Title: Default for 11.4 Differentiating Products and Quotients
Course: HELM Workbooks 2023-2024

Exercise

Here we are going to find the first derivative of \(y = x^2(\ln x) \sinh x\).

We can see that the three functions are

\[ f(x) = x^2, \hspace{1cm} g(x) = \ln x, \hspace{1cm} h(x) = \sinh x. \]

Now find the derivative of each of these functions

\( \dfrac{df}{dx} = \)

\( \dfrac{dg}{dx} = \)

\( \dfrac{dh}{dx} = \)

Finally, we have

\( \dfrac{dy}{dx} = \)
Feedback: We are having the function \(y = f(x) g(x) h(x)= x^2(\ln x) \sinh x.\) We can see that the three functions are: \( f(x) = x^2, \hspace{1cm} g(x) = \ln x, \hspace{1cm} h(x) = \sinh x. \) The derivatives of each of these functions are: \[ \dfrac{df}{dx} = [value],\hspace{1cm} \dfrac{dg}{dx} = \displaystyle[value], \hspace{1cm} \dfrac{dh}{dx} =[value]\] Finally, we have: \[\begin{align*} \dfrac{dy}{dx} &= \dfrac{df}{dx} g(x)h(x)+ f(x) \dfrac{dg}{dx} h(x) + f(x)g(x) \dfrac{dh}{dx} \\ &= [value]. \end{align*}\]

Exercise

Here we are going to find the first derivative of \(y = e^x \ln x\).

We can see that the two functions are

\[ f(x) = e^x \hspace{1cm} g(x) = \ln x \]

Now find the derivative of each of these functions

\( \dfrac{df}{dx} = \)

\( \dfrac{dg}{dx} = \)

Finally, we have

\( \dfrac{dy}{dx} = \)
Feedback: We have \(y = f(x)g(x)= e^x \ln x\), with \( f(x) = [value]\) and \(g(x) = [value] \). Then, \( \dfrac{df}{dx} = [value] \) and \( \dfrac{dg}{dx} = \displaystyle{[value]}\). So \( \dfrac{dy}{dx} = [value]([value]) + [value](\displaystyle{[value]}) = \displaystyle{[value]}\)

Exercise

Here we are going to find the first derivative of \(y(x) = \displaystyle [value].\)

We can factorise \(y(x)\) into the two factors

\[ f(x) = [value] \hspace{1cm} g(x) = [value]. \]

Now find the derivative of each of these factors.

\( \dfrac{df}{dx} = \)

\( \dfrac{dg}{dx} = \)

Finally, substitute into the product rule to determine

\( \dfrac{dy}{dx} = \)
Feedback: We have \(y = f(x)g(x)= \displaystyle [value]\), with \( f(x) = [value]\) and \(g(x) = \displaystyle [value] = x^[value]\). Then, \( \dfrac{df}{dx} = [value] \) and \( \dfrac{dg}{dx} = \displaystyle{[value]}\). So \[\begin{align*} \dfrac{dy}{dx} &= \dfrac{df}{dx} g(x)+ f(x) \dfrac{dg}{dx}\\ &= \displaystyle [value]\\ &= \displaystyle{[value]}. \end{align*}\]

Exercise

In each case find the derivative of the function

  1. \( y =[value][value] \)
\( \dfrac{dy}{dx} =\)
  2. \( y =[value][value] \)
\( \dfrac{dy}{dx} =\)
  3. \( y =\sin^2 x \)
\( \dfrac{dy}{dx} =\)
  4. \( y =[value][value] \)
\( \dfrac{dy}{dx} =\)
Feedback: 1. We have \(y = f(x)g(x)\), with \( f(x) = [value]\) and \(g(x) = [value] \). Then, \( \dfrac{df}{dx} = [value] \) and \( \dfrac{dg}{dx} = \displaystyle{[value]}\). So \( \dfrac{dy}{dx} = [value]([value]) + [value](\displaystyle{[value]}) = \displaystyle{[value]}\) 2. We have \(y = f(x)g(x)\), with \( f(x) = [value]\) and \(g(x) = [value] \). Then, \( \dfrac{df}{dx} = [value] \) and \( \dfrac{dg}{dx} = \displaystyle{[value]}\). So \( \dfrac{dy}{dx} = [value]([value]) + [value](\displaystyle{[value]}) = \displaystyle{[value]}\) 3. We have \(y = f(x)g(x)\), with \( f(x) = [value]\) and \(g(x) = [value] \). Then, \( \dfrac{df}{dx} = [value] \) and \( \dfrac{dg}{dx} = \displaystyle{[value]}\). So \( \dfrac{dy}{dx} = [value]([value]) + [value](\displaystyle{[value]}) = \displaystyle{[value]}\) 4. We have \(y = f(x)g(x)\), with \( f(x) = [value]\) and \(g(x) = [value] \). Then, \( \dfrac{df}{dx} = [value] \) and \( \dfrac{dg}{dx} = \displaystyle{[value]}\). So \( \dfrac{dy}{dx} = [value]([value]) + [value](\displaystyle{[value]}) = \displaystyle{[value]}\)

Exercise

In each case find the derivative of the function

1. \( y =\dfrac{x}{\cos (x)} \) (recall that \( \dfrac{1}{\cos(x)}=\sec(x)\)
\( \dfrac{dy}{dx} =\)
2. \( y =[value][value] \)
\( \dfrac{dy}{dx} =\)
3. Obtain the derivative of \(y = x e^{x} \tan x \) by using your answers to part (a) and (b)
\( \dfrac{dy}{dx} =\)
Feedback: 1. We have \(y = f(x)g(x)\), with \( f(x) = [value]\) and \(g(x) = \displaystyle [value] .\) Then, \( \dfrac{df}{dx} = [value] \) and \( \dfrac{dg}{dx} = \displaystyle{[value]} .\) So \[\begin{align*} \dfrac{dy}{dx} &= [value]\left([value]\right) +[value] \left( \displaystyle{[value]} \right) \\ &= \displaystyle{[value]}. \end{align*}\] 2. We have \(y = f(x)g(x)\), with \( f(x) = [value]\) and \(g(x) = [value] .\) Then, \( \dfrac{df}{dx} = [value] \) and \( \dfrac{dg}{dx} = \displaystyle{[value]}.\) So \[\begin{align*} \dfrac{dy}{dx} &= [value](\displaystyle{[value]}) + [value](\displaystyle{[value]}) \\ &= \displaystyle{[value]}. \end{align*}\] 3. The derivative of \(y = x e^{ x} \tan x = \dfrac{x}{\cos x} e^{x} \sin x \) is found by applying the product rule to the results of parts (a) and (b). So we have \(y = f(x)g(x)\), with \( f(x) = \displaystyle{[value]}\) and \(g(x) = [value] .\) Then, \( \dfrac{df}{dx} = \displaystyle{[value]} \) and \( \dfrac{dg}{dx} = \displaystyle{[value]}.\) So \[\begin{align*} \dfrac{dy}{dx} &= \displaystyle{[value]}(\displaystyle{[value]}) + [value](\displaystyle{[value]}) \\ &= \displaystyle{[value]}. \end{align*}\]

Differentiating a product

In previous Sections we have examined the process of differentiating functions. We found how to obtain the derivative of many commonly occurring functions. These are recorded in the following table (remember, arguments of trigonometric functions are assumed to be in radians).

\(y\) | \( \dfrac{dy}{dx}\)
\(x^n\) | \(nx^{n-1}\)
\(\sin ax\) | \(a \cos ax\)
\(\cos ax\) | \(-a \sin ax\)
\(\tan ax\) | \(a\sec^2 x\)
\( \sec ax \) | \(a \sec ax \tan ax\)
\(\ln ax\) | \(\dfrac{1}{x}\)
\(e^{ax}\) | \(ae^{ax}\)
\(\cosh ax\) | \(a \sinh ax\)
\(\sinh ax\) | \( a \cosh ax\)

In this Section we consider how to differentiate non-standard functions - in particular those which can be written as the product of standard functions. Being able to differentiate such functions depends upon the following Key Point.

Key Point

Product Rule

If \(y(x) = f(x)g(x)\) then \[ \dfrac{dy}{dx} = \dfrac{df}{dx}g(x) + f(x)\dfrac{dg}{dx}\text{ or }y' = f'g + fg'.\]

In slightly more compact notation this is if \(y = uv\) then \[ \dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx}\text{ or }y' = uv' + vu' .\]

It is not difficult to prove this from the limit definition of the derivative and it can be an instructive exercise to do so (hint: write the right hand side as a limit of a single ratio).  We shall concentrate on its use.

Example

Differentiate

1. \(y = x^2 \sin x \)
2. \(y = x \ln x\)

Solution

1. Writing \(f(x) = x^2\), \(g(x) = \sin x\), we have  \(\dfrac{df}{dx}= 2x; \dfrac{dg}{dx}= \cos x\)
and so \( \dfrac{dy}{dx} = 2x(\sin x) + x^2(\cos x) = x(2 \sin x + x \cos x). \)
2. Writing \(f(x) = x\), \(g(x) = \ln x\), we have  \(\dfrac{df}{dx}= 1; \dfrac{dg}{dx}= \dfrac{1}{x}\)
and so \( \dfrac{dy}{dx} = 1(\ln x) + x \left(\dfrac{1}{x}\right) = \ln x + 1.\)

The product rule with three factors

The rule for differentiating a product can be extended to any number of products. If, for example, \(y = f(x)g(x)h(x)\) then \begin{align} \dfrac{dy}{dx} &= \dfrac{df}{dx} [g(x)h(x)] + f(x)\dfrac{d}{dx}[g(x)h(x)] \\ &= \dfrac{df}{dx}g(x)h(x) + f(x) \left\{ \dfrac{dg}{dx}h(x) + g(x)\dfrac{dh}{dx} \right\} \\ &= \dfrac{df}{dx}g(x)h(x) + f(x)\dfrac{dg}{dx}h(x) + f(x)g(x)\dfrac{dh}{dx} \end{align} That is, each function in the product is differentiated in turn and the three results added together.

Example

If \(y = x e^{2x} \sin x\) then find \( \dfrac{dy}{dx}\).

Solution

Here \[ f(x) = x, g(x) = e^{2x}, h(x) = \sin x \] \[ \dfrac{df}{dx} = 1, \dfrac{dg}{dx} = 2e^{2x},\dfrac{dh}{dx} = \cos x, \] \[ \therefore \dfrac{dy}{dx} = 1(e^{2x} \sin x) + x(2e^{2x}) \sin x + xe^{2x}(\cos x)\] \[ = e^{2x}(\sin x + 2x \sin x + x \cos x) \]

Differentiating a quotient

In this Section we consider functions of the form \(y = \dfrac{f(x)}{g(x)}\). To find the derivative of such a function we make use of the following Key Point:

Key Point

Quotient Rule

If \(y = \dfrac{f(x)}{g(x)}\), then \[ \dfrac{dy}{dx} = \dfrac{ g(x) \dfrac{df}{dx} - \dfrac{dg}{dx}f(x)}{[g(x)]^2}\] or \(y' = \dfrac{ gf' - g'f}{g^2}\).

In slightly more compact notation we have that if \(y = \dfrac{u}{v}\), then \[ \dfrac{dy}{dx} = \dfrac{ v \dfrac{du}{dx} - \dfrac{dv}{dx}u}{v^2}\] or \(y' = \dfrac{ vu' - v'u}{v^2}\).

These are two equivalent versions of the quotient rule.

Example

Find the derivative of \(y = \dfrac{\ln x}{x}\).

Solution

Here \(f(x) = \ln x\) and \(g(x) = x\)

\[ \therefore \dfrac{df}{dx} = \dfrac{1}{x} \text{ and } \dfrac{dg}{dx} = 1 \] Hence \(\dfrac{dy}{dx} =\dfrac{x \left( \dfrac{1}{x} \right) - 1(\ln x) }{[x]^2} = \dfrac{1 - \ln x}{x^2}\)

Note that the product rule can also be used to differentiate this function by writing it in the form \(y = f(x)g(x) = (\ln x)(x^{-1})\).

Exercise

Here we are going to find the first derivative of \(y = \displaystyle{[value]}\) using the quotient rule.

We can see that the two functions are

\[ f(x) = [value], \hspace{1cm} g(x) = [value]. \]

Now find the derivative of each of these functions.

\( \dfrac{df}{dx} = \)

\( \dfrac{dg}{dx} = \)

Finally, we have

\( \dfrac{dy}{dx} = \)
Feedback: We have \(y = \dfrac{f(x)}{g(x)}\), with \( f(x) = [value]\) and \(g(x) = [value] \). Then, \( \dfrac{df}{dx} = [value] \) and \( \dfrac{dg}{dx} = \displaystyle{[value]}\). So \( \dfrac{dy}{dx} = \dfrac{ [value](\displaystyle{[value]}) - \displaystyle{[value]}([value])} {[value]} = \displaystyle{[value]} .\)

Exercise

Find the derivative of \(y = \displaystyle{[value]}\) using the quotient rule.

We can see that the two functions are

\[ f(x) = [value] \hspace{1cm} g(x) = [value] \]

Now find the derivative of each of these functions

\( \dfrac{df}{dx} = \)

\( \dfrac{dg}{dx} = \)

Finally, we have

\( \dfrac{dy}{dx} = \)
Feedback: We have \(y = \dfrac{f(x)}{g(x)}\), with \( f(x) = [value]\) and \(g(x) = [value] \). Then, \( \dfrac{df}{dx} = [value] \) and \( \dfrac{dg}{dx} = \displaystyle{[value]}\). So \[\begin{align*} \dfrac{dy}{dx} &= \dfrac{ ([value])(\displaystyle{[value]}) - \displaystyle{([value])}([value])} {[value]} \\ &= \displaystyle{[value]}. \end{align*}\]

Exercise

Find the derivative of \(y = \displaystyle{[value]}\) using the quotient rule.

We can see that the two functions are

\[ f(x) = [value] \hspace{1cm} g(x) = [value] \]

Now find the derivative of each of these functions

\( \dfrac{df}{dx} = \)

\( \dfrac{dg}{dx} = \)

Finally, we have

\( \dfrac{dy}{dx} = \)
Feedback: We have \(y = \dfrac{f(x)}{g(x)}\), with \( f(x) = [value]\) and \(g(x) = [value] \). Then, \( \dfrac{df}{dx} = [value] \) and \( \dfrac{dg}{dx} = \displaystyle{[value]}\). So \[\begin{align*} \dfrac{dy}{dx} &= \dfrac{ ([value])(\displaystyle{[value]}) - \displaystyle{([value])}([value])} {[value]} \\ &= \displaystyle{[value]} \\ &= \displaystyle{[value]}. \end{align*}\]

Exercise

Find the derivative of \(y = \displaystyle{[value]}\) using the quotient rule.

We can see that the two functions are

\[ f(x) = [value] \hspace{1cm} g(x) = [value] \]

Now find the derivative of each of these functions

\( \dfrac{df}{dx} = \)

\( \dfrac{dg}{dx} = \)

Finally, we have

\( \dfrac{dy}{dx} = \)
Feedback: We have \(y = \dfrac{f(x)}{g(x)}\), with \( f(x) = [value]\) and \(g(x) = [value] \). Then, \( \dfrac{df}{dx} = \displaysyle [value] \) and \( \dfrac{dg}{dx} = \displaystyle{[value]}\). So \[\begin{align*} \dfrac{dy}{dx} &= \dfrac{ \left( [value] \right) \left( \displaystyle{[value]} \right) - \displaystyle{\left( [value] \right)} \left( [value] \right)} {[value]} \\ &= \displaystyle{[value]}. \end{align*}\]