Rate of Change and the Derivative at a Point
On a road trip from Houston to Dallas, Marcus's family covers about 240 miles in 4 hours, an average of 60 mph. But a state trooper's radar doesn't care about the average. It reads your speed at one single instant, and an instant lasts zero seconds. How can "speed at an instant" mean anything? That question is the starting point of calculus, and by the end of this page you will have the answer.
Average Rate of Change
Average speed is distance divided by time. The same idea works for any function: how much did the output change per unit change in the input?
When $f$ is a position function, the ARC is the average velocity. Its units are always (units of $f$) per (unit of $x$): feet per second, followers per day, and so on.
A high school senior athlete is training for the 100 foot sprint. The distance covered (in feet) as a function of time (in seconds) is given by the expression…
A high school senior athlete is training for the 100 foot sprint. The distance covered (in feet) as a function of time (in seconds) is given by the expression $s(t) = 0.5t^2 + 2t$.
Calculate the average velocity of the athlete between the instants $t = 2$ s and $t = 6$ s. Which mathematical concept does this value correspond to?
Calculate the average rate of change (ARC) of the function $f(x) = x^2 - 4x + 5$ in the interval $[1, 4]$. Explain graphically what the obtained value represents in…
Calculate the average rate of change (ARC) of the function $f(x) = x^2 - 4x + 5$ in the interval $[1, 4]$. Explain graphically what the obtained value represents in relation to the secant line passing through the points on the curve corresponding to the endpoints of this interval.
The growth of followers of a new account on a well-known social network follows the function $S(t) = t^2 + 4t + 100$, where $t$ is the time in days since its creation…
The growth of followers of a new account on a well-known social network follows the function $S(t) = t^2 + 4t + 100$, where $t$ is the time in days since its creation and $S(t)$ is the number of followers.
- a) Calculate the average rate of change (ARC) of the number of followers in the intervals $[0, 5]$ and $[5, 10]$.
- b) Explain the meaning of the results obtained in the context of the account's growth. In which interval has the growth been faster?
The Derivative at a Point
The ARC describes a whole interval. To describe one instant, shrink the interval. Fix the point $A$ and slide $B$ toward it. The secant lines settle onto a single line, the tangent line, and the secant slopes settle onto a number. Drag $h$ toward $0$ below: the secant through $x = 1$ and $x = 1 + h$ on $y = x^2$ has slope $2 + h$.
The two forms are the same definition with $x = a + h$. Choose one and stay with it until the end of the problem.
The difference quotient $\dfrac{f(a+h)-f(a)}{h}$ gives $\dfrac{0}{0}$ if you plug in $h = 0$ right away. You must simplify first, then take the limit. Three algebra moves cover almost every case:
Given the function $f(x) = 3x^2 - 1$, use the definition of the derivative at a point via the limit: $$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$ to find the value…
Given the function $f(x) = 3x^2 - 1$, use the definition of the derivative at a point via the limit:
$$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$to find the value of the derivative at the point with x-coordinate $x = 2$.
Determine the slope of the tangent line to the graph of the function $f(x) = \sqrt{x+1}$ at the point where $x = 3$, applying the limit of the instantaneous rate of…
Determine the slope of the tangent line to the graph of the function $f(x) = \sqrt{x+1}$ at the point where $x = 3$, applying the limit of the instantaneous rate of change:
$$f'(3) = \lim_{x \to 3} \frac{f(x) - f(3)}{x - 3}$$Calculate the derivative of the function $f(x) = \dfrac{3}{x + 1}$ at the point with x-coordinate $x = 2$ strictly using the definition of the derivative via the limit:…
Calculate the derivative of the function $f(x) = \dfrac{3}{x + 1}$ at the point with x-coordinate $x = 2$ strictly using the definition of the derivative via the limit:
$$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$Limits in Disguise and the Derivative Function
Many limits are a derivative wearing a costume. If you see $\displaystyle\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$ with a specific function hiding in it, you can read off $f$ and $a$ and use a derivative instead.
If you replace the fixed number $a$ by a variable $x$, the same limit defines the derivative function $f'(x)$, which gives the slope at every point.
Once a problem does not say "use the definition," you can use the standard rules. They are developed in the next sections, but here is what you need:
Consider the rational function $f(x) = \frac{2}{x}$. Prove, applying the definition of the derivative as a limit, that the value of the instantaneous rate of change at…
Consider the rational function $f(x) = \frac{2}{x}$.
Prove, applying the definition of the derivative as a limit, that the value of the instantaneous rate of change at $x = 1$ is $f'(1) = -2$. Interpret the sign of the result obtained.
Calculate, making exclusive use of the definition of the derivative as the limit of the difference quotient, the value of the following limit: $$\lim_{h \to 0}…
Calculate, making exclusive use of the definition of the derivative as the limit of the difference quotient, the value of the following limit:
$$\lim_{h \to 0} \frac{\sin^2\left(\frac{\pi}{4} + h\right) - \frac{1}{2}}{h}$$Hint: Identify this limit with the derivative of a known function at a specific point and solve using standard differentiation rules to check your result.
Let the function be $f(x) = \frac{1}{x+1}$. Prove, applying the definition of the derivative, that its derivative function is $f'(x) = \frac{-1}{(x+1)^2}$ for any value…
Let the function be $f(x) = \frac{1}{x+1}$. Prove, applying the definition of the derivative, that its derivative function is $f'(x) = \frac{-1}{(x+1)^2}$ for any value of $x$ belonging to its domain.
From this result, determine the coordinates of the points on the graph where the tangent line is parallel to the angle bisector of the second quadrant ($y = -x$).
Tangent and Normal Lines
Because $f'(a)$ is the slope of the tangent, you already have everything you need for its equation: a point and a slope.
Drag $a$ and watch the tangent line to $y = x^2$ turn:
Consider the function $f(x) = x^2 - 6x + 5$. Determine the point on the graph where the tangent line has a slope of $m = 2$. Once the point is found, write the full…
Consider the function $f(x) = x^2 - 6x + 5$. Determine the point on the graph where the tangent line has a slope of $m = 2$. Once the point is found, write the full equation of this tangent line in point-slope form.
Find the equation of the tangent line and the normal line to the curve $f(x) = x^3 - 3x^2 + 2$ at the point where the graph intersects the y-axis ($OY$).
Find the equation of the tangent line and the normal line to the curve $f(x) = x^3 - 3x^2 + 2$ at the point where the graph intersects the y-axis ($OY$).
Find the equations of the tangent lines to the curve $y = x^2 - 4x + 5$ that pass through the external point $A(0, 0)$ (the figure shows only one of them). Note: Bear in…
Find the equations of the tangent lines to the curve $y = x^2 - 4x + 5$ that pass through the external point $A(0, 0)$ (the figure shows only one of them).
Note: Bear in mind that point A does not belong to the curve. You will need to set up the slope of the tangent at a generic point $(a, f(a))$ by equating it to the slope of the line joining that point to the origin.
Interpreting Derivatives: Graphs, Units, and Sign
A derivative is only useful if you can say what it means. There are three things to check: the sign, the size (steepness), and the units.
Compare slopes on a graph by comparing the steepness of the lines. In this graph, which bends downward as it rises, the tangent at $x = 1$ is steeper than the secant on $[1, 3]$, and the secant is steeper than the tangent at $x = 3$.
So on this graph, $f'(3) < \text{ARC}[1,3] < f'(1)$. The curve flattens as it goes right, so the average steepness over $[1,3]$ lies between the two instantaneous steepnesses.
In AP-style answers, write one full sentence: the quantity, the instant, "increasing/decreasing," the number, and the units.
Observe the following graph of a function $f(x)$. Without performing calculations, order the following values from smallest to largest, justifying your answer based on…
Observe the following graph of a function $f(x)$. Without performing calculations, order the following values from smallest to largest, justifying your answer based on the steepness of the lines:
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The average rate of change in the interval $[1, 3]$.
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The derivative of the function at the point $x = 1$.
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The derivative of the function at the point $x = 3$.
The temperature of a metal plate (in degrees Fahrenheit) varies with time $t$ (in minutes) following the model $T(t) = 20 + 4t - 0.1t^2$. Calculate the instantaneous…
The temperature of a metal plate (in degrees Fahrenheit) varies with time $t$ (in minutes) following the model $T(t) = 20 + 4t - 0.1t^2$.
Calculate the instantaneous rate of change of the temperature at the instant $t = 10$ minutes. State the units of measurement of the result and explain whether the plate is heating up or cooling down at that moment.
The temperature of an electronic component during the first hours of operation follows the function $T(t) = 20 + t \cdot e^{-t}$, where $T$ is the temperature in degrees…
The temperature of an electronic component during the first hours of operation follows the function $T(t) = 20 + t \cdot e^{-t}$, where $T$ is the temperature in degrees Fahrenheit and $t$ is the time in hours.
Calculate the instantaneous rate of change of the temperature when exactly 1 hour has elapsed ($t = 1$). Is the temperature increasing, decreasing, or neither at that instant? Justify your answer based on the sign of the derivative.
ARC Meets IRC
Often a problem puts both ideas side by side: find the ARC over an interval, then find an instant where the IRC equals it. Back on the road, Marcus's 60 mph average over the whole trip hides moments faster and slower than 60. A natural question is whether at least one instant matched the average.
For a function that is differentiable on an interval, the answer is yes. You will see this guarantee later, as the Mean Value Theorem. For now the method is straightforward: set $f'(c) = \text{ARC}$, solve, and check that $c$ lies inside the interval.
The number of followers of a new YouTube channel evolves according to the function $S(t) = t^2 + 10t + 100$, where $t$ is the time in days since its creation. a)…
The number of followers of a new YouTube channel evolves according to the function $S(t) = t^2 + 10t + 100$, where $t$ is the time in days since its creation.
- a) Calculate the average rate of change of followers between day $t = 0$ and day $t = 10$.
- b) What is the instantaneous growth of followers exactly on day $t = 5$? (Calculate $S'(5)$ using the definition).
Given the function $f(x) = \dfrac{x^2}{x - 1}$: a) Calculate the average rate of change in the interval $[2, 4]$. b) Determine if there is any point $c$ within the…
Given the function $f(x) = \dfrac{x^2}{x - 1}$:
- a) Calculate the average rate of change in the interval $[2, 4]$.
- b) Determine if there is any point $c$ within the interval $(2, 4)$ such that the instantaneous derivative $f'(c)$ matches the ARC calculated in the previous section.
A forest surveillance drone follows an approach trajectory whose height $h$ (in feet) with respect to time $t$ (in seconds) is given by the function: $$h(t) = \sqrt{t^2…
A forest surveillance drone follows an approach trajectory whose height $h$ (in feet) with respect to time $t$ (in seconds) is given by the function:
$$h(t) = \sqrt{t^2 + 144} - 10$$You are asked to:
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Calculate the Average Rate of Change (ARC) of the height in the interval $[0, 5]$. Explain its physical meaning.
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Using the definition of the derivative at a point, find the instantaneous rate of ascent at the instant $t = 5$ seconds.
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Check if there is any instant $t > 0$ at which the instantaneous velocity exactly matches the ARC calculated in the first part.
Continuity, Corners, and Vertical Tangents
The limit that defines $f'(a)$ approaches from both sides of $a$. The limit exists only if the left- and right-hand limits agree. That leads to one-sided derivatives:
$$f'(a^-) = \lim_{h\to 0^-}\frac{f(a+h)-f(a)}{h}, \qquad f'(a^+) = \lim_{h\to 0^+}\frac{f(a+h)-f(a)}{h}$$A continuous function can fail to be differentiable in three main ways: a corner, a vertical tangent, or a cusp. (A jump discontinuity fails earlier, because it is not even continuous.)
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Corner: the one-sided slopes are different finite numbers.
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Vertical tangent: the quotient tends to $+\infty$ (or $-\infty$) from both sides, so the limit is not finite. The tangent line exists, but it is the vertical line $x = a$.
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Cusp: the one-sided slopes go to $+\infty$ on one side and $-\infty$ on the other, so the graph comes to a sharp point.
Given the function $f(x) = x \cdot |x - 3|$, study its differentiability at the point $x = 3$. To do this: 1. Express the function as a piecewise function. 2. Calculate…
Given the function $f(x) = x \cdot |x - 3|$, study its differentiability at the point $x = 3$.
To do this:
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Express the function as a piecewise function.
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Calculate the one-sided derivatives $f'(3^-)$ and $f'(3^+)$ using the limit of the difference quotient.
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Geometrically interpret the result obtained: does the graph have a corner point or a smooth tangent at $x=3$? Justify your answer.
Consider the function $f(x) = \sqrt[3]{x^2}$. 1. Study the continuity of the function at the origin ($x=0$). 2. Analyze the differentiability at $x=0$ by calculating the…
Consider the function $f(x) = \sqrt[3]{x^2}$.
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Study the continuity of the function at the origin ($x=0$).
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Analyze the differentiability at $x=0$ by calculating the limit of the difference quotient.
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If the limit tends to infinity, what type of tangent line does the function present at that point? Describe the visual behavior of the graph in the vicinity of the origin.
Piecewise Functions and Parameters
When a function changes formula at $x = a$, you usually need a parameter or two to make it differentiable there. Two separate conditions must hold:
Given the piecewise function: $$f(x) = \begin{cases} x^2 + ax + b & \text{if } x < 1 \\ \ln(x) & \text{if } x \geq 1 \end{cases}$$ Calculate the values of $a$ and $b$ so…
Given the piecewise function:
$$f(x) = \begin{cases} x^2 + ax + b & \text{if } x < 1 \\ \ln(x) & \text{if } x \geq 1 \end{cases}$$Calculate the values of $a$ and $b$ so that the function is differentiable at the point $x = 1$. Remember that for a function to be differentiable, it must first be continuous.
Consider the piecewise function: $$f(x) = \begin{cases} a \cdot e^{x-2} + b \cdot x & \text{if } x \le 2 \\ \frac{x^2 - 4}{x-2} + \ln(x-1) & \text{if } x > 2…
Consider the piecewise function:
$$f(x) = \begin{cases} a \cdot e^{x-2} + b \cdot x & \text{if } x \le 2 \\ \frac{x^2 - 4}{x-2} + \ln(x-1) & \text{if } x > 2 \end{cases}$$Determine the values of the real parameters $a$ and $b$ so that the function is differentiable at the point $x = 2$. To do this, you must first ensure continuity and, subsequently, the existence of finite and equal one-sided derivatives at that point using the definition of the derivative.
Back to the trooper: your car's position over time is a function $s(t)$. Marcus's average of 60 mph is $\text{ARC}[0,4]$. The radar reads $s'(t)$ at one instant, the limit of average speeds over shorter and shorter intervals. That limit is what makes "speed at an instant" meaningful.