## Content

### The product, quotient and chain rules

We now move to some more involved properties of differentiation. To summarise, so far we have found that:

- the derivative of a constant multiple is the constant multiple of the derivative
- the derivative of a sum is the sum of the derivatives
- the derivative of a difference is the difference of the derivatives.

However, it turns out that:

- the derivative of a product \(f(x)\,g(x)\) is
*not*the product of the derivatives - the derivative of a quotient \(\dfrac{f(x)}{g(x)}\) is
*not*the quotient of the derivatives - the derivative of the composition \(f(g(x))\) is
*not*the composition of the derivatives.

The product, quotient and chain rules tell us how to differentiate in these three situations. We consider the three rules in turn.

#### The product rule

###### Theorem Product rule

Let \(f,g\) be differentiable functions. Then the derivative of their product is given by

\[ \dfrac{d}{dx} \bigl[ f(x)\,g(x) \bigr] = f(x)\,g'(x) + g(x)\,f'(x). \]The product rule is also often written as

\[ \dfrac{d}{dx} \bigl(fg\bigr) = f\,\dfrac{dg}{dx} + g\,\dfrac{df}{dx}. \]###### Proof

As before, we evaluate the limit which gives the derivative:

\[ \dfrac{d}{dx} \bigl[ f(x)\,g(x) \bigr] = \lim_{\Delta x \to 0} \dfrac{f(x+\Delta x)\,g(x+\Delta x) - f(x)\,g(x)}{\Delta x}. \]The trick is to add and subtract an extra term in the numerator, so that we can factorise and obtain some familiar-looking expressions:

\begin{align*} & f(x+\Delta x)\,g(x+\Delta x) - f(x)\,g(x) \\ ={}& f(x+\Delta x)\,g(x+\Delta x) - f(x)\,g(x + \Delta x) + f(x)\,g( x + \Delta x) - f(x)\,g(x) \\ ={}& \bigl[ f(x+\Delta x) - f(x) \bigr] g(x+\Delta x) + f(x) \bigl[ g(x+\Delta x) - g(x) \bigr]. \end{align*}We can then rewrite the limit as

\[ \lim_{\Delta x \to 0} \Bigl[ \dfrac{f(x+\Delta x) - f(x)}{\Delta x}\,g(x+\Delta x) + f(x)\,\dfrac{g(x+\Delta x) - g(x)}{\Delta x} \Bigr]. \]Now, since the limit of a sum is the sum of the limits, and the limit of a product is the product of the limits, we obtain

\[ \Bigl(\lim_{\Delta x \to 0} \dfrac{f(x+\Delta x) - f(x)}{\Delta x}\Bigr)\Bigl(\lim_{\Delta x \to 0} g(x+\Delta x)\Bigr) + f(x)\Bigl(\lim_{\Delta x \to 0} \dfrac{g(x+\Delta x) - g(x)}{\Delta x}\Bigr). \]We also used the fact that \(f(x)\) does not depend on \(\Delta x\). Recognising \(f'(x)\) and \(g'(x)\), and substituting \(\Delta x = 0\) into \(g(x+\Delta x)\), we obtain

\[ \dfrac{d}{dx} \bigl[ f(x)\,g(x) \bigr] = f'(x)\,g(x) + f(x)\,g'(x), \]which is equivalent to the desired formula.

\(\Box\)

##### Exercise 8

Starting from the fact that the derivative of \(x\) is 1, use the product rule to prove by induction on \(n\), that for all positive integers \(n\),

\[ \dfrac{d}{dx} \bigl( x^n \bigr) = nx^{n-1}. \]#### Example

Let \(f(x) = (x^3 + 2) (x^2 + 1)\). Find \(f'(x)\).

#### Solution

We could expand out \(f(x)\) and differentiate term-by-term. Alternatively, with the product rule, we obtain

\begin{align*} f'(x) &= (x^3 + 2)\,\dfrac{d}{dx}\bigl(x^2 + 1\bigr) + (x^2 + 1)\,\dfrac{d}{dx}\bigl(x^3 + 2\bigr) \\ &= (x^3 + 2)\cdot 2x + (x^2 + 1)\cdot 3x^2 \\ &= 5x^4 + 3x^2 + 4x. \end{align*}##### Exercise 9

Using the product rule, prove that in general, for a differentiable function \(f\colon \mathbb{R} \to \mathbb{R}\), the derivative of \((f(x))^2\) with respect to \(x\) is \(2\,f(x)\,f'(x)\).

##### Exercise 10

By using the product rule, prove the following 'extended product rule':

\[ \dfrac{d}{dx} \bigl[ f(x)\,g(x)\,h(x) \bigr] = f'(x)\,g(x)\,h(x) + f(x)\,g'(x)\,h(x) + f(x)\,g(x)\,h'(x). \]Generalise to the product of any number of functions.

#### The chain rule

The chain rule allows us to differentiate the composition of two functions. Recall from the module Functions II that the **composition** of two functions \(g\) and \(f\) is

We start with \(x\), apply \(g\), then apply \(f\). The chain rule tells us how to differentiate such a function.

###### Theorem Chain rule

Let \(f,g\) be differentiable functions. Then the derivative of their composition is

\[ \dfrac{d}{dx} \bigl[ f(g(x)) \bigr] = f'(g(x))\,g'(x). \]In Leibniz notation, we may write \(u = g(x)\) and \(y = f(u) = f(g(x))\); diagrammatically,

\[ x \stackrel{g}{\to} u \stackrel{f}{\to} y. \]Then the chain rule says that 'differentials cancel' in the sense that

\[ \dfrac{dy}{dx} = \dfrac{dy}{du}\,\dfrac{du}{dx}. \]###### Proof

To calculate the derivative, we must evaluate the limit

\[ \dfrac{d}{dx} \bigl[ f(g(x)) \bigr] = \lim_{\Delta x \to 0} \dfrac{f(g(x+\Delta x)) - f(g(x))}{\Delta x}. \]The trick is to multiply and divide by an extra term in the expression above, as shown, so that we obtain two expressions which both express rates of change:

\[ \dfrac{f(g(x+\Delta x)) - f(g(x))}{\Delta x} = \dfrac{f(g(x+\Delta x)) - f(g(x))}{g(x+\Delta x) - g(x)} \, \dfrac{g(x+\Delta x) - g(x)}{\Delta x}. \]We can then rewrite the desired limit as

\[ \Bigl(\lim_{\Delta x \to 0} \dfrac{f(g(x+\Delta x)) - f(g(x))}{g(x+\Delta x) - g(x)}\Bigr)\Bigl(\lim_{\Delta x \to 0} \dfrac{g(x+\Delta x) - g(x)}{\Delta x}\Bigr). \]The ratio in the first limit expresses the change in the function \(f\), from its value at \(g(x)\) to its value at \(g(x+\Delta x)\), relative to the difference between \(g(x+\Delta x)\) and \(g(x)\). So as \(\Delta x \to 0\), this first term approaches the derivative of \(f\) at the point \(g(x)\), namely \(f'(g(x))\). The second limit is clearly \(g'(x)\). We conclude that

\[ \dfrac{d}{dx} \bigl[ f(g(x)) \bigr] = f'(g(x))\,g'(x), \]as required.

\(\Box\)

The proof above is not entirely rigorous: for instance, if there are values of \(\Delta x\) close to zero such that \(g(x+\Delta x) - g(x) = 0\), then we have division by zero in the first limit. However, a fully rigorous proof is beyond the secondary school level.

The next two examples illustrate 'functional' and 'Leibniz' methods of attacking the same problem using the chain rule.

#### Example

Let \(f(x) = (x^7 - x^2)^{42}\). Find \(f'(x)\).

#### Solution

The function \(f(x)\) is the composition of the functions \(g(x) = x^7 - x^2\) and \(h(x) = x^{42}\), that is, \(f(x) = h(g(x))\). We compute

\[ g'(x) = 7x^6 - 2x, \qquad h'(x) = 42x^{41}, \]and the chain rule gives

\begin{align*} f'(x) &= h'(g(x))\,g'(x) \\ &= 42(x^7 - x^2)^{41}(7x^6 - 2x). \end{align*}#### Example

Let \(y = (x^7 - x^2)^{42}\). Find \(\dfrac{dy}{dx}\).

#### Solution

Let \(u = x^7 - x^2\), so that \(y = u^{42}\). We then have

\[ \dfrac{dy}{du} = 42u^{41}, \qquad \dfrac{du}{dx} = 7x^6 - 2x, \]and the chain rule gives

\[ \dfrac{dy}{dx} = \dfrac{dy}{du}\,\dfrac{du}{dx} = 42u^{41} (7x^6 - 2x). \]Rewriting \(u\) in terms of \(x\) gives

\[ \dfrac{dy}{dx} = 42(x^7 - x^2)^{41}(7x^6 - 2x). \]##### Exercise 12

In exercise 9, we proved that the derivative of \((f(x))^2\) with respect to \(x\) is \(2\,f(x)\,f'(x)\). Re-prove this fact using the chain rule.

##### Exercise 13

Prove the following 'extended chain rule':

\[ \dfrac{d}{dx} \bigl[ f(g(h(x))) \bigr] = f'(g(h(x)))\,g'(h(x))\,h'(x). \]Generalise to the composition of any number of functions.

The following exercise shows how, if you know the derivative of \(x^n\) for a positive number \(n\), you can find the derivative of \(x^{-n}\).

##### Exercise 14

Let \(g(x) = x^n\), where \(n\) is positive. Using the facts \(g'(x) = nx^{n-1}\) and \(\dfrac{d}{dx} \Biggl(\dfrac{1}{x}\Biggr) = -\dfrac{1}{x^2}\) and the chain rule, calculate \(\dfrac{d}{dx} \Biggl(\dfrac{1}{x^n}\Biggr)\).

#### The quotient rule

###### Theorem Quotient rule

Let \(f,g\) be differentiable functions. Then the derivative of their quotient is

\[ \dfrac{d}{dx} \Biggl(\dfrac{f(x)}{g(x)}\Biggr) = \dfrac{g(x)\,f'(x) - f(x)\,g'(x)}{(g(x))^2}. \]Alternatively, we can write

\[ \dfrac{d}{dx} \Biggl( \dfrac{f}{g} \Biggr) = \dfrac{g\,\dfrac{df}{dx} - f\,\dfrac{dg}{dx}}{g^2}. \]#### Example

Let \(f(x) = \dfrac{x^2 + 1}{x^2 - 1}\). What is \(f'(x)\)?

#### Solution

Using the quotient rule, we have

\begin{align*} f'(x) &= \dfrac{(x^2 - 1)\,\dfrac{d}{dx}(x^2 + 1) - (x^2 + 1)\,\dfrac{d}{dx}(x^2 - 1)}{(x^2 - 1)^2} \\ &= \dfrac{(x^2 - 1)\cdot 2x - (x^2 + 1)\cdot 2x}{(x^2 - 1)^2} = \dfrac{-4x}{(x^2 - 1)^2}. \end{align*}#### Example

Let \(f(x) = \dfrac{x}{\sqrt{x^2+1}}\). Find \(f'(x)\).

#### Solution

We first apply the quotient rule:

\[ f'(x) = \dfrac{\sqrt{x^2+1}\,\dfrac{d}{dx}(x) - x\,\dfrac{d}{dx}\sqrt{x^2+1}}{x^2 + 1}. \]To differentiate \(\sqrt{x^2 + 1}\), we use the chain rule:

\begin{align*} \dfrac{d}{dx} (x^2 + 1)^{\frac{1}{2}} &= \dfrac{1}{2}(x^2 + 1)^{-\frac{1}{2}}\,\dfrac{d}{dx}(x^2 + 1) \\ &= \dfrac{1}{2}(x^2+1)^{-\frac{1}{2}}(2x) \\ &= x (x^2 + 1)^{-\frac{1}{2}}. \end{align*}Now returning to \(f'(x)\), we obtain

\begin{align*} f'(x) &= \dfrac{\sqrt{x^2 + 1} \cdot 1 - x \cdot x (x^2 + 1)^{-\frac{1}{2}}}{x^2 + 1} \\ &= \dfrac{(x^2 + 1) - x^2}{(x^2 + 1)^{\frac{3}{2}}} \\ &= (x^2 + 1)^{-\frac{3}{2}}. \end{align*}The quotient rule can be proved using the product and chain rules, as the next two exercises show.

##### Exercise 15

Let \(g\) be a differentiable function. Using the chain rule, show that

\[ \dfrac{d}{dx} \Biggl( \dfrac{1}{g(x)} \Biggr) = -\dfrac{g'(x)}{(g(x))^2}. \](This is a generalisation of exercise 14.)