The derivative is a powerful tool but is admittedly awkward given its reliance on limits. Fortunately, one thing mathematicians are good at is abstraction. For instance, instead of continually finding derivatives at a point, we abstracted and found the derivative function.
Let's practice abstraction on linear functions, \(y=mx+b\text{.}\) What is \(y'\text{?}\) Without limits, recognize that linear function are characterized by being functions with a constant rate of change (the slope). The derivative, \(y'\text{,}\) gives the instantaneous rate of change; with a linear function, this is constant, \(m\text{.}\) Thus \(y'=m\text{.}\)
Let's abstract once more. Let's find the derivative of the general quadratic function, \(f(x) = ax^2+bx+c\text{.}\) Using the definition of the derivative, we have: \begin{align*} \fp(x) \amp = \lim_{h\to 0}\frac{a(x+h)^2+b(x+h)+c-(ax^2+bx+c)}{h}\\ \amp =\lim_{h\to 0}\frac{ax^2+2ahx+ah^2+bx+bh+c-ax^2-bx-c)}{h}\\ \amp = \lim_{h\to 0} \frac{ah^2+2ahx+bh}{h}\\ \amp = \lim_{h\to 0} ah+2ax+b\\ \amp = 2ax+b. \end{align*}
So if \(y = 6x^2+11x-13\text{,}\) we can immediately compute \(y' = 12x+11\text{.}\)
In this section (and in some sections to follow) we will learn some of what mathematicians have already discovered about the derivatives of certain functions and how derivatives interact with arithmetic operations. We start with a theorem.
\(\lzoo{x}{c} = 0\text{,}\) where \(c\) is a constant.
\(\lzoo{x}{x^n}= nx^{n-1}\text{,}\) where \(n\) is an integer, \(n>0\text{.}\)
\(\lzoo{x}{\sin(x)} = \cos(x)\)
\(\lzoo{x}{\cos(x)} = {-\sin(x)}\)
\(\lzoo{x}{e^x} = e^x\)
\(\lzoo{x}{\ln(x)} = \frac{1}{x}\text{,}\) for \(x\gt 0\text{.}\)
This theorem starts by stating an intuitive fact: constant functions have zero rate of change as they are constant. Therefore their derivative is \(0\) (they change at the rate of \(0\)). The theorem then states some fairly amazing things. The Power Rule states that the derivatives of Power Functions (of the form \(y=x^n\)) are very straightforward: multiply by the power, then subtract \(1\) from the power. We see something incredible about the function \(y=e^x\text{:}\) it is its own derivative. We also see a new connection between the sine and cosine functions.
One special case of the Power Rule is when \(n=1\text{,}\) i.e., when \(f(x) = x\text{.}\) What is \(\fp(x)\text{?}\) According to the Power Rule, \begin{equation*} \fp(x) = \lzoo{x}{x} = \lzoo{x}{x^1} = 1\cdot x^0 = 1. \end{equation*}
In words, we are asking “At what rate does \(f\) change with respect to \(x\text{?}\)” Since \(f\) is \(x\text{,}\) we are asking “At what rate does \(x\) change with respect to \(x\text{?}\)” The answer is: \(1\text{.}\) They change at the same rate. We can also interpret the derivative as the slope of the tangent line to the function at a point \((c,f(c))\text{.}\) Since \(f(x)=x\) is a linear function with constant slope \(1\text{,}\) we can say that the derivative of \(f(x)=x\) is \(\fp(x)=1\text{.}\)
Let's practice using this theorem.
Let \(f(x)=x^3\text{.}\)
Find \(\fp(x)\text{.}\)
Find the equation of the line tangent to the graph of \(f\) at \(x=-1\text{.}\)
Use the tangent line to approximate \((-1.1)^3\text{.}\)
Sketch \(f\text{,}\) \(\fp\) and the tangent line from 2 on the same axis.
Theorem 2.3.1 gives useful information, but we will need much more. For instance, using the theorem, we can easily find the derivative of \(y=x^3\text{,}\) but it does not tell how to compute the derivative of \(y=2x^3\text{,}\) \(y=x^3+\sin(x)\) nor \(y=x^3\sin(x)\text{.}\) The following theorem helps with the first two of these examples (the third is answered in the next section).
Let \(f\) and \(g\) be differentiable on an open interval \(I\) and let \(c\) be a real number. Then:
\begin{align*} \lzoo{x}{f(x) \pm g(x)} \amp= \lzoo{x}{f(x)} \pm \lzoo{x}{g(x)}\\ \amp= \fp(x)\pm g'(x) \end{align*}
\begin{align*} \lzoo{x}{c\cdot f(x)} \amp= c\cdot\lzoo{x}{f(x)}\\ \amp = c\cdot\fp(x)\text{.} \end{align*}
Theorem 2.3.4 allows us to find the derivatives of a wide variety of functions. It can be used in conjunction with the Power Rule to find the derivatives of any polynomial. Recall in Example 2.1.17 that we found, using the limit definition, the derivative of \(f(x) = 3x^2+5x-7\text{.}\) We can now find its derivative without expressly using limits: \begin{align*} \lzoo{x}{3x^2+5x+7} \amp = 3\lzoo{x}{x^2} + 5\lzoo{x}{x} + \lzoo{x}{7}\\ \amp = 3\cdot 2x+5\cdot 1+ 0\\ \amp = 6x+5. \end{align*}
We were a bit pedantic here, showing every step. Normally we would do all the arithmetic and steps in our head and readily find \(\lzoo{x}{3x^2+5x+7}= 6x+5\text{.}\)
Let \(f(x) = \sin(x) + 2x+1\text{.}\) Approximate \(f(3)\) using an appropriate tangent line.
The derivative of a function \(f\) is itself a function, therefore we can take its derivative. The following definition gives a name to this concept and introduces its notation.
Let \(y=f(x)\) be a differentiable function on \(I\text{.}\)
The second derivative of \(f\) is: \begin{equation*} \fp'(x) = \lzoo{x}{\fp(x)} = \lzoo{x}{\lz{y}{x}} = \lzn{y}{x}{2}=y''. \end{equation*}
The third derivative of \(f\) is: \begin{equation*} \fp''(x) = \lzoo{x}{\fp'(x)} = \lzoo{x}{\lzn{y}{x}{2}} = \lzn{y}{x}{3}=y'''. \end{equation*}
The \(n\)th derivative of \(f\) is: \begin{equation*} f^{(n)}(x) = \lzoo{x}{f^{(n-1)}(x)} = \lzoo{x}{\lzn{y}{x}{n-1}} = \lzn{y}{x}{n}=y^{(n)}. \end{equation*}
Definition 2.3.8 comes with the caveat “Where the corresponding limits exist.” With \(f\) differentiable on \(I\text{,}\) it is possible that \(\fp\) is not differentiable on all of \(I\text{,}\) and so on.
In general, when finding the fourth derivative and on, we resort to the \(f^{(4)}(x)\) notation, not \(\fp'''(x)\text{;}\) after a while, too many ticks is too confusing.
Let's practice using this new concept.
Find the first four derivatives of the following functions:
\(f(x) = 4x^2\)
\(f(x) = \sin(x)\)
\(f(x) = 5e^x\)
What do higher order derivatives mean? What is the practical interpretation?
Our first answer is a bit wordy, but is technically correct and beneficial to understand. That is,
The second derivative of a function \(f\) is the rate of change of the rate of change of \(f\text{.}\)
One way to grasp this concept is to let \(f\) describe a position function. Then, as stated in Key Idea 2.2.3, \(\fp\) describes the rate of position change: velocity. We now consider \(\fp'\text{,}\) which describes the rate of velocity change. Sports car enthusiasts talk of how fast a car can go from \(0\) to 60 mph; they are bragging about the acceleration of the car.
We started this chapter with amusement-park riders free-falling with position function \(f(t) = -16t^2+150\text{.}\) It is easy to compute \(\fp(t)=-32t\) ft/s and \(\fp'(t) = -32\) (ft/s)/s. We may recognize this latter constant; it is the acceleration due to gravity. In keeping with the unit notation introduced in the previous section, we say the units are “feet per second per second.” This is usually shortened to “feet per second squared,” written as “ft/s\(^2\text{.}\)”
It can be difficult to consider the meaning of the third, and higher order, derivatives. The third derivative is “the rate of change of the rate of change of the rate of change of \(f\text{.}\)” That is essentially meaningless to the uninitiated. In the context of our position/velocity/acceleration example, the third derivative is the “rate of change of acceleration,” commonly referred to as “jerk.”
Make no mistake: higher order derivatives have great importance even if their practical interpretations are hard (or “impossible”) to understand. The mathematical topic of series makes extensive use of higher order derivatives.
Terms and Concepts
In the following exercises, compute the derivative of the given function.
In the following exercises, compute the first four derivatives of the given function.
In the following exercises, find the equations of the tangent and normal lines to the graph of the function at the given point.
Review