1.10: The Derivative as a Function

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Note to the Instructor (click to expand)

In this section, focus on the limit definition of the derivative as a function (as opposed to at a number). Make sure that students learn how to graph the derivative given the graph of the original function and that they know (at the very least) Lagrange and Leibniz notations. Connect derivatives and continuity (differentiability implies continuity—not the other way around). Finally, introduce higher-order derivatives. Derivative tricks and shortcuts are not to be used in this section!

Core Prerequisites (i.e., "Understanding the Lesson"): The following prerequisite topics (which have not already been listed as prerequisites in previous sections of this text) are required to understand the core concepts in this lesson.

(click to expand)

• Graphing Functions
  • Intervals of Increase and Decrease

Operational Prerequisites (i.e., "Completing the Homework"): The following prerequisite topics (which have not already been listed as prerequisites in previous sections of this text) are needed to complete the homework.

(click to expand)

• Simplifying Expressions
  • Adding and Subtracting Polynomials

The following is a list of learning objectives for this section.

Learning Objectives (click to expand)

• Define the derivative function of a given function.
• Graph a derivative function from the graph of a given function.
• State the connection between derivatives and continuity.
• Describe three conditions when a function has no derivative.
• Explain the meaning of a higher-order derivative.

As we have seen, the derivative of a function at a given point gives us the rate of change or slope of the tangent line to the function at that point. We obtain the velocity if we differentiate a position function at a given time. Knowing the function's derivative at every point produces valuable information about the function's behavior. However, finding the derivative at even a handful of values using the techniques of the preceding section would quickly become quite tedious. In this section, we define the derivative function and learn a process for finding it.

Derivative Functions

The derivative function gives the derivative of a function at each point in the domain of the original function for which the derivative is defined. We can formally define a derivative function as follows.

Definition: Derivative Function

Let \(f\) be a function. The derivative function, denoted by \(f^{\prime}\), is the function whose domain consists of those values of \(x\) such that the following limit exists:\[f^{\prime}(x)=\lim_{h \to 0}\dfrac{f(x+h)-f(x)}{h}. \label{derdef}\]

A function \(f(x)\) is said to be differentiable at \(a\) if \(f^{\prime}(a)\) exists. More generally, a function is said to be differentiable on \(S\) if it is differentiable at every point in an open set \(S\), and a differentiable function is one in which \(f^{\prime}(x)\) exists on its domain.

In the next few examples we use Equation \ref{derdef} to find the derivative of a function.

Example \(\PageIndex{1}\): Finding the Derivative of a Square-Root Function

Find the derivative of \(f(x)=\sqrt{x}\).

Solution

Start directly with the definition of the derivative function.

Substitute \(f(x+h)=\sqrt{x+h}\) and \(f(x)=\sqrt{x}\) into \(f^{\prime}(x)= \displaystyle \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}\).\[\begin{array}{rcl} f^{\prime}(x) & = & \displaystyle \lim_{h \to 0}\dfrac{\sqrt{x+h}-\sqrt{x}}{h} \\[16pt] & = & \displaystyle\lim_{h \to 0}\dfrac{\left( \sqrt{x+h}-\sqrt{x} \right)}{h} \cdot \dfrac{\left( \sqrt{x+h}+\sqrt{x} \right)}{\left( \sqrt{x+h}+\sqrt{x} \right)} \\[16pt] & = & \displaystyle\lim_{h \to 0}\dfrac{h}{h\left(\sqrt{x+h}+\sqrt{x}\right)} \\[16pt] & = & \displaystyle\lim_{h \to 0}\dfrac{\cancelto{1}{h}}{\cancel{h}\left(\sqrt{x+h}+\sqrt{x}\right)} \\[16pt] & = & \displaystyle\lim_{h \to 0}\dfrac{1}{\sqrt{x+h}+\sqrt{x}} \\[16pt] & \overset{\text{D.S.}}{=} & \dfrac{1}{2\sqrt{x}} \\ \end{array}\nonumber\]

Example \(\PageIndex{2}\): Finding the Derivative of a Quadratic Function

Find the derivative of the function \(f(x)=x^2-2x\).

Solution

Follow the same procedure, but without multiplying by the conjugate.

Substitute \(f(x+h)=(x+h)^2-2(x+h)\) and \(f(x)=x^2-2x\) into \(f^{\prime}(x)= \displaystyle \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}\).\[\begin{array}{rcl} f^{\prime}(x) & = & \displaystyle \lim_{h \to 0}\dfrac{((x+h)^2-2(x+h))-(x^2-2x)}{h} \\[16pt] & = & \displaystyle\lim_{h \to 0}\dfrac{x^2+2xh+h^2-2x-2h-x^2+2x}{h} \\[16pt] & = & \displaystyle\lim_{h \to 0}\dfrac{2xh-2h+h^2}{h} \\[16pt] & = & \displaystyle\lim_{h \to 0}\dfrac{h(2x-2+h)}{h} \\[16pt] & = & \displaystyle\lim_{h \to 0}\dfrac{\cancelto{1}{h}(2x-2+h)}{\cancelto{1}{h}} \\[16pt] & = & \displaystyle\lim_{h \to 0}(2x-2+h) \\[16pt] & \overset{\text{D.S.}}{=} & 2x-2 \\ \end{array}\nonumber\]

Checkpoint \(\PageIndex{2}\)

Find the derivative of \(f(x)=x^2\).

Answer

\(f^{\prime}(x)=2x\)

Notations for the Derivative

There is no single uniform notation for differentiation. Instead, various notations for the derivative of a function or variable have been proposed by various mathematicians. The usefulness of each notation varies with the context, and it is sometimes advantageous to use more than one notation in a given context.

Lagrange's Notation: \(f^{\prime}(x)\)

One of the most common modern notations for differentiation is named after Joseph Louis Lagrange, even though Euler invented it and Lagrange just popularized it. In Lagrange's notation, a prime mark denotes a derivative. If \(y = f(x)\) is a function, then its derivative evaluated at \(x\) is written\[\displaystyle f^{\prime}(x) \quad \text{(a common alternative form is } y^{\prime} \text{)}\nonumber\] Lagrange's notation first appeared in print in 1749.

If the derivative is evaluated at \(x =a\), then the value is written as \(f^{\prime}(a)\) or \(y^{\prime}(a)\).

Leibniz's Notation: \(\dfrac{dy}{dx}\)

The original notation employed by Gottfried Leibniz is still frequently used throughout mathematics. It is prevalent when the equation \(y = f(x)\) is regarded as a functional relationship between dependent and independent variables \(y\) and \(x\). Leibniz's notation makes this relationship explicit by writing the derivative as\[\displaystyle \dfrac{dy}{dx} \quad \left(\text{common alternative forms are }\dfrac{df(x)}{dx}, \dfrac{df}{dx}(x), \text{ and } \dfrac{d}{dx}f(x) \right)\nonumber\]In Example \(\PageIndex{2}\), we could have expressed the derivative as \(\frac{dy}{dx}=2x-2\). We could have conveyed the same information by writing \(\frac{d}{dx}\left(x^2-2x\right)=2x-2\).

To understand this notation better, recall that the derivative of a function at a point is the limit of the slopes of secant lines as the secant lines approach the tangent line. The slopes of these secant lines are often expressed in the form \(\frac{\Delta y}{\Delta x}\) where \(\Delta y\) is the difference in the \(y\) values corresponding to the difference in the \(x\) values, which are expressed as \(\Delta x\) (Figure \(\PageIndex{1}\)). Thus the derivative, which can be thought of as the instantaneous rate of change of \(y\) with respect to \(x\), is expressed as\[\displaystyle \dfrac{dy}{dx}= \lim_{\Delta x \to 0}\dfrac{\Delta y}{\Delta x}.\nonumber\]

Figure \(\PageIndex{1}\): The derivative is expressed as \(\frac{dy}{dx}=\displaystyle\lim_{\Delta x \to 0}\frac{\Delta y}{\Delta x}\).

If the derivative is evaluated at \(x = a\), then the value is written as \(\frac{dy}{dx}\Big|_{x = a}\) or \(\frac{df}{dx}\Big|_{x = a}\).

Caution: Careful Use of Leibniz's Notation

When using Leibniz's notation to find the derivative of \(y = f(x)\) at \(x = a\), it is incorrect to write \(\frac{df(a)}{dx}\). The reason for this will be explored in the next chapter.

Euler's Notation: \(D\)

Leonhard Euler's notation uses a differential operator suggested by Louis François Antoine Arbogast, denoted as \(D\). When applied to a function \(f(x)\), it is defined by\[(Df)(x)\nonumber\] Euler's notation is useful in a subject called differential equations; however, we will not be using it in this text.

Newton's Notation: \(\dot{y}\)

Isaac Newton's notation for differentiation (also called the dot notation, fluxions, or sometimes, crudely, the flyspeck notation for differentiation) places a dot over the dependent variable. That is if \(y\) is a function of \(t\), then the derivative of \(y\) with respect to \(t\) is\[\dot{y}\nonumber\]This notation is popular in physics and mathematical physics. It also appears in areas of mathematics connected with physics, such as differential equations. As with Euler's notation, we will not be using Newton's notation in this text.

Graphing a Derivative

We have already discussed how to graph a function, so given the equation of a function or the equation of a derivative function, we could graph it. Given both, we would expect to see a correspondence between the graphs of these two functions since \(f^{\prime}(x)\) gives the rate of change of a function \(f(x)\) (or slope of the tangent line to \(f(x)\)).

In Example \(\PageIndex{1}\), we found that for \(f(x)=\sqrt{x}\), \(f^{\prime}(x)=\frac{1}{2\sqrt{x}}\). If we graph these functions on the same axes, as in Figure \(\PageIndex{2}\), we can use the graphs to understand the relationship between these two functions. First, we notice that \(f(x)\) is increasing over its entire domain, which means that the slopes of its tangent lines at all points are positive. Consequently, we expect \(f^{\prime}(x)>0\) for all values of \(x\) in its domain. Furthermore, as \(x\) increases, the slopes of the tangent lines to \(f(x)\) are decreasing. We expect to see a corresponding decrease in \(f^{\prime}(x)\). We also observe that \(f(0)\) is undefined and that \(\displaystyle \lim_{x \to 0^+}f^{\prime}(x)=+\infty\), corresponding to a vertical tangent to \(f(x)\) at \(0\).

Figure \(\PageIndex{2}\): The derivative \(f^{\prime}(x)\) is positive everywhere because the function \(f(x)\) is increasing.

In Example \(\PageIndex{2}\), we found that for \(f(x)=x^2-2x,\; f^{\prime}(x)=2x-2\). The graphs of these functions are shown in Figure \(\PageIndex{3}\). Observe that \(f(x)\) is decreasing for \(x<1\). For these same values of \(x\), \(f^{\prime}(x)<0\). For values of \(x>1\), \(f(x)\) is increasing and \(f^{\prime}(x)>0\). Also, \(f(x)\) has a horizontal tangent at \(x=1\) and \(f^{\prime}(1)=0\).

Figure \(\PageIndex{3}\): The derivative \(f^{\prime}(x)<0\) where the function \(f(x)\) is decreasing and \(f^{\prime}(x)>0\) where \(f(x)\) is increasing. The derivative is zero where the function has a horizontal tangent

Interactive Element: Reading \(f^{\prime}\) off the Slopes of \(f\)

The following Interactive Element might help you visualize how the graph of the derivative is built from the slopes of the original function.

Interact: Drag the orange point along the upper curve \(f\); watch the red tangent rotate and the green point on \(f^{\prime}\) ride up and down.

Observation: The height of the point on \(f^{\prime}\) always equals the slope of the tangent to \(f\) at the same \(x\). Where \(f\) has a horizontal tangent, \(f^{\prime}\) meets the \(x\)-axis; where \(f\) rises, \(f^{\prime}>0\); where \(f\) falls, \(f^{\prime}<0\).

Example \(\PageIndex{3}\): Sketching a Derivative Using a Function

Use the following graph of \(f(x)\) to sketch a graph of \(f^{\prime}(x)\).

Figure \(\PageIndex{4}\)

Solution

The solution is shown in the following graph. Observe that \(f(x)\) is increasing and \(f^{\prime}(x)>0\) on \((-2,3)\). Also, \(f(x)\) is decreasing and \(f^{\prime}(x)<0\) on \((-\infty,-2)\) and on \((3,+\infty)\). Also note that \(f(x)\) has horizontal tangents at \(-2\) and \(3\), and \(f^{\prime}(-2)=0\) and \(f^{\prime}(3)=0\).

Figure \(\PageIndex{5}\)

Checkpoint \(\PageIndex{3}\)

Sketch the graph of \(f(x)=x^2-4\). On what interval is the graph of \(f^{\prime}(x)\) above the \(x\)-axis?

Answer

\((0,+\infty)\)

Derivatives and Continuity

Now that we can graph a derivative let's examine the behavior of the graphs. First, we consider the relationship between differentiability and continuity. We will see that if a function is differentiable at a point, it must be continuous there; however, a continuous function need not be differentiable at that point. A function may be continuous at a point and fail to be differentiable at the point for one of several reasons.

Theorem: Differentiability Implies Continuity

Let \(f(x)\) be a function and \(a\) be in its domain. If \(f(x)\) is differentiable at \(a\), then \(f\) is continuous at \(a\).

Proof

If \(f(x)\) is differentiable at \(a\), then \(f^{\prime}(a)\) exists and, if we let \(h = x - a\), we have \(x = a + h\), and as \(h=x-a\to 0\), we can see that \(x\to a\).

Then\[f^{\prime}(a) = \lim_{h\to 0}\dfrac{f(a+h)-f(a)}{h}\nonumber\]can be rewritten as\[f^{\prime}(a)=\displaystyle \lim_{x \to a}\dfrac{f(x)-f(a)}{x-a}.\nonumber\]We want to show that \(f(x)\) is continuous at \(a\) by showing that \(\displaystyle \lim_{x \to a}f(x)=f(a)\). Thus,\[\begin{array}{rcl} \displaystyle \lim_{x \to a}f(x) & = & \displaystyle \lim_{x \to a}\;\big(f(x)-f(a)+f(a)\big) \\[16pt] & = & \displaystyle \lim_{x \to a}\left(\dfrac{f(x)-f(a)}{x-a} \cdot (x-a)+f(a)\right) \\[16pt] & = & \left(\displaystyle \lim_{x \to a}\dfrac{f(x)-f(a)}{x-a}\right) \cdot \left( \displaystyle \lim_{x \to a}\;(x-a)\right)+\displaystyle \lim_{x \to a} f(a) \\[16pt] & = & f^{\prime}(a) \cdot 0+f(a) \\[16pt] & = & f(a). \\ \end{array}\nonumber\]Therefore, since \(f(a)\) is defined and \(\displaystyle \lim_{x \to a}f(x)=f(a)\), we conclude that \(f\) is continuous at \(a\).

Q.E.D.

We have just proven that differentiability implies continuity, but now we consider whether continuity implies differentiability. To determine an answer to this question, we examine the function \(f(x)=|x|\). This function is continuous everywhere; however, \(f^{\prime}(0)\) is undefined. This observation leads us to believe that continuity does not imply differentiability. Let's explore further. For \(f(x)=|x|\),\[f^{\prime}(0)=\displaystyle \lim_{x \to 0}\dfrac{f(x)-f(0)}{x-0}= \lim_{x \to 0}\dfrac{|x|-|0|}{x-0}= \lim_{x \to 0}\dfrac{|x|}{x}.\nonumber\]This limit does not exist because\[\displaystyle \lim_{x \to 0^-}\dfrac{|x|}{x}=-1 \quad \text{and} \quad \displaystyle \lim_{x \to 0^+}\dfrac{|x|}{x}=1.\nonumber\]See Figure \(\PageIndex{6}\).

Figure \(\PageIndex{6}\): The function \(f(x)=|x|\) is continuous at \(0\) but is not differentiable at \(0\).

Interactive Element: Why \(|x|\) Has No Tangent at the Corner

The following Interactive Element might help you visualize why a continuous function can still fail to be differentiable at a sharp corner.

Interact: Drag the orange point along \(y=|x|\) from the right side across to the left side.

Observation: From the right the line has slope \(+1\); from the left it has slope \(-1\). No single straight line is tangent at the corner, so \(f^{\prime}(0)\) does not exist even though \(f\) is continuous there.

Let's consider some additional situations in which a continuous function fails to be differentiable. Consider the function \(f(x)=\sqrt[3]{x}\):\[f^{\prime}(0)=\displaystyle \lim_{x \to 0}\dfrac{\sqrt[3]{x}-0}{x-0}=\displaystyle \lim_{x \to 0}\dfrac{1}{\sqrt[3]{x^2}}=+\infty.\nonumber\]Thus \(f^{\prime}(0)\) does not exist. A quick look at the graph of \(f(x)=\sqrt[3]{x}\) clarifies the situation. The function has a vertical tangent line at \(0\) (Figure \(\PageIndex{7}\)).

Figure \(\PageIndex{7}\): The function \(f(x)=\sqrt[3]{x}\) has a vertical tangent at \(x=0\). It is continuous at \(0\) but is not differentiable at \(0\).

The function \(f(x)=\begin{cases} x\sin\left(\frac{1}{x}\right), & & \text{ if } x \neq 0\\0, & & \text{ if } x=0\end{cases}\) also has a derivative that exhibits interesting behavior at \(0\).

We see that\[f^{\prime}(0)=\displaystyle \lim_{x \to 0}\dfrac{x\sin\left(1/x\right)-0}{x-0}= \lim_{x \to 0}\sin\left(\dfrac{1}{x}\right).\nonumber\]This limit does not exist, essentially because the slopes of the secant lines continuously change direction as they approach zero (Figure \(\PageIndex{8}\)).

Figure \(\PageIndex{8}\): The function \(f(x)=\begin{cases} x\sin\left(\frac{1}{x}\right), & & \text{ if } x \neq 0\\0, & & \text{ if } x=0\end{cases}\) is not differentiable at \(0\).

In summary:

• We observe that if a function is not continuous, it cannot be differentiable since every differentiable function must be continuous. However, if a function is continuous, it may still fail to be differentiable.
• We saw that \(f(x)=|x|\) failed to be differentiable at \(0\) because the limits of the slopes of the tangent lines on the left and right were not the same. Visually, this resulted in a sharp corner on the function's graph at \(0\).. From this, we conclude that to be differentiable at a point, a function must be "smooth" at that point.
• As we saw in the example of \(f(x)=\sqrt[3]{x}\), a function fails to be differentiable at a point where there is a vertical tangent line.
• As we saw with \(f(x)=\begin{cases}x\sin\left(\frac{1}{x}\right), & & \text{ if } x \neq 0\\0, & &\text{ if } x=0\end{cases}\) a function may fail to be differentiable at a point in more complicated ways as well.

Example \(\PageIndex{4}\): A Piecewise Function that is Continuous and Differentiable

A toy company wants to design a track for a toy car that starts along a parabolic curve and then converts to a straight line (Figure \(\PageIndex{9}\)). The function that describes the track is to have the form \(f(x)=\begin{cases}\frac{1}{10}x^2+bx+c, & & \text{ if }x<-10\\-\frac{1}{4}x+\frac{5}{2}, & & \text{ if } x \geq -10\end{cases}\) where \(x\) and \(f(x)\) are in inches. For the car to move smoothly along the track, the function \(f(x)\) must be both continuous and differentiable at \(-10\). Find values of \(b\) and \(c\) that make \(f(x)\) both continuous and differentiable.

Figure \(\PageIndex{9}\): The function must be continuous and differentiable for the car to move smoothly along the track.

Solution

For the function to be continuous at \(x=-10\), \(\displaystyle \lim_{x \to 10^-}f(x)=f(-10)\). Thus, since\[\displaystyle \lim_{x \to -10^-}f(x)=\dfrac{1}{10}(-10)^2-10b+c=10-10b+c\nonumber\]and \(f(-10)=5\), we must have \(10-10b+c=5\). Equivalently, we have \(c=10b-5\).

For the function to be differentiable at \(-10\),\[f^{\prime}(10)=\displaystyle \lim_{x \to -10}\dfrac{f(x)-f(-10)}{x+10}\nonumber\]must exist. Since \(f(x)\) is defined using different rules on the right and the left, we must evaluate this limit from the right and the left and then set them equal to each other:\[\begin{array}{rcl} \displaystyle \lim_{x \to -10^-}\dfrac{f(x)-f(-10)}{x+10} & = & \displaystyle \lim_{x \to -10^-}\dfrac{\frac{1}{10}x^2+bx+c-5}{x+10} \\[16pt] & = & \displaystyle \lim_{x \to -10^-}\dfrac{\frac{1}{10}x^2+bx+(10b-5)-5}{x+10} \\[16pt] & = & \displaystyle \lim_{x \to -10^-}\dfrac{x^2-100+10bx+100b}{10(x+10)} \\[16pt] & = & \displaystyle \lim_{x \to -10^-}\dfrac{(x+10)(x-10+10b)}{10(x+10)} \\[16pt] & = & b-2 \\ \end{array}\nonumber\]We also have\[\begin{array}{rcl} \displaystyle \lim_{x \to -10^+}\dfrac{f(x)-f(-10)}{x+10} & = & \displaystyle \lim_{x \to -10^+}\dfrac{-\frac{1}{4}x+\frac{5}{2}-5}{x+10} \\[16pt] & = & \displaystyle \lim_{x \to -10^+}\dfrac{-(x+10)}{4(x+10)} \\[16pt] & = & -\dfrac{1}{4} \\ \end{array}\nonumber\]This gives us \(b-2=-\frac{1}{4}\). Thus \(b=\frac{7}{4}\) and \(c=10(\frac{7}{4})-5=\frac{25}{2}\).

Checkpoint \(\PageIndex{4}\)

Find values of a and b that make \(f(x)=\begin{cases}ax+b, & & \text{ if } x<3\\x^2, & & \text{ if } x \geq 3\end{cases}\) both continuous and differentiable at \(3\).

Answer

\(a=6\) and \(b=-9\)

Higher-Order Derivatives

The derivative of a function is itself a function, so we can find the derivative of a derivative. For example, the derivative of a position function is the rate of change of position or velocity. The derivative of velocity is the rate of change of velocity, which is acceleration. The new function obtained by differentiating the derivative is called the second derivative. Furthermore, we can continue taking derivatives to obtain the third, fourth, and so on. Collectively, these are referred to as higher-order derivatives. The notation for the higher-order derivatives of \(y=f(x)\) can be expressed in any of the following forms:\[\begin{array}{c} f^{\prime\prime}(x),\; f^{\prime\prime\prime}(x),\; f^{(4)}(x),\; \ldots \; ,\; f^{(n)}(x) \\ y^{\prime\prime}(x),\; y^{\prime\prime\prime}(x),\; y^{(4)}(x),\; \ldots \; ,\; y^{(n)}(x) \\ \dfrac{d^2y}{dx^2},\;\dfrac{d^3y}{dy^3},\;\dfrac{d^4y}{dy^4},\; \ldots \;,\;\dfrac{d^ny}{dy^n}. \\ \end{array}\nonumber\]It is interesting to note that the notation for \(\frac{d^2y}{dx^2}\) may be viewed as an attempt to express \(\frac{d}{dx}\left(\frac{dy}{dx}\right)\) more compactly.

Analogously, \(\frac{d}{dx}\left(\frac{d}{dx}\left(\frac{dy}{dx}\right)\right)=\frac{d}{dx}\left(\frac{d^2y}{dx^2}\right)=\frac{d^3y}{dx^3}\).

Example \(\PageIndex{5}\): Finding a Second Derivative

For \(f(x)=2x^2-3x+1\), find \(f^{\prime\prime}(x)\).

Solution

First find \(f^{\prime}(x)\).

Substitute \(f(x)=2x^2-3x+1\) and \(f(x+h)=2(x+h)^2-3(x+h)+1\) into \(f^{\prime}(x)=\displaystyle \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}\).\[\begin{array}{rcl} f^{\prime}(x) & = & \displaystyle \lim_{h \to 0}\dfrac{(2(x+h)^2-3(x+h)+1)-(2x^2-3x+1)}{h} \\[16pt] & = & \displaystyle \lim_{h \to 0}\dfrac{4xh+2h^2-3h}{h} \\[16pt] & = & \displaystyle \lim_{h \to 0}(4x+2h-3) \\[16pt] & \overset{\text{D.S.}}{=} & 4x-3 \\ \end{array}\nonumber\]Next, find \(f^{\prime\prime}(x)\) by taking the derivative of \(f^{\prime}(x)=4x-3\).\[\begin{array}{rcl} f^{\prime\prime}(x) & = & \displaystyle \lim_{h \to 0}\dfrac{f^{\prime}(x+h)-f^{\prime}(x)}{h} \\[16pt] & = & \displaystyle \lim_{h \to 0}\dfrac{(4(x+h)-3)-(4x-3)}{h} \\[16pt] & = & \displaystyle \lim_{h \to 0}4 \\[16pt] & \overset{\text{D.S.}}{=} & 4 \\ \end{array}\nonumber\]

Checkpoint \(\PageIndex{5}\)

Find \(f^{\prime\prime}(x)\) for \(f(x)=x^2\).

Answer

\(f^{\prime\prime}(x)=2\)

Example \(\PageIndex{6}\): Finding Acceleration

The position of a particle along a coordinate axis at time \(t\) (in seconds) is given by \(s(t)=3t^2-4t+1\) (in meters). Find the function that describes its acceleration at time \(t\).

Solution

Since \(v(t)=s^{\prime}(t)\) and \(a(t)=v^{\prime}(t)=s^{\prime\prime}(t)\), we begin by finding the derivative of \(s(t)\):\[\begin{array}{rcl} s^{\prime}(t) & = & \displaystyle \lim_{h \to 0}\dfrac{s(t+h)-s(t)}{h} \\[16pt] & = & \displaystyle \lim_{h \to 0}\dfrac{3(t+h)^2-4(t+h)+1-(3t^2-4t+1)}{h} \\[16pt] & = & 6t-4 \\ \end{array}\nonumber\]Next,\[\begin{array}{rcl} s^{\prime\prime}(t) & = & \displaystyle \lim_{h \to 0}\dfrac{s^{\prime}(t+h)-s^{\prime}(t)}{h} \\[16pt] & = & \displaystyle \lim_{h \to 0}\dfrac{6(t+h)-4-(6t-4)}{h} \\[16pt] & = & 6 \\ \end{array}\nonumber\]Thus, \(a=6 \;\text{m/s}^2\).

Checkpoint \(\PageIndex{6}\)

For \(s(t)=t^3\), find \(a(t)\).

Answer

\(a(t)=6t\)

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Success in Mathematics: Why Study Mathematics? (Part 1)

Until modern times, only a tiny percent of people had the opportunity to be "liberally educated." These people were not studying to learn a job or trade - they wanted to learn what other humans had thought and created. They were also trying to develop new knowledge. Menaechmus (350 B.C.E) was this kind of student. As he studied geometry, he discovered the properties of conic sections. Others built on his knowledge. Apollonius (350 B.C.E) wrote an eight-chapter book about conic sections.

Today we use our knowledge of conic sections to predict orbits, build telescopes, locate ships at sea, make "eyes" for robots, and collect solar energy. But we also study conic sections as an accomplishment of our human ancestors. In developing their equations, we also develop logic and creative problem-solving skills. Think about the football coach who has his team doing agility drills and running through tires. Why? When has a football team ever had to run through tires during a game? Similarly, with the right attitude, you can develop your ability to reason and think in your life outside of college by doing mathematics. You are responsible for recognizing that learning and studying may be valuable even when it is not immediately practical.

One problem with teaching mathematics in the K-12 system - and I see it as a major difficulty - is that there is virtually nothing the pupils learn that has a non-trivial application in today's world. The most a teacher can tell a student who inquires, entirely reasonably, "How is this useful?" is that almost all mathematics finds uses, in many cases important ones, and that what they learn in school leads on to mathematics that definitely is used.

- Keith Devlin, Standford University

By "non-trivial," Dr. Devlin means that many real-life applications require mathematics beyond what is studied in high school.

1. Younger students often ask, " When will I use this math?" Their teachers often reply, "You need to learn this math to go on to the next course. In college, you can take math that is very powerful for doing all kinds of practical things." Do you think this is the best reply to give to the student? Explain.
2. I've often pondered the best mixture of application and theory in a mathematics course. My love of mathematics lies in theory; however, my research in mathematics is in applications. As a student, which do you prefer? Explain.

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Success in College: Identifying Your Learning Style(s)

Visual/Verbal Learners

These people learn best when information is presented visually and in a written form. In a classroom setting, they prefer professors who use visual aids (i.e., whiteboard or PowerPoint presentation) to list the essential points of a lecture to provide them with an outline to follow. They benefit from information obtained from textbooks and class notes. These learners like to study by themselves in quiet environments. They visualize information in their "minds' eyes" to remember something. The online environment is especially appropriate for visual/verbal learners because most of the information for a course is presented in written form.

Visual/Nonverbal Learners

These people learn best when information is presented visually and in a picture or design format. In a classroom setting, they benefit from professors who supplement their lectures with materials such as film, video, maps, and diagrams. They relate well to information obtained from the images and charts in textbooks. They tend to prefer to work alone in quiet environments. They visualize an image of something in their mind when trying to remember it. These learners may also be artistic and enjoy visual art and design. The online environment is well suited for this learner because graphical representations of information can help them remember concepts and ideas. Graphical information can be presented using charts, tables, graphs, and images.

Auditory/Verbal Learners

These people learn best when information is presented aurally. They benefit from listening to lectures and participating in group discussions in a classroom setting. They also benefit from obtaining information from audio recordings. When trying to remember something, they often repeat it aloud and mentally "hear" how the professor explained the information to them. They learn best when interacting with others in a listening/speaking activity. Online learning environments can complement these learners' styles. Although most information is presented visually (either written or graphically), group participation and collaborative activities are accomplished well online. In addition, streaming audio and computer conferencing can be incorporated into an online course to best meet the learning style of these students.

Tactile/Kinesthetic Learners

These people learn best when doing a physical "hands-on" activity. In the classroom, they prefer to learn new materials in a lab setting where they can touch and manipulate materials. They learn best in physically active learning situations. They benefit from professors who use in-class demonstrations, hands-on learning experiences, and fieldwork outside the classroom. Online environments can provide learning opportunities for tactile/kinesthetic learners. Simulations with 3-Dimensional graphics can replicate physical demonstrations. Lab sessions can be conducted at predetermined locations or homes and then discussed online. Also, outside fieldwork can be incorporated into the coursework, with ample online discussion preceding and following the experience. Finally, the online environment is well suited for the presentation and discussion of either group or individual projects and activities.

Take the learning style survey at Learning Styles.

1. What was your most extreme score, and what was it about?
2. Of all the results, which surprised you the most?