## 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

$(f \circ g)(x) = f(g(x)).$

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 11

Find the derivative of $$(x^2+7)^{100}$$ with respect to $$x$$.

##### 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.)

##### Exercise 16

Using the previous exercise and the product rule, find the derivative of $$f(x) \cdot \dfrac{1}{g(x)}$$, and hence prove the quotient rule.

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