The Product and Quotient Rules.
To simplify differentiation of more functions, we would like to be able to deal with products, quotients, compositions and inverses, much as we did with limits and continuity.
Warning: none of these derivatives are given by rules quite as simple as for limits. Only sums, differences and constant multiples work that simply.
Example 3.2.1. A: a cautionary one.
Compare
- the derivative of the product \(x \cdot x^2=x^3 \)
- the product of the derivatives of \(x \) and \(x^2 \text{.}\)
The Power Rules tells us that the derivative of this product is \(3x^2 \text{;}\) the product of the derivatives of \(x \) and \(x^2 \) is \(1 \cdot 2x = 2x \text{:}\) not the same!
The Derivative of a Product of Functions.
The Leibniz notation is nice here. Let \(u=f(x) \text{,}\) \(v=g(x) \text{,}\) and compute the derivative of the product \(uv\) using the formula
\begin{equation}\frac{d(uv)}{dx} = \lim_{\Delta x \to 0} \frac{\Delta (uv)}{\Delta x}\label{deriv-uv}\tag{3.2.1}\end{equation}
What is \(\Delta (uv) \text{?}\)
First look at what \(\Delta u \) and \(\Delta v \) are: \(\Delta u = f(x+\Delta x) - f(x) \text{,}\) so \(f(x+\Delta x) = u+\Delta u \text{,}\) and likewise \(g(x+\Delta x) = v+\Delta v \text{.}\)
Next, \(\Delta (uv)\) is the change in the value of the product \(f(x)g(x)\) as the argument changes from \(x \) to \(x+\Delta x\text{:}\)
\begin{equation*}\Delta (uv) = f(x+\Delta x)g(x+\Delta x) - f(x)g(x)= (u+\Delta u)(v+\Delta v) - uv= u \Delta v + v \Delta u + \Delta u \Delta v\end{equation*}
That is,
\begin{equation*}\Delta (uv) = u \Delta v + v \Delta u + \Delta u \Delta v\end{equation*}
The difference quotient in (3.2.1) is thus
\begin{equation}\frac{\Delta (uv)}{\Delta x}= \frac{u \Delta v + v \Delta u + \Delta u \Delta v}{\Delta x}= \frac{u \Delta v}{\Delta x}+\frac{v \Delta u}{\Delta x}+\frac{\Delta u \Delta v}{\Delta x}= u \frac{\Delta v}{\Delta x}+v \frac{\Delta u}{\Delta x} + \frac{\Delta u}{\Delta x} \frac{\Delta v}{\Delta x} \Delta x\label{diff-quot}\tag{3.2.2}\end{equation}
The two difference quotients here have limits as \(\Delta x \to 0\text{,}\) as do the factors multiplying them: \(\displaystyle \frac{\Delta u}{\Delta x} \to \frac{du}{dx} \text{,}\) \(\displaystyle \frac{\Delta v}{\Delta x} \to \frac{dv}{dx}\text{,}\) \(\Delta x \to 0\text{,}\) and \(u\) and \(v\) do not vary as \(\Delta x \to 0\text{.}\) Thus we can compute the limit as \(\Delta x \to 0\) in Equation (3.2.2):
Theorem 3.2.2. 1. The Product Rule for Derivatives.
\begin{equation*}\frac{d(uv)}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}, \text{ or } (fg)' = f' g + f g'.\end{equation*}
Example 3.2.3. (Example 1 in the text).
- If \(f(x)=xe^x\text{,}\) calculate its derivative \(f'(x)\text{.}\)
- Compute the second and third derivatives \(f''\) and \(f'''\) of this function.
- Compute all the derivatives \(f^{(n)}\) of this function.
Example 3.2.4. (Example 2 in the text).
Differentiate (compute the derivative of) the function \(f(t)=\sqrt{t}(a+bt).\) Do this two ways, with and without the Product Rule. Hint: as usual, it can help to rewrite roots as powers, and division by powers and roots as negative powers.
Sometimes, you only know the values of a function and its derivative at one point, like having measurements of the position and velocity of an object at one time. This can be enough to compute the derivative of another function got from the first with a product or such:
Example 3.2.5. (Example 3 in the text).
If \(f(x)=\sqrt{x}g(x)\text{,}\) calculate \(f'(x)\) in terms of \(x\) and \(g'(x)\text{,}\) and use this to find \(f'(4)\) given that \(g(4)=2\) and \(g'(4)=3\text{.}\)
The Derivative of the Reciprocal of a Function.
To deal with division, start with the simplest case, \(f(x) = 1/g(x)\text{,}\) and use a common strategy:
Rephrase a new problem in terms of a problem we have already solved.
In this case, we can restate the situation in terms of a product, and use the Product Rule. First, clear the denominator, getting \(f(x) g(x) = 1\text{.}\) Then use the product rule to get
\begin{equation*}f'(x) g(x) + f(x) g'(x) = 0.\end{equation*}
Solve for \(f'\) and substitute in \(f(x) = 1/g(x)\text{:}\)
\begin{equation*}f'(x) = -g'(x) f(x)/g(x) = -g'(x)/[g(x)]^2,\end{equation*}
or
\begin{equation*}\left(\frac{1}{g}\right)' = -\frac{g'}{g^2}.\end{equation*}
The minus sign goes with the fact that an increase in \(g\) will cause a decrease in \(1/g\text{.}\)
The Quotient Rule for Derivatives.
Now it is east to get a rule for the derivative of any quotient, by combining the product and reciprocal rules:\begin{equation*}\left( \frac{f}{g} \right)' = \left( f \cdot \frac{1}{g} \right)' = f' \cdot \frac{1}{g} + f \cdot \left(\frac{1}{g}\right)' = \frac{f'}{g} - f \cdot \frac{g'}{g^2} = \frac{f' g - f g'}{g^2}.\end{equation*}
Theorem 3.2.6. The Quotient Rule for Derivatives.
\begin{equation*}\left(\frac{f}{g}\right)' = \frac{f' g - f g'}{g^2}, \quad \text{or} \quad\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{\displaystyle \frac{d}{dx}[f(x)] \cdot g(x) - f(x) \cdot \frac{d}{dx}[g(x)]}{[g(x)]^2}\end{equation*}
OR
\begin{equation*}\begin{split}\left(\frac{f}{g}\right)' \amp= \frac{f' g - f g'}{g^2}, \quad \text{or} \quad\\\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] \amp= \frac{\displaystyle \frac{d}{dx}[f(x)] \cdot g(x) - f(x) \cdot \frac{d}{dx}[g(x)]}{[g(x)]^2}\end{split}\end{equation*}
Example 3.2.7. (Example 4 of the text, modified).
- Differentiate \(\displaystyle y= \frac{x^2+x-2}{x^3+6}.\)
- Find the equation of the tangent line to this curve at point \(P(-1,-2/5).\)
See Example 5 in the text.
Differentiation Facts So Far.
- \(\displaystyle \displaystyle \frac{d}{dx}(c) = 0\)
- \(\displaystyle \displaystyle \frac{d}{dx}(x^n) = n x^{n-1}\)
- \(\displaystyle \displaystyle \frac{d}{dx}(e^x) = e^x\)
- \(\displaystyle \displaystyle \frac{d}{dx}(a^x) = (\ln a)a^x\)
- \(\displaystyle \displaystyle \frac{d}{dx}(cf) = c \frac{df}{dx}\)
- \(\displaystyle \displaystyle \frac{d}{dx}(f+g) = \frac{df}{dx} + \frac{dg}{dx}\)
- \(\displaystyle \displaystyle \frac{d}{dx}(f-g) = \frac{df}{dx} - \frac{dg}{dx}\)
- \(\displaystyle \displaystyle \frac{d}{dx}(fg) = \frac{df}{dx}g + f \frac{dg}{dx}\)
- \(\displaystyle \displaystyle \frac{d}{dx}\left(\frac{f}{g}\right) = \frac{\displaystyle\frac{df}{dx}g - f \frac{dg}{dx}}{g^2}\)
What is missing so far? Mostly,
- Trigonometric and logarithmic functions, and
- compositions and inverses of functions.
These gaps will be filled in the next four sections.
