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The limit of the difference quotient, why it implies continuity, and the three ways it can fail to exist.
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By the end of this lesson you will be able to compute a derivative from the definition, prove that differentiability implies continuity, and identify the three standard ways the limit can fail — a corner, a vertical tangent and oscillation. You will also be able to read the derivative as the coefficient of the best linear approximation, which is the reading the rest of the course and every later course depends on.
The epsilon-delta limit, the algebra of limits, and continuity. The derivative is one particular limit and nothing more; every rule of calculus is a theorem about it, proved with the tools of the last unit.
The difference quotient at a point is the change in the function divided by the change in the input. A corner is a point where the two one-sided quotients converge to different numbers. A function is smooth on a set when derivatives of every order exist there.
$f$ is differentiable at $c$ when $$f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}$$ exists. The quotient is undefined at $h = 0$, which is exactly the situation a limit is built for: the deleted neighbourhood is not a technicality here but the whole point.
Differentiable implies continuous. For $h \ne 0$, $$f(c+h) - f(c) = \frac{f(c+h)-f(c)}{h} \cdot h \to f'(c) \cdot 0 = 0.$$ The converse fails: $|x|$ is continuous at $0$ with one-sided quotients $1$ and $-1$.
The approximation reading. Rearranged, differentiability says $$f(c+h) = f(c) + f'(c)h + E(h), \qquad \frac{E(h)}{h} \to 0.$$ The error is small relative to $h$, and no other slope achieves that. This is the definition that survives into several variables, where difference quotients cannot be formed.
The rules are theorems. Sum, product, quotient and chain rules are proofs about this limit, each an application of the algebra of limits with a middle term inserted. The product rule's proof is the splitting move from unit 1 once more.
Another way: picture
The secant through the graph at $c$ and at $c + h$ has slope equal to the difference quotient. Differentiability says the secants settle on one line as the second point slides in, and the tangent is that limiting line rather than a line touching at one point — a definition that would admit far too much.
Another way: steps
To differentiate from the definition:
Step 3 is where the indeterminate form disappears, and it is available only because the limit never inspects $h = 0$.
A corner. $|x|$ at $0$: both one-sided limits of the quotient exist and disagree. The function is continuous; the derivative is not there.
A vertical tangent. $x^{1/3}$ at $0$: the quotient is $h^{-2/3}$, which grows without bound. The limit does not exist as a real number, though the graph has a perfectly definite direction.
Oscillation. $x\sin(1/x)$ with value $0$ at $0$: the quotient is $\sin(1/h)$, which oscillates for ever. Continuous, and no one-sided derivatives either.
Multiply by one more factor of $x$ and the picture changes: $x^2\sin(1/x)$ has quotient $h\sin(1/h) \to 0$, so it is differentiable at $0$, with derivative $0$. Its derivative elsewhere contains $\cos(1/x)$, which has no limit at $0$ — so the derivative exists everywhere and is discontinuous at $0$. Differentiability does not give a continuous derivative, and that example is worth keeping.
| Rule | The identity it rests on |
|---|---|
| sum | the quotient of a sum is the sum of the quotients |
| product | $\dfrac{f(c+h)g(c+h) - f(c)g(c)}{h} = f(c+h)\dfrac{g(c+h)-g(c)}{h} + g(c)\dfrac{f(c+h)-f(c)}{h}$ |
| quotient | the same insertion, applied to $1/g$ after bounding $g$ away from $0$ |
| chain | the quotient for the composite, multiplied and divided by the inner increment |
The product rule's identity is the splitting move: insert $\pm f(c+h)g(c)$, then use that $f$ is continuous at $c$ — which it is, because it is differentiable. The chain rule's proof needs care where the inner increment vanishes, which is why a careful statement defines an auxiliary function rather than dividing.
None of these is new mathematics. Each is the algebra of limits applied to an identity, and that is the sense in which this course proves calculus rather than using it.
Letting $h$ be zero. The cancellation of $h$ happens while $h \ne 0$; the limit is taken afterwards. Setting $h = 0$ first produces $0/0$ and no information.
Averaging one-sided derivatives. At a corner the two one-sided limits exist and differ, and there is no derivative. Their average is a number with no claim on the definition.
Thinking continuity is nearly enough. There are functions continuous everywhere and differentiable nowhere. Continuity is a strictly weaker condition and not a near miss.
Expecting the derivative to be continuous. It need not be, and $x^2\sin(1/x)$ shows it. The functions whose derivatives are continuous have their own name for a reason.
Let $f(x) = 1/x$ and $c \ne 0$. The quotient is $\dfrac{1}{h}\left(\dfrac{1}{c+h} - \dfrac{1}{c}\right)$.
Write it down with the increment non-zero.
Combining over a common denominator gives $\dfrac{-1}{c(c+h)}$, after the factor $h$ cancels.
Cancel while the increment is non-zero.
Letting $h \to 0$ gives $-1/c^2$, using that the denominator's limit $c^2$ is non-zero.
The quotient rule's hypothesis, checked.
Let $f(x) = x^2$ for rational $x$ and $0$ for irrational $x$. At $0$ the quotient is $h$ or $0$.
Both cases, computed.
Either way it is at most $|h|$ in size, so it tends to $0$: the derivative at $0$ is $0$.
Squeeze, with no case distinction needed.
Anywhere else $f$ is not even continuous, so it is not differentiable. One point of differentiability, and no interval of it.
Differentiability is a pointwise notion.
The quotient at $0$ is $h|h|/h = |h|$, which tends to $0$, so $f'(0) = 0$.
Compute the quotient and take the limit.
Away from $0$, $f(x) = x^2$ for $x > 0$ and $-x^2$ for $x < 0$, so $f'(x) = 2|x|$.
Differentiate on each piece.
And $2|x| \to 0 = f'(0)$, so the derivative is continuous at $0$ — this function is smooth enough to be differentiable once with a continuous derivative, and not twice, since $f''$ has a jump at $0$. Each extra derivative is a genuinely new requirement.
Put the steps of computing the derivative of $f(x) = 3x^2$ at a point $c$, from the definition, into order.
Number the steps in order (write the number in the box):
Let $f(x) = 9x^2 + 3x$. Compute $f'(1)$ from the definition.
Answer:
Match each function, at the point $8$ or $0$ as stated, to what it shows.
| differentiable, and so continuous | continuous but not differentiable | differentiable, with a derivative that is not continuous | not continuous, and so not differentiable | |
|---|---|---|---|---|
| $x^2$ at $8$ | ||||
| $|x|$ at $0$ | ||||
| $x^2\sin(1/x)$, with value $0$ at $0$ | ||||
| the function equal to $0$ below $8$ and $1$ from $8$ on, at $8$ |
Build the proof that a function differentiable at $c$ is continuous at $c$.
This task has no paper form; do it on a device.
Here is an argument that $f(x) = 5|x|$ is differentiable at $0$. Mark the one step that does not follow.
This task has no paper form; do it on a device.
$f$ is differentiable at $8$. In what sense is $f(8) + f'(8)h$ the *best* linear approximation to $f(8 + h)$?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Put the steps of computing the derivative of $f(x) = 4x^2$ at a point $c$, from the definition, into order.
Number the steps in order (write the number in the box):
You can differentiate from the definition, prove that differentiability implies continuity, and name what goes wrong at a corner. Say in your own words why the derivative does not have to be continuous. Next: the theorem that turns a fact about the derivative into a fact about the function.
10. Your turn: is $f(x) = x|x|$ differentiable at $0$, and is $f'$ continuous there?, step 3