Secant lines closing in on a tangent line — and the limit that turns a slope into a rate of change.
Now that we have both a conceptual understanding of a limit and the practical ability to compute limits, the foundation is in place. Most mathematicians and historians agree that calculus was developed independently by the Englishman Isaac Newton (1643–1727) and the German Gottfried Leibniz (1646–1716). Crediting them with developing calculus really means that they were the first to understand the relationship between the derivative and the integral.
Both benefited from the work of predecessors such as Barrow, Fermat and Cavalieri. Their relationship appears to have been amicable at first; in later years a bitter controversy erupted over whose work took precedence. Although it seems likely that Newton arrived at the ideas first, we are indebted to Leibniz for the notation we commonly use today.
Secant Lines and Tangent Slopes
We use the slope of a secant line to a function at a point $(a,f(a))$ to estimate the rate of change — the rate at which one variable changes in relation to another. Let $a+h$ be a value of $x$ close to $a$. Draw the line through $(a,f(a))$ and $(a+h,f(a+h))$: this is a secant, and its slope is the difference quotient with increment $h$,
As the values of $x$ approach $a$, the slopes of the secant lines give better and better estimates of the rate of change of the function at $a$ — and the secant lines themselves approach the tangent line, which is the limit of the secant lines. Equivalently, as $h$ gets closer to $0$ the secants approach the tangent. The slope of the tangent line at $a$ is the rate of change of the function at $a$.
Interactive 1 — Secant Lines Becoming a Tangent
increment $h$—
secant slope—
tangent slope (the limit)—
📘 Definition — Slope of the tangent line
Let $f(x)$ be a function defined in an open interval containing $a$. The slope of the tangent line to the graph of $f$ at the point $(a,f(a))$ is
$$m_{\tan}=\lim_{h\to0}\frac{f(a+h)-f(a)}{h},$$
provided this limit exists. The tangent line is the line through $(a,f(a))$ with this slope.
✏️ Example 1 — The slope of a tangent line
Find the slope of the line tangent to the graph of $f(x)=x^2$ at $x=3$.
💡 Show solution▶
Step 1 Write the difference quotient at $a=3$:
$$m_{\tan}=\lim_{h\to0}\frac{f(3+h)-f(3)}{h}=\lim_{h\to0}\frac{(3+h)^2-3^2}{h}.$$
Step 2 Expand and simplify:
$$=\lim_{h\to0}\frac{9+6h+h^2-9}{h}=\lim_{h\to0}\frac{h(6+h)}{h}.$$
Step 3 Cancel $h$ (legal because $h\neq0$ in the limit) and evaluate:
$$=\lim_{h\to0}(6+h)=6.$$
The Derivative of a Function at a Point
The type of limit we just computed occurs in many applications across many disciplines: velocity and acceleration in physics, marginal profit in business, growth rates in biology. It occurs so frequently that the value gets a special name — the derivative. The process of finding it is called differentiation.
📘 Definition — Derivative of the function at a point
Let $f(x)$ be a function defined in an open interval containing $a$. The derivative of $f(x)$ at $a$, denoted $f'(a)$, is
$$f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h},$$
provided this limit exists.
✏️ Example 2 — Finding the derivative at a point
For $f(x)=3x^2-4x+1$, find $f'(2)$ using the definition of the derivative at a point.
💡 Show solution▶
Step 1 Substitute into the definition:
$$f'(2)=\lim_{h\to0}\frac{\left[3(2+h)^2-4(2+h)+1\right]-\left[3(2)^2-4(2)+1\right]}{h}.$$
Step 2 Expand the bracket: $3(4+4h+h^2)-8-4h+1=3h^2+8h+5$, and the second bracket is $5$. So the numerator is $3h^2+8h$:
$$=\lim_{h\to0}\frac{3h^2+8h}{h}=\lim_{h\to0}\frac{h(3h+8)}{h}.$$
Step 3 Cancel and evaluate:
$$=\lim_{h\to0}(3h+8)=8.$$
Velocities and Rates of Change
The difference quotient and its limit have a physical meaning. If $s(t)$ is the position of an object moving along a coordinate axis, the average velocity over the time interval between $a$ and $t$ is the difference quotient
$$v_{\text{ave}}=\frac{s(t)-s(a)}{t-a}=\frac{\text{change in position}}{\text{elapsed time}},$$
which is the slope of the secant line through $(a,s(a))$ and $(t,s(t))$ on the position graph. As $t$ approaches $a$ the interval shrinks to a single instant, and the average velocities approach the instantaneous velocity at $a$ — the slope of the tangent line:
Nothing in this argument depends on $s$ being a position. For any quantity $y=f(x)$ the same two ideas appear side by side:
Average rate of change on $[a,a+h]$
$$\frac{\Delta y}{\Delta x}=\frac{f(a+h)-f(a)}{h}$$
A single number describing a whole interval · the slope of a secant line · computed from two values of $f$.
Instantaneous rate of change at $a$
$$f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$
A single number describing one instant · the slope of the tangent line · the limit of the average rates as $h\to0$.
📘 Definition — Instantaneous rate of change
The instantaneous rate of change of a function $f(x)$ at a value $a$ is its derivative $f'(a)$.
⚠️ Two forms of the same definition
$$f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h} \qquad\text{and}\qquad f'(a)=\lim_{t\to a}\frac{f(t)-f(a)}{t-a}$$
are the same thing written with $t=a+h$. Use whichever makes the algebra easier: the $h$-form suits expanding brackets, the $t$-form suits factoring (for example, $t^2-a^2=(t-a)(t+a)$ cancels the $t-a$ in the denominator).
Units. A rate of change is a ratio, so its units are always (units of the output) per (unit of the input). The derivative carries exactly the same units as the average rate it is the limit of.
Quantity $y=f(x)$
Input $x$
Meaning of $f'(a)$
Units of $f'(a)$
position $s(t)$ in metres
time $t$ in seconds
velocity at time $a$
m/s
temperature $T(t)$ in °F
time $t$ in hours
heating (or cooling) rate
°F/h
cost $C(q)$ in dollars
items produced $q$
marginal cost
dollars per unit
population $P(t)$
time $t$ in years
growth rate
people per year
The sign. If $f'(a)>0$ the quantity is increasing at $a$; if $f'(a)<0$ it is decreasing; if $f'(a)=0$ it is momentarily neither. The size $|f'(a)|$ tells how fast. For motion, a negative velocity means the position coordinate is decreasing — a dropped stone has negative velocity because its height is going down, not because it is slowing down.
An illustration in words. Suppose $T(t)$ is the temperature in a house, in °F, $t$ hours after 9 p.m., and we are told $T'(3)=-1.6$. This says that at midnight the temperature is falling at $1.6$ degrees per hour: if that rate stayed the same for the next hour, the house would cool by about $1.6^\circ$F. By contrast, a statement such as "the temperature dropped $10^\circ$F between 9 p.m. and 3 a.m." describes an average rate, $-\tfrac{10}{6}\approx-1.67$ °F/h, over a whole interval; it says nothing definite about any single moment. Computing velocities and other rates of change in applications is the subject of the section on derivatives as rates of change.
The Derivative as a Function
Finding the derivative at even a handful of values with the method above would quickly become tedious. Knowing the derivative at every point is far more valuable — so we define the derivative function.
📘 Definition — The derivative function
Let $f$ be a function. The derivative function, denoted $f'$, is the function whose domain consists of those values of $x$ for which the following limit exists:
$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}.$$
A function $f(x)$ is differentiable at $a$ if $f'(a)$ exists; it is differentiable on a set $S$ if it is differentiable at every point of the open set $S$.
✏️ Example 3 — Derivative of a square-root function
Find the derivative of $f(x)=\sqrt{x}$.
💡 Show solution▶
Step 1 Start directly from the definition:
$$f'(x)=\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt{x}}{h}.$$
Step 2 Multiply by the conjugate $\sqrt{x+h}+\sqrt{x}$ over itself:
$$=\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\cdot\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}=\lim_{h\to0}\frac{h}{h\left(\sqrt{x+h}+\sqrt{x}\right)}.$$
Step 3 Cancel $h$ and evaluate:
$$=\lim_{h\to0}\frac{1}{\sqrt{x+h}+\sqrt{x}}=\frac{1}{2\sqrt{x}}.$$
✏️ Example 4 — Derivative of a quadratic function
Find the derivative of $f(x)=x^2-2x$.
💡 Show solution▶
Step 1 Same procedure, no conjugate needed:
$$f'(x)=\lim_{h\to0}\frac{\left[(x+h)^2-2(x+h)\right]-\left[x^2-2x\right]}{h}.$$
Step 3 Factor out $h$, cancel, and evaluate:
$$=\lim_{h\to0}\frac{h(2x+h-2)}{h}=\lim_{h\to0}(2x+h-2)=2x-2.$$
Several notations express the same derivative. For $y=f(x)=x^2-2x$ we may write $y'=2x-2$, or $\dfrac{dy}{dx}=2x-2$, or $\dfrac{d}{dx}\left(x^2-2x\right)=2x-2$.
$f'(x)$
$\dfrac{dy}{dx}$
$y'$
$\dfrac{d}{dx}\big(f(x)\big)$
In place of $f'(a)$ we may also write $\left.\dfrac{dy}{dx}\right|_{x=a}$. The $\dfrac{dy}{dx}$ notation is called Leibniz notation and is very common in engineering and physics.
Graphing a Derivative
Since $f'(x)$ gives the slope of the tangent to $f(x)$, the two graphs must correspond. For $f(x)=\sqrt{x}$ we found $f'(x)=\dfrac{1}{2\sqrt{x}}$. The function is increasing over its whole domain, so its tangent slopes are positive and $f'(x)>0$ everywhere. As $x$ increases the tangents get less steep, and correspondingly $f'(x)$ decreases. Also $f'(0)$ is undefined and $\displaystyle\lim_{x\to0^+}f'(x)=+\infty$, matching the vertical tangent of $f$ at $0$.
For $f(x)=x^2-2x$ we found $f'(x)=2x-2$. Observe that $f$ is decreasing for $x<1$, and for those same values $f'(x)<0$; $f$ is increasing for $x>1$ and there $f'(x)>0$; and $f$ has a horizontal tangent at $x=1$, where $f'(1)=0$.
Interactive 2 — The Derivative as a Function
The left panel shows the tangent sliding along $f$. The right panel plots that slope as a point — and traces out the graph of $f'$.
$f$ with its tangent at $x$
$f'$ being traced out
$x$—
$f(x)$—
slope $f'(x)$—
$f$ is…—
✏️ Example 5 — Sketching a derivative from a graph
Use the graph of $f(x)$ below to sketch a graph of $f'(x)$.
Example 5 — $f$ has horizontal tangents at $x=-2$ and $x=3$.
💡 Show solution▶
Step 1 Read off where $f$ increases: on $(-2,3)$. There the tangent slopes are positive, so $f'(x)>0$ on $(-2,3)$.
Step 2 Read off where $f$ decreases: on $(-\infty,-2)$ and on $(3,+\infty)$. There $f'(x)<0$.
Step 3 The horizontal tangents at $x=-2$ and $x=3$ give $f'(-2)=0$ and $f'(3)=0$ — the two zeros of $f'$.
Step 4 So $f'$ is a downward parabola crossing the axis at $-2$ and $3$, negative outside and positive between. Select the third option in Interactive 2 to watch that curve being drawn.
Example 5, solution — $f'$ is positive on $(-2,3)$, negative outside, and zero at $-2$ and $3$.
Let $f(x)$ be a function and $a$ a point in its domain. If $f(x)$ is differentiable at $a$, then $f$ is continuous at $a$.
⚠️ The converse is false
A function that is continuous at a point need not be differentiable there. A function may be continuous at a point and fail to be differentiable for several reasons: a corner (the one-sided difference quotients settle on different numbers), a cusp, or a vertical tangent (the difference quotients grow without bound). Exercise VII below tests corners, and Exercise X tests all three cases.
Exercises
I
For $f(x)=x^2+3x+2$, find $f'(1)$.
Difference quotient as $h\to0$
$h$
quotient
$h \to 0^-$—
$h \to 0^+$—
$f'(a)$—
💡 Show solution▶
Step 1 $f(1)=1+3+2=6$, and
$$f(1+h)=(1+h)^2+3(1+h)+2=1+2h+h^2+3+3h+2=h^2+5h+6.$$
Step 2 Form the difference quotient:
$$\frac{f(1+h)-f(1)}{h}=\frac{h^2+5h}{h}=h+5\quad(h\neq0).$$
Step 3 Let $h\to0$: $f'(1)=5$.
II. Slope of the tangent line at $x=a$
II
Use the definition $m_{\tan}=\displaystyle\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$ to find the slope of the tangent line to the graph of $f$ at $x=a$.
1 $f(x)=3-4x$, $a=2$
2 $f(x)=x^2+x$, $a=2$
3 $f(x)=1-x-x^2$, $a=0$
4 $f(x)=2-3x^2$, $a=-2$
5 $f(x)=x^5+6$, $a=-1$
Secant closing in on the tangent
increment $h$—
secant slope—
tangent slope—
💡 Show solutions▶
1 $\dfrac{f(2+h)-f(2)}{h}=\dfrac{\left[3-4(2+h)\right]-(-5)}{h}=\dfrac{-4h}{h}=-4$, so $m_{\tan}=-4$. (Every secant of a line is the line itself, so the tangent slope equals the slope of the line.)
2 $f(2)=6$, so $\dfrac{(2+h)^2+(2+h)-6}{h}=\dfrac{h^2+5h}{h}=h+5\to5$. Thus $m_{\tan}=5$.
3 $f(0)=1$, so $\dfrac{\left[1-h-h^2\right]-1}{h}=-1-h\to-1$. Thus $m_{\tan}=-1$.
4 $f(-2)=-10$, so $\dfrac{\left[2-3(-2+h)^2\right]-(-10)}{h}=\dfrac{12h-3h^2}{h}=12-3h\to12$. Thus $m_{\tan}=12$.
5 $\dfrac{(-1+h)^5+6-5}{h}$. Expanding, $(-1+h)^5=-1+5h-10h^2+10h^3-5h^4+h^5$, so the quotient is $5-10h+10h^2-5h^3+h^4\to5$ (here $f(-1)=5$). Thus $m_{\tan}=5$.
II
Continued.
6 $f(x)=\dfrac{-3}{x-1}$, $a=4$
7 $f(x)=\dfrac{2}{x+3}$, $a=-4$
8 $f(x)=\dfrac{3}{x^2}$, $a=3$
9 $f(x)=\sqrt{x+8}$, $a=1$
Secant closing in on the tangent
increment $h$—
secant slope—
tangent slope—
💡 Show solutions▶
6 $f(4)=-1$. The quotient is
$$\frac{\dfrac{-3}{3+h}+1}{h}=\frac{\dfrac{-3+3+h}{3+h}}{h}=\frac{1}{3+h}\longrightarrow\frac13.$$
So $m_{\tan}=\tfrac13$.
7 $f(-4)=-2$. The quotient is
$$\frac{\dfrac{2}{-1+h}+2}{h}=\frac{\dfrac{2+2(-1+h)}{-1+h}}{h}=\frac{2}{-1+h}\longrightarrow-2.$$
So $m_{\tan}=-2$.
8 $f(3)=\tfrac13$. The quotient is
$$\frac{\dfrac{3}{(3+h)^2}-\dfrac13}{h}=\frac{9-(3+h)^2}{3h(3+h)^2}=\frac{-6h-h^2}{3h(3+h)^2}=\frac{-(6+h)}{3(3+h)^2}\longrightarrow\frac{-6}{27}=-\frac29.$$
So $m_{\tan}=-\tfrac29$.
9 $f(1)=3$. Multiply by the conjugate:
$$\frac{\sqrt{9+h}-3}{h}=\frac{(9+h)-9}{h\left(\sqrt{9+h}+3\right)}=\frac{1}{\sqrt{9+h}+3}\longrightarrow\frac16.$$
So $m_{\tan}=\tfrac16$.
III. $f'(a)$ from the definition
III
For the following functions $y=f(x)$, find $f'(a)$ using the definition of the derivative at a point.
1 $f(x)=5x+4$, $a=-1$
2 $f(x)=-7x+1$, $a=3$
3 $f(x)=x^2+9x$, $a=2$
4 $f(x)=3x^2-x+2$, $a=1$
5 $f(x)=\sqrt{x}$, $a=4$
Difference quotient as $h\to0$
$h$
quotient
$h \to 0^-$—
$h \to 0^+$—
$f'(a)$—
💡 Show solutions▶
1 $\dfrac{\left[5(-1+h)+4\right]-(-1)}{h}=\dfrac{5h}{h}=5$, so $f'(-1)=5$. (For any line $y=mx+b$ the derivative is $m$ everywhere.)
2 $\dfrac{\left[-7(3+h)+1\right]-(-20)}{h}=\dfrac{-7h}{h}=-7$, so $f'(3)=-7$.
3 $f(2)=22$ and $f(2+h)=(2+h)^2+9(2+h)=h^2+13h+22$, so the quotient is $\dfrac{h^2+13h}{h}=h+13\to13$.
4 $f(1)=4$ and $f(1+h)=3(1+h)^2-(1+h)+2=3h^2+5h+4$, so the quotient is $3h+5\to5$.
7 $f(4)=\tfrac12$. Combine and use the conjugate:
$$\frac{\dfrac{1}{\sqrt{4+h}}-\dfrac12}{h}=\frac{2-\sqrt{4+h}}{2h\sqrt{4+h}}=\frac{4-(4+h)}{2h\sqrt{4+h}\left(2+\sqrt{4+h}\right)}=\frac{-1}{2\sqrt{4+h}\left(2+\sqrt{4+h}\right)}\longrightarrow\frac{-1}{2(2)(4)}=-\frac{1}{16}.$$
9 $f(-1)=-\tfrac14$ and $f(-1+h)=\dfrac{1}{h-4}$, so
$$\frac{\dfrac{1}{h-4}+\dfrac14}{h}=\frac{4+(h-4)}{4h(h-4)}=\frac{1}{4(h-4)}\longrightarrow-\frac{1}{16}.$$
10 $f(1)=1$ and $f(1+h)=\dfrac{1}{(1+h)^3}$, so
$$\frac{\dfrac{1}{(1+h)^3}-1}{h}=\frac{1-(1+h)^3}{h(1+h)^3}=\frac{-3h-3h^2-h^3}{h(1+h)^3}=\frac{-3-3h-h^2}{(1+h)^3}\longrightarrow-3.$$
IV–VI. Derivative functions
IV
Sketch the graph of $f(x)=x^2-4$. On what interval is the graph of $f'(x)$ above the $x$-axis?
Watch $f'$ being traced
$f$ with its tangent
$f'$ being traced
$x$—
$f(x)$—
$f'(x)$—
$f$ is…—
💡 Show solution▶
Step 1 The graph of $f(x)=x^2-4$ is the parabola $y=x^2$ shifted down $4$ units, with vertex $(0,-4)$ and $x$-intercepts $\pm2$.
Step 2 From the definition, $f'(x)=2x$ (the constant $-4$ contributes nothing to the difference quotient).
Step 3 $f'(x)=2x>0$ exactly when $x>0$, so the graph of $f'$ lies above the $x$-axis on $(0,\infty)$ — which is precisely where $f$ is increasing.
V
Using the definition of a derivative, find the derivative of $f(x)=x^2-8$. VI. Using the definition of a derivative, find the derivative of $f(x)=x^2-6x$.
💡 Show solutions▶
V $$f'(x)=\lim_{h\to0}\frac{\left[(x+h)^2-8\right]-\left[x^2-8\right]}{h}=\lim_{h\to0}\frac{2xh+h^2}{h}=\lim_{h\to0}(2x+h)=2x.$$
VI $$f'(x)=\lim_{h\to0}\frac{\left[(x+h)^2-6(x+h)\right]-\left[x^2-6x\right]}{h}=\lim_{h\to0}\frac{2xh+h^2-6h}{h}=\lim_{h\to0}(2x+h-6)=2x-6.$$
Check $f'(x)=2x-6$ is zero at $x=3$, exactly where the parabola $y=x^2-6x$ has its vertex and a horizontal tangent. Use the mini-lab above to watch it.
VII. Differentiable at $x=1$?
VII
Using the definition of a derivative, check whether each function is differentiable at $x=1$.
1 $f(1)=2$ and both branches give $2$ at $x=1$, so $f$ is continuous there. From the left,
$$\lim_{h\to0^-}\frac{2\sqrt{1+h}-2}{h}=\lim_{h\to0^-}\frac{2}{\sqrt{1+h}+1}=1;$$
from the right, $\dfrac{3(1+h)-1-2}{h}=3$. Since $1\neq3$, $f$ is not differentiable at $1$.
2 $f(1)=1$ and the right branch also gives $1$, so $f$ is continuous. From the left, $\dfrac{-(1+h)^2+2-1}{h}=-2-h\to-2$; from the right, $\dfrac{(1+h)-1}{h}=1$. Since $-2\neq1$, not differentiable.
3 $f(1)=3$ and the left branch is the constant $3$, so $f$ is continuous at $1$. From the left the quotient is $\dfrac{3-3}{h}=0$; from the right it is $\dfrac{3(1+h)-3}{h}=3$. Since $0\neq3$, not differentiable.
4 $f(1)=2$ and $\dfrac21=2$, so $f$ is continuous. From the left, $\dfrac{2(1+h)-2}{h}=2$; from the right,
$$\frac{\dfrac{2}{1+h}-2}{h}=\frac{-2}{1+h}\longrightarrow-2.$$
Since $2\neq-2$, not differentiable.
The moral All four functions are continuous at $x=1$ and none is differentiable there. Continuity is necessary but not sufficient — each graph has a corner, where the left and right difference quotients settle on different numbers.
VIII. $f'(a)$ from the definition, and reading a limit as a derivative
VIII
For parts 1–3, use the definition of the derivative at a point to find $f'(a)$.
1 $f(x)=x^3$, $a=2$
2 $f(x)=\dfrac{3}{x+1}$, $a=2$
3 $f(x)=\sqrt{2x+1}$, $a=4$
For parts 4–6, each limit is the derivative $f'(a)$ of some function $f$ at some number $a$. Identify $f$ and $a$, then evaluate the limit.
1 $f(2)=8$ and $(2+h)^3=8+12h+6h^2+h^3$, so
$$\frac{f(2+h)-f(2)}{h}=\frac{12h+6h^2+h^3}{h}=12+6h+h^2\longrightarrow12.$$
Thus $f'(2)=12$.
2 $f(2)=\tfrac33=1$. Combine over $3+h$:
$$\frac{\dfrac{3}{3+h}-1}{h}=\frac{\dfrac{3-(3+h)}{3+h}}{h}=\frac{-h}{h(3+h)}=\frac{-1}{3+h}\longrightarrow-\frac13.$$
Thus $f'(2)=-\tfrac13$.
3 $f(4)=\sqrt9=3$. Multiply by the conjugate:
$$\frac{\sqrt{9+2h}-3}{h}\cdot\frac{\sqrt{9+2h}+3}{\sqrt{9+2h}+3}=\frac{(9+2h)-9}{h\left(\sqrt{9+2h}+3\right)}=\frac{2}{\sqrt{9+2h}+3}\longrightarrow\frac{2}{6}=\frac13.$$
Thus $f'(4)=\tfrac13$.
4 Match with $\dfrac{f(a+h)-f(a)}{h}$: $f(x)=\sqrt{x}$ and $a=4$, since $f(4)=2$. (Equally valid: $f(x)=\sqrt{4+x}$ with $a=0$.) Conjugate:
$$\frac{\sqrt{4+h}-2}{h}=\frac{(4+h)-4}{h\left(\sqrt{4+h}+2\right)}=\frac{1}{\sqrt{4+h}+2}\longrightarrow\frac14.$$
5 This is the $t$-form $\displaystyle\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$ with $f(x)=x^5$ and $a=1$ (note $f(1)=1$). Factor $x^5-1=(x-1)\left(x^4+x^3+x^2+x+1\right)$:
$$\lim_{x\to1}\left(x^4+x^3+x^2+x+1\right)=5.$$
6 Here $f(x)=\dfrac1x$ and $a=3$. Combine over $3(3+h)$:
$$\frac{\dfrac{3-(3+h)}{3(3+h)}}{h}=\frac{-h}{3h(3+h)}=\frac{-1}{3(3+h)}\longrightarrow-\frac19.$$
IX. Derivative functions from the definition
IX
Use the definition $f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$ to find the derivative function. State the domain of $f'$ in parts 2 and 3.
1 $f(x)=2x^2-5x+3$
2 $f(x)=\dfrac{3}{x-2}$
3 $f(x)=\sqrt{3x+1}$
4 $f(x)=x^3-12x$. Then find every $x$ at which $f'(x)=0$ (where the graph is momentarily flat).
2 Combine over $(x+h-2)(x-2)$:
$$\frac{\dfrac{3}{x+h-2}-\dfrac{3}{x-2}}{h}=\frac{3(x-2)-3(x+h-2)}{h(x+h-2)(x-2)}=\frac{-3h}{h(x+h-2)(x-2)}=\frac{-3}{(x+h-2)(x-2)}.$$
Letting $h\to0$: $f'(x)=\dfrac{-3}{(x-2)^2}$, with domain $x\neq2$ (the same as the domain of $f$).
3 Conjugate:
$$\frac{\sqrt{3x+3h+1}-\sqrt{3x+1}}{h}=\frac{(3x+3h+1)-(3x+1)}{h\left(\sqrt{3x+3h+1}+\sqrt{3x+1}\right)}=\frac{3}{\sqrt{3x+3h+1}+\sqrt{3x+1}}.$$
Letting $h\to0$: $f'(x)=\dfrac{3}{2\sqrt{3x+1}}$. The domain of $f$ is $x\ge-\tfrac13$, but $f'$ exists only for $x>-\tfrac13$ — at $x=-\tfrac13$ the tangent is vertical.
4 $(x+h)^3-12(x+h)-\left(x^3-12x\right)=3x^2h+3xh^2+h^3-12h$, so
$$f'(x)=\lim_{h\to0}\left(3x^2+3xh+h^2-12\right)=3x^2-12.$$
We need $f'(x)=0$: $3x^2=12$, so $x=-2$ and $x=2$.
X. Continuity and differentiability
X
Decide whether each function is continuous at the given point and whether it is differentiable there. Justify with one-sided difference quotients.
1 $f(2)=0$ and $\displaystyle\lim_{x\to2}|x-2|=0$, so $f$ is continuous. The quotient is $\dfrac{|h|}{h}$, which equals $-1$ for $h<0$ and $1$ for $h>0$. The one-sided limits $-1$ and $1$ differ, so $f$ has a corner and is not differentiable at $2$.
2 $f$ is continuous at $0$ (cube roots are continuous everywhere). The quotient is
$$\frac{h^{1/3}-0}{h}=\frac{1}{h^{2/3}}\longrightarrow+\infty\quad\text{from both sides},$$
since $h^{2/3}>0$. The limit does not exist as a number: there is a vertical tangent, so $f$ is not differentiable at $0$.
3 $f(1)=1$ and $\displaystyle\lim_{x\to1^+}(2x-1)=1$, so $f$ is continuous. From the left, $\dfrac{(1+h)^2-1}{h}=2+h\to2$; from the right, $\dfrac{2(1+h)-1-1}{h}=2$. Both equal $2$, so $f$ is differentiable at $1$, with $f'(1)=2$ — the pieces join smoothly.
4 $f(0)=0$ but $\displaystyle\lim_{x\to0^-}(x+1)=1\neq0$, so $f$ is not continuous at $0$. By Theorem 2.1 (contrapositive: not continuous $\Rightarrow$ not differentiable) it is not differentiable either. Indeed the left quotient $\dfrac{(h+1)-0}{h}\to-\infty$ as $h\to0^-$.
5 Continuous at $0$, with $f(0)=0$. The quotient is $\dfrac{h^{2/3}}{h}=\dfrac{1}{h^{1/3}}$, which $\to-\infty$ as $h\to0^-$ and $\to+\infty$ as $h\to0^+$. The graph has a cusp, so $f$ is not differentiable at $0$.
The moral Parts 1, 2 and 5 show the three ways a continuous function can fail to be differentiable — corner, vertical tangent, cusp. Part 3 shows that a piecewise function can still be differentiable at the join, and part 4 shows that a discontinuity rules out differentiability at once.
🔬 Research–Teaching NexusResearch-based · optional, about 30 minutes
Research question. How close does an average velocity get to the instantaneous velocity of a falling object as the time interval shrinks?
Why it matters. Sports scientists, crash investigators and robotics engineers estimate velocity from video frames in exactly this way.
What to do.
Film a ball dropped from about 2 m beside a measuring tape, using your phone's slow-motion mode (e.g. 240 frames per second).
Step through the video and record the height at several frames. Each frame is $\frac{1}{240}$ s apart (check your phone's frame rate).
Choose a time $t_0$. Compute average velocities $\frac{\Delta s}{\Delta t}$ over shrinking intervals starting at $t_0$.
Compare the limit with the model $v=g\,t$, $g\approx9.8$ m/s$^2$ ($t$ = time since release).
Hand in: A half-page note: your table of average velocities, your estimate of the instantaneous velocity, and one source of error.
Going further. Anywhere: use any free video-analysis app (e.g. Tracker) with any falling object.
Instructor note: Students should notice that very small intervals can make the estimate worse, because measurement error in $\Delta s$ is divided by a tiny $\Delta t$.
🧪 Self-check quiz
0/5 correct
Q1. The slope of the tangent to $f$ at $a$ is defined as:
Q2. The limit $\displaystyle\lim_{h\to0}\frac{(2+h)^3-8}{h}$ is $f'(a)$ for:
Q3. Using the definition, the derivative of $f(x)=x^2+1$ is found from which limit?
Q4. If $f'(x)<0$ on an interval, then on that interval $f$ is:
Q5. $f(x)=\left\{\begin{array}{cl} 3 & \quad x<1 \\[2pt] 3x & \quad x\ge1 \end{array}\right.$ at $x=1$ is: