2.4: Arithmetic of Derivatives - a Differentiation Toolbox
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So far, we have evaluated derivatives only by applying Definition 2.2.1 to the function at hand and then computing the required limits directly. It is quite obvious that as the function being differentiated becomes even a little complicated, this procedure quickly becomes extremely unwieldy. It is many orders of magnitude more efficient to have access to
- a list of derivatives of some simple functions and
- a collection of rules for breaking down complicated derivative computations into sequences of simple derivative computations.
This is precisely what we did to compute limits. We started with limits of simple functions and then used “arithmetic of limits” to computed limits of complicated functions.
We have already started building our list of derivatives of simple functions. We have shown, in Examples 2.2.2, 2.2.3, 2.2.5 and 2.2.9, that
\begin{align*} \frac{\mathrm{d} }{\mathrm{d} x} 1 &= 0 & \frac{\mathrm{d} }{\mathrm{d} x} x &= 1 & \frac{\mathrm{d} }{\mathrm{d} x} x^2 &= 2x &\frac{\mathrm{d} }{\mathrm{d} x} \sqrt{x} &= \frac{1}{2\sqrt{x}} \end{align*}
We'll expand this list later.
We now start building a collection of tools that help reduce the problem of computing the derivative of a complicated function to that of computing the derivatives of a number of simple functions. In this section we give three derivative “rules” as three separate theorems. We'll give the proofs of these theorems in the next section and examples of how they are used in the following section.
As was the case for limits, derivatives interact very cleanly with addition, subtraction and multiplication by a constant. The following result actually follows very directly from the first three points of Theorem 1.4.3.
Let \(f(x),g(x)\) be differentiable functions and let \(c \in \mathbb{R}\) be a constant. Then
\begin{align*} \frac{\mathrm{d} }{\mathrm{d} x} \big\{ f(x)+g(x) \big\} &= f'(x)+g'(x)\\ \frac{\mathrm{d} }{\mathrm{d} x} \big\{ f(x)-g(x)\big\} &= f'(x)-g'(x)\\ \frac{\mathrm{d} }{\mathrm{d} x} \big\{ c f(x) \big\} &= c f'(x) \end{align*}
That is, the derivative of the sum is the sum of the derivatives, and so forth.
Following this we can combine the three statements in this lemma into a single rule which captures the “linearity of differentiation”.
Again, let \(f(x),g(x)\) be differentiable functions, let \(\alpha, \beta \in \mathbb{R}\) be constants and define the “linear combination”
\begin{align*} S(x) &= \alpha f(x) + \beta g(x). \end{align*}
Then the derivative of \(S(x)\) at \(x=a\) exists and is
\begin{align*} \frac{\mathrm{d} S}{\mathrm{d} x} = S'(x) &= \alpha f'(x) + \beta g'(x). \end{align*}
Note that we can recover the three rules in the previous lemma by setting \(\alpha=\beta=1\) or \(\alpha=1, \beta=-1\) or \(\alpha=c\text{,}\) \(\beta=0\text{.}\)
Unfortunately, the derivative does not act quite as simply on products or quotients. The rules for computing derivatives of products and quotients get their own names and theorems:
Let \(f(x),g(x)\) be differentiable functions, then the derivative of the product \(f(x)g(x)\) exists and is given by
\begin{align*} \frac{\mathrm{d} }{\mathrm{d} x} \big\{ f(x) \, g(x) \big\} &= f'(x) \, g(x) + f(x) \, g'(x). \end{align*}
Before we proceed to the derivative of the ratio of two functions, it is worth noting a special case of the product rule when \(g(x)=f(x)\text{.}\) In fact, since this is a useful special case, let us call it a corollary 1:
Let \(f(x)\) be a differentiable function, then the derivative of its square is:
\begin{align*} \frac{\mathrm{d} }{\mathrm{d} x} \big\{ f(x)^2 \big\} &= 2\, f(x)\, f'(x) \end{align*}
With a little work this can be generalised to other powers — but that is best done once we understand how to compute the derivative of the composition of two functions. That requires the chain rule (see Theorem 2.9.2 below). But before we get to that, we need to see how to take the derivative of a quotient of two functions.
Let \(f(x), g(x)\) be differentiable functions. Then the derivative of their quotient is
\begin{align*} \frac{\mathrm{d} }{\mathrm{d} x} \left\{ \frac{f(x)}{g(x)} \right\} &= \frac{f'(x) \, g(x) - f(x) \, g'(x)}{g(x)^2}. \end{align*}
This derivative exists except at points where \(g(x)=0\text{.}\)
There is a useful special case of this theorem which we obtain by setting \(f(x)=1\text{.}\) In that case, the quotient rule tells us how to compute the derivative of the reciprocal of a function.
Let \(g(x)\) be a differentiable function. Then the derivative of the reciprocal of \(g\) is given by
\begin{align*} \frac{\mathrm{d} }{\mathrm{d} x} \left\{ \frac{1}{g(x)} \right\} &= -\frac{g'(x)}{g(x)^2} \end{align*}
and exists except at those points where \(g(x)=0\text{.}\)
So we have covered, sums, differences, products and quotients. This allows us to compute derivatives of many different functions — including polynomials and rational functions. However we are still missing trigonometric functions (for example), and a rule for computing derivatives of compositions. These will follow in the near future, but there are a couple of things to do before that — understand where the above theorems come from, and practice using them.
Exercises
Stage 1
True or false: \(\frac{\mathrm{d} }{\mathrm{d} x}\{f(x)+g(x)\}=f'(x)+g'(x)\) when \(f\) and \(g\) are differentiable functions.
True or false: \(\frac{\mathrm{d} }{\mathrm{d} x}\{f(x)g(x)\}=f'(x)g'(x)\) when \(f\) and \(g\) are differentiable functions.
True or false: \(\frac{\mathrm{d} }{\mathrm{d} x}\left\{\dfrac{f(x}{g(x)}\right\}=\dfrac{f'(x)}{g(x)}-\dfrac{f(x)g'(x)}{g^2(x)}\) when \(f\) and \(g\) are differentiable functions.
Let \(f\) be a differentiable function. Use at least three different rules to differentiate \(g(x)=3f(x)\text{,}\) and verify that they all give the same answer.
Stage 2
Differentiate \(f(x)=3x^5+4x^{2/3}\text{.}\)
Use the product rule to differentiate \(f(x)=(2x+5)(8\sqrt{x}-9x)\text{.}\)
Find the equation of the tangent line to the graph of \(y=x^3\) at \(x=\dfrac{1}{2}\text{.}\)
A particle moves along the \(x\)--axis so that its position at time \(t\) is given by \(x=t^3-4t^2+1\).
- At \(t=2\text{,}\) what is the particle's speed?
- At \(t=2\text{,}\) in what direction is the particle moving?
- At \(t=2\text{,}\) is the particle's speed increasing or decreasing?
Calculate and simplify the derivative of \(\dfrac{2x-1}{2x+1}\)
What is the slope of the graph \(y=\left(\dfrac{3x+1}{3x-2}\right)^2\) when \(x=1\text{?}\)
Find the equation of the tangent line to the curve \(f(x)=\dfrac{1}{\sqrt{x}+1}\) at the point \(\left(1,\frac{1}{2}\right)\text{.}\)
Stage 3
A town is founded in the year 2000. After \(t\) years, it has had \(b(t)\) births and \(d(t)\) deaths. Nobody enters or leaves the town except by birth or death (whoa). Give an expression for the rate the population of the town is growing.
Find all points on the curve \(y=3x^2\) where the tangent line passes through \((2,9)\text{.}\)
Evaluate \(\displaystyle \lim_{y\rightarrow 0}\left( \dfrac{\sqrt{100180+y}-\sqrt{100180}}{y}\right)\) by interpreting the limit as a derivative.
A rectangle is growing. At time \(t=0\text{,}\) it is a square with side length 1 metre. Its width increases at a constant rate of 2 metres per second, and its length increases at a constant rate of 5 metres per second. How fast is its area increasing at time \(t \gt 0\text{?}\)
Let \(f(x)=x^2g(x)\) for some differentiable function \(g(x)\text{.}\) What is \(f'(0)\text{?}\)
Verify that differentiating \(f(x)=\dfrac{g(x)}{h(x)}\) using the quotient rule gives the same answer as differentiating \(f(x)=\dfrac{g(x)}{k(x)}\cdot\dfrac{k(x)}{h(x)}\) using the product rule and the quotient rule.