Chain Rule

Chain Rule

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

Chain Rule

The chain rule is best used when we have a function too complex to differentiate in one fell swoop. We look to turn one function into a function of a function – it sounds worse, but I guarantee it will make your job much easier!

A Level AQA Edexcel OCR

Adapting to \dfrac{dx}{dy}

For equations of the form \textcolor{blue}{x} = f(\textcolor{limegreen}{y}), we can calculate \dfrac{d\textcolor{blue}{x}}{d\textcolor{limegreen}{y}} in the same way as usual, and find its reciprocal to find an expression for \dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}}. We can then couple this with Integration to find an expression for \textcolor{limegreen}{y} = f^{-1}(\textcolor{blue}{x}), but we’ll not worry about this for the moment.

The most important thing to remember is that \dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}} \times \dfrac{d\textcolor{blue}{x}}{d\textcolor{limegreen}{y}} = 1, so we have

\dfrac{d\textcolor{blue}{x}}{d\textcolor{limegreen}{y}} = \dfrac{1}{\left( \dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}}\right) }

A LevelAQAEdexcelOCR

How to Use the Chain Rule

Let’s say our complex function f(\textcolor{blue}{x}) can be split into a function of a function, i.e. f(\textcolor{blue}{x}) = g(h(\textcolor{blue}{x})).

Then we have the Chain Rule:

\dfrac{df(\textcolor{blue}{x})}{d\textcolor{blue}{x}} = \dfrac{dg(h(\textcolor{blue}{x}))}{d\textcolor{blue}{x}} = \dfrac{dg(h(\textcolor{blue}{x}))}{dh(\textcolor{blue}{x})} \times \dfrac{dh(\textcolor{blue}{x})}{d\textcolor{blue}{x}}

That might still seem a little confusing, so you may see it given where y=g(u) and u = h(x):

\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}

A LevelAQAEdexcelOCR

Connecting Rates of Change

You may see examples where there are several variables linked, for example distance, speed and acceleration or length, surface area and volume. If you know the rate of change of one variable, and the equations linking the variables, then you can use the chain rule to find the rates of change of the other variables.

Watch out for hidden derivatives given as words, such as “rate” or “per“.

Once you understand what the question is asking the calculations are usually pretty straightforward.

A LevelAQAEdexcelOCR
A Level AQA Edexcel OCR

Example 1: Using the Chain Rule

Let f(\textcolor{blue}{x}) = \sqrt{2\textcolor{blue}{x} - \dfrac{1}{\textcolor{blue}{x}^2}}. Find the derivative of f(\textcolor{blue}{x}) with respect to \textcolor{blue}{x}.

[4 marks]

Take g(\textcolor{red}{u}) = \sqrt{\textcolor{red}{u}} and \textcolor{red}{u} = h(\textcolor{blue}{x}) = 2\textcolor{blue}{x} - \dfrac{1}{\textcolor{blue}{x}^2}, so f(\textcolor{blue}{x}) = g(h(\textcolor{blue}{x})).

Then,

\dfrac{dg(\textcolor{red}{u})}{d\textcolor{red}{u}} = \dfrac{1}{2\sqrt{\textcolor{red}{u}}} = \dfrac{1}{2\sqrt{2\textcolor{blue}{x} - \dfrac{1}{\textcolor{blue}{x}^2}}} and \dfrac{d\textcolor{red}{u}}{d\textcolor{blue}{x}} =\dfrac{dh(\textcolor{blue}{x})}{d\textcolor{blue}{x}} = 2 + \dfrac{2}{\textcolor{blue}{x}^3} = \dfrac{2\textcolor{blue}{x}^3 + 2}{\textcolor{blue}{x}^3}.

Multiplying the two gives

\dfrac{df(\textcolor{blue}{x})}{d\textcolor{blue}{x}} = \left( \dfrac{2\textcolor{blue}{x}^3 + 2}{\textcolor{blue}{x}^3}\right) \left( \dfrac{1}{2\sqrt{2\textcolor{blue}{x} - \dfrac{1}{\textcolor{blue}{x}^2}}}\right) = \dfrac{\textcolor{blue}{x}^3 + 1}{\textcolor{blue}{x}^3 \sqrt{2\textcolor{blue}{x} - \dfrac{1}{\textcolor{blue}{x}^2}}}

 

A LevelAQAEdexcelOCR

Example 2: Connecting Rates of Change

The surface area of a cube of width x \text{ cm}, length 2x \text{ cm} and height 4x \text{ cm} decreases at a constant rate of 0.07 \text{ cm}^2 \text{s}^{-1}. Find \textcolor{purple}{\dfrac{dx}{dt}} at the point where \textcolor{orange}{x = 5 \text{ cm}}.

[3 marks]

The surface area of the cuboid is:

A = 2 \times (x \times 2x) + 2 \times (x \times 4x) + 2 \times (2x \times 4x)  = 28x^2

So,

\dfrac{dA}{dx} = 56x

A decreases at a constant rate of 0.07 \text{ cm}^2\text{s}^{-1}, so we can write this as

\dfrac{dA}{dt} = - 0.07

Now, use the chain rule to find \textcolor{purple}{\dfrac{dx}{dt}}:

\begin{aligned} \textcolor{purple}{\dfrac{dx}{dt}} &= \dfrac{dx}{dA} \times \dfrac{dA}{dt} \\[1.2em] &= \dfrac{1}{\left( \dfrac{dA}{dx} \right)} \times \dfrac{dA}{dt} \\[1.2em] &= \dfrac{1}{56x} \times (-0.07) \\[1.2em] &= - \dfrac{0.00125}{x} \end{aligned}

When \textcolor{orange}{x = 5},

\textcolor{purple}{\dfrac{dx}{dt}} = - \dfrac{0.00125}{x} = - \dfrac{0.00125}{\textcolor{orange}{5}} = -0.00025 \text{ cms}^{-1}

A LevelAQAEdexcelOCR

Example Questions

Let g(u) = \sin u and u = h(x) = \dfrac{1}{x}. Then we have y = g(h(x)).

\dfrac{dy}{dx} = \dfrac{dg(u)}{du} \times \dfrac{du}{dx}

\dfrac{dg(u)}{du} = \cos u = \cos \dfrac{1}{x} and \dfrac{du}{dx} = \dfrac{-1}{x^2}

so, \dfrac{dy}{dx} = \dfrac{-\cos \dfrac{1}{x}}{x^2}.

Set x = 3y^2 - 5y + 2 = 0 to get y = \dfrac{2}{3} and y = 1.

\dfrac{dx}{dy} = 6y - 5

Therefore,

\dfrac{dy}{dx} = \dfrac{1}{6y - 5}

When y = \dfrac{2}{3}, \dfrac{dy}{dx} = -1.

Since the metal is cubic, we have V = x^3.

Therefore, \dfrac{dV}{dx} = 3x^2.

We also have

\dfrac{dx}{dt} = 0.1

so

\dfrac{dV}{dt} = \dfrac{dV}{dx} \times \dfrac{dx}{dt} = 3x^2 \times 0.1 = 0.3x^2

Additional Resources

MME

Exam Tips Cheat Sheet

A Level
MME

Formula Booklet

A Level

You May Also Like...

A Level Maths Revision Cards

The best A level maths revision cards for AQA, Edexcel, OCR, MEI and WJEC. Maths Made Easy is here to help you prepare effectively for your A Level maths exams.

£14.99
View Product

A Level Maths – Cards & Paper Bundle

A level maths revision cards and exam papers for Edexcel. Includes 2022 predicted papers based on the advance information released in February 2022! MME is here to help you study from home with our revision cards and practise papers.

From: £22.99
View Product

Transition Maths Cards

The transition maths cards are a perfect way to cover the higher level topics from GCSE whilst being introduced to new A level maths topics to help you prepare for year 12. Your ideal guide to getting started with A level maths!

£8.99
View Product