 2 x - 5 ) ( x^{2} - 5 x + 7 )^{7 - 1}$

$\frac{dy}{dx}=7(2x-5)(x^{2}-5x+7)^{6}$

(b) Find the derivative of $y=\sqrt{\mathrm{sin}x}$.

First use laws of indices to write as a power

$y = ( \sin x )^{1 \frac{1}{2}}$

This is $y = [ \mathrm{f} ( x ) ]^{n}$ with $n=\frac{1}{2}$ and $\mathrm{f} ( x ) = \sin x$

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

$\frac{ \mathrm{d} }{ \mathrm{d} x} ( \sin x ) = \cos x$

Now use $\frac{ \mathrm{d} y}{ \mathrm{d} x} = n \mathrm{f}^{' ( x ) [ \mathrm{f} ( x ) ]^{n} - 1}$

$\begin{gathered} \frac{ \mathrm{d} y}{ \mathrm{d} x} & = ( 1 \frac{1}{2} ) ( \cos x ) ( \sin x )^{1 \frac{1}{2} - 1} \\ = \frac{\cos x}{2} ( \sin x )^{- 1 \frac{1}{2}} \end{gathered}$

And by laws of indices $( \sin x )^{- 1 \frac{1}{2}} = \frac{1}{( \sin x )^{1 \frac{1}{2}}} = \frac{1}{ \sqrt{\sin x} }$

$\frac{dy}{dx}=\frac{\mathrm{cos}x}{2\sqrt{\mathrm{sin}x}}$

(c) Find the derivative of $y=e^{\mathrm{cos}x}$.

This is not a 'power of a function', so we need to use the general chain rule method

Start by identifying $y$ and $u$

$y=e^{u}\,\,u=\mathrm{cos}x$

Differentiate

$\frac{dy}{du}=e^{u}\,\,\frac{du}{dx}=-\mathrm{sin}x$

Substitute into $\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}$

$\frac{ \mathrm{d} y}{ \mathrm{d} x} = ( \mathrm{e}^{u} ) \times ( - \sin x )$

Substitute the expression for $u$ back in

$\frac{ \mathrm{d} y}{ \mathrm{d} x} = ( \mathrm{e}^{\cos x} ) \times ( - \sin x )$

$\frac{dy}{dx}=-(\mathrm{sin}x)e^{\mathrm{cos}x}$