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The derivative of a function at a point

The instantaneous rate of change of a function is an idea that sits at the foundation of calculus.

Introduction

The instantaneous rate of change of a function is an idea that sits at the foundation of calculus. It is a generalization of the notion of instantaneous velocity and measures how fast a particular function is changing at a given input. If the original function represents the position of a moving object, this instantaneous rate of change is precisely the instantaneous velocity of the object. In other contexts, instantaneous rate of change could measure the number of cells added to a bacteria culture per day, the number of additional gallons of gasoline consumed per mile by increasing a car's velocity one mile per hour, or the number of dollars added to a mortgage payment for each percentage point increase in interest rate. The instantaneous rate of change can also be interpreted geometrically on the function's graph, and this connection is fundamental to many of the main ideas in calculus.

Recall that for a moving object with position function \(s\), its average velocity on the time interval \(t = a\) to \(t = a+h\) is given by the quotient \[\begin{aligned}\end{aligned}\].

In a similar way, we make the following definition for an arbitrary function \(y = f(x)\).

It is essential to understand how the average rate of change of \(f\) on an interval is connected to its graph.

Exploration

The Derivative of a Function at a Point

Just as we defined instantaneous velocity in terms of average velocity, we now define the instantaneous rate of change of a function at a point in terms of the average rate of change of the function \(f\) over related intervals. This instantaneous rate of change of \(f\) at \(a\) is called the derivative of \(f\) at \(a\), and is denoted by \(f'(a)\).

Aloud, we read the symbol \(f'(a)\) as either \(f\)-prime at \(a\) or the derivative of \(f\) evaluated at \(x = a\). Much of our work in Chapters 1-3 will be devoted to understanding, computing, applying, and interpreting derivatives. For now, we observe the following important things.

We first consider the derivative at a given value as the slope of a certain line.

When we compute an instantaneous rate of change, we allow the interval \([a,a+h]\) to shrink as \(h \to 0\). We can think of one endpoint of the interval as sliding towards the other. In particular, provided that \(f\) has a derivative at \((a,f(a))\), the point \((a+h,f(a+h))\) will approach \((a,f(a))\) as \(h \to 0\). Because the process of taking a limit is a dynamic one, it can be helpful to use computing technology to visualize it. One option is an interactive graphic in which the user is able to control the point that is moving. For a helpful collection of examples, consider the work of David Austin of Grand Valley State University, and this particularly relevant example. For interactives that have been built in Geogebra You can even consider building your own examples; the fantastic program Geogebra is available for free download and is easy to learn and use. , see Marc Renault's library via Shippensburg University, with this example being especially fitting for our work in this section.

The following example demonstrates several key ideas involving the derivative of a function.

Condensed — the full section is in Boelkins, Active Calculus.

Summary

\(f\)\([a,b]\)\(AV_{[a,b]} = \frac{f(b)-f(a)}{b-a}\)units of \(f(x)\) per unit of \(x\)\((a,f(a))\)\((b,f(b))\)\(y = f(x)\)\([a,a+h]\)\([a,b]\)\(AV_{[a,b]} = \frac{f(a+h)-f(a)}{h}\)
\(x\)\(f\)\(x = a\)\(f'(a)\)the derivative of \(f\) evaluated at \(a\)\(f\)-prime at \(a\)\[\begin{aligned}\end{aligned}\]\(x = a\)\([a,a+h]\)\(h \to 0\)units of \(f(x)\) per unit of \(x\)
\(f'(a)\)\(f\)\(x\)\(x = a\)\(y = f(x)\)\((a,f(a))\)\(f'(a)\)slope of the curve\((a,f(a))\)
Limits allow us to move from the rate of change over an interval to the rate of change at a single point.

Practice (11)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

When we are talking about slope, students from American schools often say the words riseand run. Explain in your own words what riseand run mean. How do we combine riseand run to determine slope?
In Definition, what does \(h\) represent in the computation of the average rate of change of \(f\)?
In Definition, what does the quantity \(f(a + h) - f(a)\) represent in the computation of average rate of change?
Consider the graph of \(y = f(x)\) provided in Figure.
1. On the graph of \(y = f(x)\), sketch and label the following quantities:
  - \(y = f(x)\)\([-3,-1]\)\(y = f(x)\)\([0,2]\)
  - \(y = f(x)\)\(x = -3\)\(y = f(x)\)\(x = 0\)
2. \(f\)\([-3,-1]\)\([0,2]\)
3. \(f\)\(x = -3\)\(x = 0\)
Жауап беріңіз
1. The requested secant lines are drawn in green, with the noted tangent lines in magenta.
2. First, \(AV_{[-3,-1]} = \frac{f(-1)-f(-3)}{-1-(-3)} \approx \frac{2.3}{2} = 1.15\). Similarly, \(AV_{[0,2]} = \frac{f(2)-f(0)}{2-0} \approx -\frac{0.8}{2} = -0.4\). These values are the respective slopes of the green secant lines plotted in (a).
3. \(f'(-3) \approx 3\) and \(f'(0) \approx -\frac{1}{2}\), as these are estimates of the slopes of the tangent lines plotted in magenta.
For each of the following prompts, sketch a graph on the provided axes in Figure of a function that has the stated properties.
1. \(y = f(x)\) such that
  - \(f\)\([-3,0]\)\(-2\)\(f\)\([1,3]\)
  - \(f\)\(x = -1\)\(-1\)\(f\)\(x = 2\)
2. \(y = g(x)\) such that
  - \(\frac{g(3)-g(-2)}{5} = 0\)\(\frac{g(1)-g(-1)}{2} = -1\)
  - \(g'(2) = 1\)\(g'(-1) = 0\)
Жауап беріңіз
1. For this problem, you should draw a graph so that the slope of the secant line from \((-3,f(-3))\) to \((0,f(0))\) is \(-2\), and similarly the slope of the secant line from \((1,f(1))\) to \((3,f(3))\) is \(0.5\), while the slopes of the tangent lines at \(x = -1\) and \(x = 2\) are \(-1\) and \(1\), respectively. For instance, you could let \(f(-3) = 3\) and have \(f\) pass through the points \((-3,3)\), \((-1,-2)\), \((0,-3)\), \((1,-2)\), and \((3,-1)\) and draw the desired tangent lines accordingly.
2. For this problem, you should draw a graph so that the slope of the secant line from \((-2,g(-2))\) to \((3,f(3))\) is \(0\), and similarly the slope of the secant line from \((-1,g(-1))\) to \((1,g(1))\) is \(-1\), while the slopes of the tangent lines to \(y = g(x)\) at \(x=-1\) and \(x = 2\) are \(0\) and \(1\), respectively. For instance, you could draw a function \(g\) that passes through the points \((-2,3)\), \((-1,2)\), \((1,0)\), \((2,0)\), and \((3,3)\) in such a way that the tangent line at \((-1,2)\) is horizontal and the tangent line at \((2,0)\) has slope \(1\).
Suppose that the population, \(P\), of China (in billions) can be approximated by the function \(P(t) = 1.15(1.014)^t\) where \(t\) is the number of years since the start of 1993.
1. According to the model, what was the total change in the population of China between January 1, 1993 and January 1, 2000? What will be the average rate of change of the population over this time period? Is this average rate of change greater or less than the instantaneous rate of change of the population on January 1, 2000? Explain and justify, being sure to include proper units on all your answers.
2. According to the model, what is the average rate of change of the population of China in the ten-year period starting on January 1, 2012?
3. Write an expression involving limits that, if evaluated, would give the exact instantaneous rate of change of the population on today's date. Then estimate the value of this limit (discuss how you chose to do so) and explain the meaning (including units) of the value you have found.
4. \(y = P(t)\)\(t\)
Жауап беріңіз
1. First, \(P(7)-P(0) = 115(1.014)^7-115(1.014)^0 \approx 0.1175\) billion people is the total population change. It follows that the average rate of change over this time period is \(AV_{[0,7]}=\frac{0.1175}{7} \approx 0.01679\) billion people per year. The instantaneous rate of change of the population at \(t = 7\) is \[\begin{aligned}\end{aligned}\], and approximating this quantity with small values of \(h\), it follows that \(P'(7) \approx 0.1762\). Hence, the instantaneous rate of change at \(t = 7\) is greater than the average rate of change of \(P\)on the interval \([0,7]\).
2. Using the formula for average rate of change, \[\begin{aligned}\end{aligned}\] billion people/year.
3. We will say that today's date is July 1, 2015, which means that \(t = 22.5\), so we want to compute \(P'(22.5)\). The limit definition tells us that \[\begin{aligned}\end{aligned}\]. Estimating this limit by using small values of \(h\), we determine that \(P'(22.5) \approx 0.02186\), which is measured in billions of people per year.
4. We want the equation of the tangent line that has slope \(P'(22.5) \approx 0.02186\) and that passes through the point \((22.5, P(22.5))\). Using point-slope form, it follows that the line is given by \[\begin{aligned}\end{aligned}\].

The goal of this problem is to compute the value of the derivative at a point for several different functions, where for each one we do so in three different ways, and then to compare the results to see that each produces the same value.

For each of the following functions, use the limit definition of the derivative to compute the value of \(f'(a)\) using three different approaches: strive to use the algebraic approach first (to compute the limit exactly), then test your result using numerical evidence (with small values of \(h\)), and finally plot the graph of \(y = f(x)\) near \((a,f(a))\) along with the appropriate tangent line to estimate the value of \(f'(a)\) visually. Compare your findings among all three approaches; if you are unable to complete the algebraic approach, still work numerically and graphically.

\(f(x) = x^2 - 3x\)\(a = 2\)
\(f(x) = \frac{1}{x}\)\(a = 1\)
\(f(x) = \sqrt{x}\)\(a = 1\)
\(f(x) = 2 - |x-1|\)\(a = 1\)
\(f(x) = \sin(x)\)\(a = \frac{\pi}{2}\)

Жауап беріңіз

The table below gives values of the difference quotient \(\frac{f(a+h)-f(a)}{h}\) for \(a=2\) and several small values of \(h\). This table indicates that \(f'(2)\) is approximately \(1\).

\(h\)	\(-0.0001\)	\(-0.001\)	\(-0.01\)	\(0.01\)	\(0.001\)	\(0.0001\)
\(\frac{f(2+h)-f(2)}{h}\)	\(0.9999\)	\(0.9990\)	\(0.9900\)	\(1.0100\)	\(1.0010\)	\(1.0001\)

The tangent line to \(f\) at the point \((2,f(2))\) as shown in the following appears to have a slope that is approximately 1.

Finally, we verify these approximations using an algebraic approach. The definition of the derivative of \(f\) at \(x=a\) shows that \[\begin{aligned}f'(2) &= \lim_{h \to 0} \frac{f(2+h)-f(2)}{h} \\ &= \lim_{h \to 0} \frac{\left[(2+h)^2 - 3(2+h)\right] - (-2)}{h} \\ &= \lim_{h \to 0} \frac{\left[4+4h+h^2 - 6 - 3h\right] + 2}{h} \\ &= \lim_{h \to 0} \frac{h^2 + h}{h} \\ &= \lim_{h \to 0} \frac{h(h+1)}{h} \\ &= \lim_{h \to 0} h+1 \\ &= 1\end{aligned}\] This confirms our conclusions using the numeric and geometric approaches.

The following table gives values of the difference quotient \(\frac{f(a+h)-f(a)}{h}\) for \(a=1\) and several small values of \(h\). This table indicates that \(f'(1)\) is approximately \(-1\).

\(h\)	\(-0.0001\)	\(-0.001\)	\(-0.01\)	\(0.01\)	\(0.001\)	\(0.0001\)
\(\frac{f(1+h)-f(1)}{h}\)	\(-1.00010\)	\(-1.001001\)	\(-1.010101\)	\(-0.990099\)	\(-0.999001\)	\(-0.999900\)

The tangent line to \(f\) at the point \((1,f(1))\) as shown in the following figure appears to have a slope that is approximately \(-1\).

Finally, we verify these approximations using an algebraic approach. The definition of the derivative of \(f\) at \(x=a\) shows that \[\begin{aligned}f'(1) &= \lim_{h \to 0} \frac{f(1+h)-f(1)}{h} \\ &= \lim_{h \to 0} \frac{\frac{1}{1+h} - 1}{h} \\ &= \lim_{h \to 0} \frac{\frac{1}{1+h} - 1}{h} \left(\frac{1+h}{1+h} \right) \\ &= \lim_{h \to 0} \frac{1-(1+h)}{h(1+h)} \\ &= \lim_{h \to 0} \frac{-h)}{h(1+h)} \\ &= \lim_{h \to 0} -\frac{1}{1+h} \\ &= -1\end{aligned}\]. This confirms our conclusions using the numeric and geometric approaches.

The table below gives values of the difference quotient \(\frac{f(a+h)-f(a)}{h}\) for \(a=1\) and several small values of \(h\). This table indicates that \(f'(1)\) is approximately \(0.5\).

\(h\)	\(-0.0001\)	\(-0.001\)	\(-0.01\)	\(0.01\)	\(0.001\)	\(0.0001\)
\(\frac{f(1+h)-f(1)}{h}\)	\(0.50001300\)	\(0.50012510\)	\(0.50125629\)	\(0.49875620\)	\(0.49987500\)	\(0.49999000\)

The tangent line to \(f\) at the point \((1,f(1))\) as shown in the figure below appears to have a slope that is approximately \(0.5\).

Finally, we verify these approximations using an algebraic approach. The definition of the derivative of \(f\) at \(x=a\) shows that \[\begin{aligned}f'(1) &= \lim_{h \to 0} \frac{f(1+h)-f(1)}{h} \\ &= \lim_{h \to 0} \frac{\sqrt{1+h}-\sqrt{1}}{h} \\ &= \lim_{h \to 0} \frac{\sqrt{1+h}-1}{h} \\ &= \lim_{h \to 0} \frac{\sqrt{1+h}-1}{h} \left(\frac{\sqrt{1+h}+1}{\sqrt{1+h}+1} \right) \\ &= \lim_{h \to 0} \frac{(1+h)-1}{h(\sqrt{1+h}+1)} \\ &= \lim_{h \to 0} \frac{h}{h(\sqrt{1+h}+1)} \\ &= \frac{1}{\sqrt{1+h}+1} \\ &= \frac{1}{2}\end{aligned}\]. This confirms our conclusions using the numeric and geometric approaches.

The data in the following table indicates that \(\frac{f(1+h)-f(1)}{h}\) has values of \(1\) for \(h \lt 0\) and values of \(-1\) for \(h \gt 0\). This implies that \(f'(1)\) does not exist.

\(h\)	\(-0.0001\)	\(-0.001\)	\(-0.01\)	\(0.01\)	\(0.001\)	\(0.0001\)
\(\frac{f(1+h)-f(1)}{h}\)	\(1\)	\(1\)	\(1\)	\(-1\)	\(-1\)	\(-1\)

The graph of \(f\) in the figure below appears to show a sharp corner at the point \((1,f(1))\), also implying that there is no tangent line to \(f\)at \((1,f(1))\).

Now we verify these approximations using an algebraic approach. Notice first that \[\begin{aligned}\frac{f(1+h)-f(1)}{h} &= \frac{[2-|(1+h)-1|] - 2}{h} \\ &= \frac{(2-|h|)-2}{h} \\ &= \frac{-|h|}{h}\end{aligned}\] The definition of the absolute value function tells us that \(|h| = h\) whenever \(h \ge 0\), while \(|h| = -h\) whenever \(h \lt 0\). Thus, for \(h \gt 0\), we have \[\begin{aligned}\end{aligned}\] and for \(h \lt 0\), we have \[\begin{aligned}\end{aligned}\]. This confirms that \(\lim_{h \to 0} \frac{f(1+h)-f(1)}{h}\) does not exist.

The data in the following table indicate that \(\frac{f(\frac{\pi}{2}+h)-f(\frac{\pi}{2})}{h}\) approaches \(0\) as \(h \to 0\).

\(h\)	\(-0.0001\)	\(-0.001\)	\(-0.01\)	\(0.01\)	\(0.001\)	\(0.0001\)
\(\frac{f(1+h)-f(1)}{h}\)	\(0.00005\)	\(0.0005\)	\(0.005\)	\(-0.005\)	\(-0.0005\)	\(-0.00005\)

The graph of \(f\) in the figure below shows a horizontal tangent line at the point \((\frac{\pi}{2},f(\frac{\pi}{2}))\), which also suggests that \(f'(\frac{\pi}{2}) = 0\).

Finally, we attempt to verify these approximations using an algebraic approach. Notice first that \[\begin{aligned}\frac{f(\frac{\pi}{2}+h)-f(\frac{\pi}{2})}{h} &= \frac{\sin(\frac{\pi}{2}+h) - \sin \frac{\pi}{2}}{h} \\ &= \frac{\sin(\frac{\pi}{2}+h) - 1}{h}\end{aligned}\] Using the sum of two angles identity for the sine function, it follows that \[\begin{aligned}\end{aligned}\]. Hence, \[\begin{aligned}f'(\frac{\pi}{2}) &= \lim_{h \to 0} \frac{f(\frac{\pi}{2}+h)-f(\frac{\pi}{2})}{h} \\ &= \lim_{h \to 0} \frac{\cos(h) - 1}{h}\end{aligned}\] At this point, we encounter a limit that we are unable to evaluate further using algebraic means. Like the earlier limit above, we can estimate the limit using small values of \(h\). Doing so confirms that this limit also approaches \(0\), and we hence conclude that \(f'(\frac{\pi}{2}) = 0\).

Consider the graph of \(y = f(x)\) provided in Figure.
1. On the graph of \(y = f(x)\), sketch and label the following quantities:
  - \(y = f(x)\)\([-3,-1]\)\(y = f(x)\)\([0,2]\)
  - \(y = f(x)\)\(x = -3\)\(y = f(x)\)\(x = 0\)
2. \(f\)\([-3,-1]\)\([0,2]\)
3. \(f\)\(x = -3\)\(x = 0\)
Жауап беріңіз
1. The requested secant lines are drawn in green, with the noted tangent lines in magenta.
2. First, \(AV_{[-3,-1]} = \frac{f(-1)-f(-3)}{-1-(-3)} \approx \frac{2.3}{2} = 1.15\). Similarly, \(AV_{[0,2]} = \frac{f(2)-f(0)}{2-0} \approx -\frac{0.8}{2} = -0.4\). These values are the respective slopes of the green secant lines plotted in (a).
3. \(f'(-3) \approx 3\) and \(f'(0) \approx -\frac{1}{2}\), as these are estimates of the slopes of the tangent lines plotted in magenta.
For each of the following prompts, sketch a graph on the provided axes in Figure of a function that has the stated properties.
1. \(y = f(x)\) such that
  - \(f\)\([-3,0]\)\(-2\)\(f\)\([1,3]\)
  - \(f\)\(x = -1\)\(-1\)\(f\)\(x = 2\)
2. \(y = g(x)\) such that
  - \(\frac{g(3)-g(-2)}{5} = 0\)\(\frac{g(1)-g(-1)}{2} = -1\)
  - \(g'(2) = 1\)\(g'(-1) = 0\)
Жауап беріңіз
1. For this problem, you should draw a graph so that the slope of the secant line from \((-3,f(-3))\) to \((0,f(0))\) is \(-2\), and similarly the slope of the secant line from \((1,f(1))\) to \((3,f(3))\) is \(0.5\), while the slopes of the tangent lines at \(x = -1\) and \(x = 2\) are \(-1\) and \(1\), respectively. For instance, you could let \(f(-3) = 3\) and have \(f\) pass through the points \((-3,3)\), \((-1,-2)\), \((0,-3)\), \((1,-2)\), and \((3,-1)\) and draw the desired tangent lines accordingly.
2. For this problem, you should draw a graph so that the slope of the secant line from \((-2,g(-2))\) to \((3,f(3))\) is \(0\), and similarly the slope of the secant line from \((-1,g(-1))\) to \((1,g(1))\) is \(-1\), while the slopes of the tangent lines to \(y = g(x)\) at \(x=-1\) and \(x = 2\) are \(0\) and \(1\), respectively. For instance, you could draw a function \(g\) that passes through the points \((-2,3)\), \((-1,2)\), \((1,0)\), \((2,0)\), and \((3,3)\) in such a way that the tangent line at \((-1,2)\) is horizontal and the tangent line at \((2,0)\) has slope \(1\).
Suppose that the population, \(P\), of China (in billions) can be approximated by the function \(P(t) = 1.15(1.014)^t\) where \(t\) is the number of years since the start of 1993.
1. According to the model, what was the total change in the population of China between January 1, 1993 and January 1, 2000? What will be the average rate of change of the population over this time period? Is this average rate of change greater or less than the instantaneous rate of change of the population on January 1, 2000? Explain and justify, being sure to include proper units on all your answers.
2. According to the model, what is the average rate of change of the population of China in the ten-year period starting on January 1, 2012?
3. Write an expression involving limits that, if evaluated, would give the exact instantaneous rate of change of the population on today's date. Then estimate the value of this limit (discuss how you chose to do so) and explain the meaning (including units) of the value you have found.
4. \(y = P(t)\)\(t\)
Жауап беріңіз
1. First, \(P(7)-P(0) = 115(1.014)^7-115(1.014)^0 \approx 0.1175\) billion people is the total population change. It follows that the average rate of change over this time period is \(AV_{[0,7]}=\frac{0.1175}{7} \approx 0.01679\) billion people per year. The instantaneous rate of change of the population at \(t = 7\) is \[\begin{aligned}\end{aligned}\], and approximating this quantity with small values of \(h\), it follows that \(P'(7) \approx 0.1762\). Hence, the instantaneous rate of change at \(t = 7\) is greater than the average rate of change of \(P\)on the interval \([0,7]\).
2. Using the formula for average rate of change, \[\begin{aligned}\end{aligned}\] billion people/year.
3. We will say that today's date is July 1, 2015, which means that \(t = 22.5\), so we want to compute \(P'(22.5)\). The limit definition tells us that \[\begin{aligned}\end{aligned}\]. Estimating this limit by using small values of \(h\), we determine that \(P'(22.5) \approx 0.02186\), which is measured in billions of people per year.
4. We want the equation of the tangent line that has slope \(P'(22.5) \approx 0.02186\) and that passes through the point \((22.5, P(22.5))\). Using point-slope form, it follows that the line is given by \[\begin{aligned}\end{aligned}\].

\(f(x) = x^2 - 3x\)\(a = 2\)
\(f(x) = \frac{1}{x}\)\(a = 1\)
\(f(x) = \sqrt{x}\)\(a = 1\)
\(f(x) = 2 - |x-1|\)\(a = 1\)
\(f(x) = \sin(x)\)\(a = \frac{\pi}{2}\)

Жауап беріңіз

The table below gives values of the difference quotient \(\frac{f(a+h)-f(a)}{h}\) for \(a=2\) and several small values of \(h\). This table indicates that \(f'(2)\) is approximately \(1\).

\(h\)	\(-0.0001\)	\(-0.001\)	\(-0.01\)	\(0.01\)	\(0.001\)	\(0.0001\)
\(\frac{f(2+h)-f(2)}{h}\)	\(0.9999\)	\(0.9990\)	\(0.9900\)	\(1.0100\)	\(1.0010\)	\(1.0001\)

The tangent line to \(f\) at the point \((2,f(2))\) as shown in the following appears to have a slope that is approximately 1.

The following table gives values of the difference quotient \(\frac{f(a+h)-f(a)}{h}\) for \(a=1\) and several small values of \(h\). This table indicates that \(f'(1)\) is approximately \(-1\).

\(h\)	\(-0.0001\)	\(-0.001\)	\(-0.01\)	\(0.01\)	\(0.001\)	\(0.0001\)
\(\frac{f(1+h)-f(1)}{h}\)	\(-1.00010\)	\(-1.001001\)	\(-1.010101\)	\(-0.990099\)	\(-0.999001\)	\(-0.999900\)

The tangent line to \(f\) at the point \((1,f(1))\) as shown in the following figure appears to have a slope that is approximately \(-1\).

The table below gives values of the difference quotient \(\frac{f(a+h)-f(a)}{h}\) for \(a=1\) and several small values of \(h\). This table indicates that \(f'(1)\) is approximately \(0.5\).

\(h\)	\(-0.0001\)	\(-0.001\)	\(-0.01\)	\(0.01\)	\(0.001\)	\(0.0001\)
\(\frac{f(1+h)-f(1)}{h}\)	\(0.50001300\)	\(0.50012510\)	\(0.50125629\)	\(0.49875620\)	\(0.49987500\)	\(0.49999000\)

The tangent line to \(f\) at the point \((1,f(1))\) as shown in the figure below appears to have a slope that is approximately \(0.5\).

Finally, we verify these approximations using an algebraic approach. The definition of the derivative of \(f\) at \(x=a\) shows that \[\begin{aligned}f'(1) &= \lim_{h \to 0} \frac{f(1+h)-f(1)}{h} \\ &= \lim_{h \to 0} \frac{\sqrt{1+h}-\sqrt{1}}{h} \\ &= \lim_{h \to 0} \frac{\sqrt{1+h}-1}{h} \\ &= \lim_{h \to 0} \frac{\sqrt{1+h}-1}{h} \left(\frac{\sqrt{1+h}+1}{\sqrt{1+h}+1} \right) \\ &= \lim_{h \to 0} \frac{(1+h)-1}{h(\sqrt{1+h}+1)} \\ &= \lim_{h \to 0} \frac{h}{h(\sqrt{1+h}+1)} \\ &= \frac{1}{\sqrt{1+h}+1} \\ &= \frac{1}{2}\end{aligned}\]. This confirms our conclusions using the numeric and geometric approaches.

The data in the following table indicates that \(\frac{f(1+h)-f(1)}{h}\) has values of \(1\) for \(h \lt 0\) and values of \(-1\) for \(h \gt 0\). This implies that \(f'(1)\) does not exist.

\(h\)	\(-0.0001\)	\(-0.001\)	\(-0.01\)	\(0.01\)	\(0.001\)	\(0.0001\)
\(\frac{f(1+h)-f(1)}{h}\)	\(1\)	\(1\)	\(1\)	\(-1\)	\(-1\)	\(-1\)

The graph of \(f\) in the figure below appears to show a sharp corner at the point \((1,f(1))\), also implying that there is no tangent line to \(f\)at \((1,f(1))\).

The data in the following table indicate that \(\frac{f(\frac{\pi}{2}+h)-f(\frac{\pi}{2})}{h}\) approaches \(0\) as \(h \to 0\).

\(h\)	\(-0.0001\)	\(-0.001\)	\(-0.01\)	\(0.01\)	\(0.001\)	\(0.0001\)
\(\frac{f(1+h)-f(1)}{h}\)	\(0.00005\)	\(0.0005\)	\(0.005\)	\(-0.005\)	\(-0.0005\)	\(-0.00005\)

The graph of \(f\) in the figure below shows a horizontal tangent line at the point \((\frac{\pi}{2},f(\frac{\pi}{2}))\), which also suggests that \(f'(\frac{\pi}{2}) = 0\).

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

P(A),\ P(A \mid B)

probability, conditional probability
Chance of A; chance of A given that B happened.

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

Euler's number
2.71828…, the base whose exponential is its own derivative.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: The derivative of a function at a point

How is the average rate of change of a function on a given interval defined, and what does this quantity measure?
How is the instantaneous rate of change of a function at a particular point defined? How is the instantaneous rate of change linked to average rate of change?
What is the derivative of a function at a given point? What does this derivative value measure? How do we interpret the derivative value graphically?
How are limits used formally in the computation of derivatives?

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

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Parts of this page are adapted from Boelkins, Active Calculus (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.

Келесіде Calculus

Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests