13 Further differentiation
So far we have looked at differentiating powers of \(x\) when they are added together. This section introduces differentiating \(e^x\) and \(\ln{x}\), then goes on to look at how to differentiate, functions inside functions, products of functions (when functions are multiplied together) and quotients of functions (when functions are divided by each other).
13.1 Standard results
We can now expand our table of derivatives. Here are all the rules from the last differentiation along with some new ones.
Try some differentiation with some fractional powers:
13.2 The chain rule
The chain rule is used when we have functions inside other functions.
If we have a function of the form \(y=f(g(x))\), sometimes described as a function of a function, to calculate its derivative we need to use the chain rule:
\[ \frac{dy}{dx} = \frac{du}{dx} \times \frac{dy}{du} \]
This can be split up into steps:
Let \(u=g(x)\); Rewrite \(y\) in terms of \(u\), such that \(y=f(u)\); Calculate \(\frac{du}{dx}\) and \(\frac{dy}{du}\); Write \(\frac{dy}{dx}\) as a product of \(\frac{du}{dx}\) and \(\frac{dy}{du}\); Make sure \(\frac{dy}{dx}\) is only in terms of \(x\). Ensure any \(u\) terms have been replaced using the initial substitution.
Following this process, we must first identify \(g(x)\). Since the function is of the form \(y=f(g(x))\), we are looking for the ‘inner’ function.
So, for \(y=-(4x^2+1)^4\), \[ g(x)=4x^2+1. \]
If we now set \(u=g(x)\), we can rewrite \(y\) in terms of \(u\) such that \(y=f(u)\):
\[ y=-u^4 \]
Next, we calculate the two derivatives \(\frac{du}{dx}\) and \(\frac{dy}{du}\):
\[ \frac{du}{dx}=8x, \quad \frac{dy}{du}=-4u^3 \]
Plugging these into the chain rule:
\[ \begin{aligned} \frac{dy}{dx} &= \frac{du}{dx} \times \frac{dy}{du}, \\ &= 8x \times-4u^3, \\ &= -32xu^3. \end{aligned} \]
Finally, we need to express \(\frac{dy}{dx}\) only in terms of \(x\), so we must replace the \(u\) term using the initial substitution \(u=4x^2+1\):
\[ \frac{dy}{dx} =-32x(4x^2+1)^3. \]
Phew! Time for a cup of tea, or maybe some more questions…
A firm’s revenue is related to the quantity it sells. Suppose revenue is given by:
\[ R = (100 - 2q)^3 \]
where \(q\) is quantity. This is a simplified example, but it captures a common problem in economics: the variable \(q\) appears inside a linear expression \((100 - 2q)\), and that whole expression is then raised to a power. You cannot apply the basic power rule directly to \(q\) because \(q\) is not the base; it is inside the brackets.
To find the marginal revenue (the derivative of revenue with respect to quantity) we need the chain rule. The chain rule is used whenever one function is inside another.
Let \(u = 100 - 2q\), so that \(R = u^3\). Then:
\[ \frac{dR}{dq} = \frac{dR}{du} \times \frac{du}{dq} = 3u^2 \times (-2) = -6(100 - 2q)^2 \]
The marginal revenue tells us how much extra revenue the firm gets from selling one more unit. Because revenue depends on quantity through the composite function \((100 - 2q)^3\), the chain rule is essential.
13.3 The product rule
If we have a function of the form \(y=u(x)v(x)\), to calculate its derivative we need to use the product rule:
\[ \dfrac{dy}{dx} = u(x) \times \dfrac{dv}{dx} + v(x) \times\dfrac{du}{dx}. \]
This can be split up into steps:
Identify the functions \(u(x)\) and \(v(x)\); Calculate their derivatives \(\tfrac{du}{dx}\) and \(\tfrac{dv}{dx}\); Substitute these into the formula for the product rule to obtain an expression for \(\tfrac{dy}{dx}\); Simplify \(\tfrac{dy}{dx}\) where possible.
13.4 The quotient rule
If we have a function of the form \(y=\tfrac{u(x)}{v(x)}\), to calculate its derivative we need to use the quotient rule:
\[ \dfrac{dy}{dx} = \dfrac{v(x) \times \frac{du}{dx} - u(x) \times\frac{dv}{dx}}{[v(x)]^2}\,. \]
This can be split up into steps:
Identify the functions \(u(x)\) and \(v(x)\); Calculate their derivatives \(\tfrac{du}{dx}\) and \(\tfrac{dv}{dx}\); Substitute these into the formula for the quotient rule to obtain an expression for \(\tfrac{dy}{dx}\); Simplify \(\tfrac{dy}{dx}\) where possible.
Following this process, we must first identify \(u(x)\) and \(v(x)\).
For example if: