# 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

## 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 Level

## 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 Level

## 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 Level
A Level

## 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 Level

## 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 Level

## 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

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

A Level

A Level

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