What Is The Differentiability Of A Function?
The differentiability of a function at a point $x = a$ means that its derivative exists there, defined by the limit of the difference quotient:
$$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$
The function is differentiable at $a$ only when this limit exists and is finite. Because $h$ can approach $0$ from either side, the definition carries a hidden condition: the left-hand derivative and the right-hand derivative must both exist and be equal.
$$f'{-}(a) = \lim{h \to 0^{-}} \frac{f(a+h) - f(a)}{h}, \qquad f'{+}(a) = \lim{h \to 0^{+}} \frac{f(a+h) - f(a)}{h}$$
If $f'{-}(a) = f'{+}(a)$ and the common value is finite, then $f'(a)$ exists and equals that value. If the two disagree, or either runs off to infinity, the function is not differentiable at $a$.
A function is differentiable on an interval when it is differentiable at every point of that interval. For a closed interval $[a, b]$, the endpoints are tested with the appropriate one-sided derivative only.
Geometrically, $f'(a)$ is the slope of the tangent line to the graph at the point $\big(a, f(a)\big)$. So the differentiability of a function at a point is exactly the statement that the graph has one well-defined, non-vertical tangent line there.
How Do You Check Differentiability At A Point?
To check differentiability at $x = a$, evaluate the difference-quotient limit directly, or compute the left- and right-hand derivatives and confirm they match. The worked limit below shows the direct method.
Example 1: Show that $f(x) = x^2$ is differentiable at $x = 2$, and find $f'(2)$.
Apply the definition with $a = 2$:
$$f'(2) = \lim_{h \to 0} \frac{(2+h)^2 - 2^2}{h}$$
Expand the numerator:
$$= \lim_{h \to 0} \frac{4 + 4h + h^2 - 4}{h} = \lim_{h \to 0} \frac{4h + h^2}{h}$$
Cancel $h$ (valid because $h \neq 0$ inside the limit) and take the limit:
$$= \lim_{h \to 0} (4 + h) = 4$$
The limit exists and is finite, so $f$ is differentiable at $2$.
Final answer: $f'(2) = 4$. This matches the power-rule result $f'(x) = 2x$ evaluated at $x = 2$, and it is the slope of the tangent line to $y = x^2$ at the point $(2, 4)$.
What Is The Link Between Differentiability And Continuity?
Differentiability is the stronger condition. The core theorem of the topic states it precisely.
Theorem. If $f$ is differentiable at $x = a$, then $f$ is continuous at $x = a$.
Here is the short proof. Assume $f'(a)$ exists. Consider the gap between $f(x)$ and $f(a)$ and rewrite it:
$$\lim_{x \to a} \big[ f(x) - f(a) \big] = \lim_{x \to a} \left[ \frac{f(x) - f(a)}{x - a} \cdot (x - a) \right]$$
The first factor tends to $f'(a)$ and the second tends to $0$, so the product tends to $f'(a) \cdot 0 = 0$:
$$\lim_{x \to a} \big[ f(x) - f(a) \big] = 0 \quad \Longrightarrow \quad \lim_{x \to a} f(x) = f(a)$$
That last statement is the definition of continuity at $a$. So differentiable forces continuous.
The converse is false, and this is the fact that trips most students. A function can be continuous at a point and still not differentiable there. The standard witness is the absolute value function $f(x) = |x|$, which is continuous everywhere but has no derivative at $x = 0$. Continuity is necessary for differentiability, never sufficient.
What Are The Four Ways Differentiability Fails?
A function fails to be differentiable at a point in one of four ways. Three of them can happen even while the function stays continuous.
Discontinuity. If $f$ is not continuous at $a$, it cannot be differentiable at $a$ (the theorem above, read backwards). A jump such as the unit step $f(x) = 0$ for $x < 0$ and $f(x) = 1$ for $x \ge 0$ has no derivative at $0$.
Corner. The left and right slopes both exist but disagree. For $f(x) = |x|$ at $0$, the one-sided derivatives are $-1$ and $+1$.
Cusp. The one-sided slopes both run to infinity with opposite signs, pinching the graph to a point. The model case is $f(x) = x^{2/3}$ at $0$.
Vertical tangent. The slope runs to $+\infty$ (or $-\infty$) from both sides, so the tangent line is vertical. The model case is $f(x) = x^{1/3}$ at $0$.
Example 2: The corner of $f(x) = |x|$ at $x = 0$.
Split the absolute value by sign. For $h \to 0^{-}$, $|h| = -h$; for $h \to 0^{+}$, $|h| = h$:
$$f'{-}(0) = \lim{h \to 0^{-}} \frac{|h| - 0}{h} = \lim_{h \to 0^{-}} \frac{-h}{h} = -1, \qquad f'{+}(0) = \lim{h \to 0^{+}} \frac{h}{h} = +1$$
The one-sided derivatives are $-1$ and $+1$. They disagree, so $f'(0)$ does not exist, even though $|x|$ is continuous at $0$.
Example 3: The cusp of $f(x) = x^{2/3}$ at $x = 0$.
The difference quotient at $0$ is $\dfrac{h^{2/3} - 0}{h} = h^{-1/3}$:
$$f'{+}(0) = \lim{h \to 0^{+}} h^{-1/3} = +\infty, \qquad f'{-}(0) = \lim{h \to 0^{-}} h^{-1/3} = -\infty$$
Both one-sided slopes are infinite with opposite signs. The limit is not finite, so $f$ is not differentiable at $0$, though it is continuous there since $0^{2/3} = 0$. By contrast, $f(x) = x^{1/3}$ gives $h^{-2/3} \to +\infty$ from both sides, a vertical tangent rather than a cusp.
Table: The four ways the differentiability of a function fails at a point.
Failure type | Example at $x = 0$ | What the derivative does | Still continuous? |
|---|---|---|---|
Discontinuity | unit step function | one-sided limits disagree | No |
Corner | $f(x) = \lvert x \rvert$ | slopes $-1$ and $+1$ disagree | Yes |
Cusp | $f(x) = x^{2/3}$ | slopes $\to -\infty$ and $+\infty$ | Yes |
Vertical tangent | $f(x) = x^{1/3}$ | slope $\to +\infty$ both sides | Yes |
How Do You Check A Piecewise Function For Differentiability?
A piecewise function is differentiable at a join only when two conditions both hold at that point: the pieces meet continuously, and their one-sided derivatives agree. Checking continuity alone is not enough, and this is the single most common exam slip on the topic.
Example 4: Find $a$ and $b$ so that the function below is differentiable at $x = 1$.
$$f(x) = \begin{cases} x^2 & x \le 1 \ ax + b & x > 1 \end{cases}$$
Step 1, match the values (continuity). The left piece gives $f(1) = 1^2 = 1$; the right piece approaches $a(1) + b = a + b$. Continuity needs:
$$a + b = 1$$
Step 2, match the one-sided derivatives. The left piece has derivative $2x$, which is $2$ at $x = 1$; the right piece has derivative $a$. Differentiability needs:
$$a = 2$$
Step 3, solve. Substituting $a = 2$ into $a + b = 1$ gives $b = -1$.
Final answer: $a = 2$ and $b = -1$. With these values the pieces meet at $(1, 1)$ and share the slope $2$, so $f$ is differentiable at $x = 1$. Skipping Step 2 would have accepted any pair with $a + b = 1$, most of which produce a corner.
Why Does The Differentiability Of A Function Matter Geometrically?
Behind the algebra sits one geometric idea: a differentiable function is locally straight. Zoom far enough into a differentiable point and the graph becomes indistinguishable from its tangent line.
The derivative $f'(a)$ is the slope of that tangent line, so differentiability is the guarantee that a single, non-vertical tangent exists. See tangents and normals for the tangent-line construction.
A corner refuses to settle on one slope, because two different straight lines meet there. A cusp and a vertical tangent do have a limiting direction, but it is either doubled back on itself or straight up, neither of which is a finite slope.
Local straightness is why calculus can approximate a curved function by a line near a point. That approximation, $f(x) \approx f(a) + f'(a)(x - a)$, is the foundation of tangent-line estimates, Newton's method, and the whole idea of calculus as the study of local rates of change.
This is the reason continuity is not enough. A continuous graph can still have a kink, and a kink has no single direction to approximate with. Differentiability is continuity plus the promise of a definite direction.
Who Shaped The Study Of Differentiability?
The derivative was used for decades before anyone defined it carefully, and the gap between using it and defining it is one of the better stories in mathematics.
Two earlier chapters set this up:
Isaac Newton (1643–1727, England) and Gottfried Wilhelm Leibniz (1646–1716, Germany) independently built the derivative in the 1670s and 1680s, Newton through "fluxions" and Leibniz through the $\frac{dy}{dx}$ notation still used today. Their followers argued for a century over who was first.
Augustin-Louis Cauchy (1789–1857, France) put the derivative on the limit definition used in this article, replacing vague talk of "infinitely small" quantities with the precise $\lim_{h \to 0}$ statement that makes differentiability a testable condition.
Where Is The Differentiability Of A Function Used In The Real World?
The differentiability of a function is the mathematical test for "smooth enough to have a rate of change," and that question shows up across science and engineering.
Physics and motion. Velocity is the derivative of position and acceleration is the derivative of velocity. A point where position is non-differentiable would model an instantaneous, infinite jerk, which is why real motion is modelled with differentiable paths.
Engineering and design. Roads, railway tracks, and roller-coaster rails are built from curves joined to be differentiable at the seams, so a vehicle feels no sudden sideways snap. A corner in the track would be a corner in the math.
Computer graphics and animation. Smooth camera moves and character paths use spline curves stitched together to be differentiable, which is exactly what keeps motion from looking jerky on screen.
Economics. Marginal cost and marginal revenue are derivatives of the cost and revenue functions, and optimisation of profit relies on those functions being differentiable so the maximum can be found where the derivative is zero.
Machine learning. Training by gradient descent needs a differentiable loss function, because the gradient that guides each step is a derivative that has to exist.
One test, "does the derivative exist here," decides whether these models behave, which is why differentiability is checked before the calculus even begins.
What Are The Most Common Mistakes With Differentiability?
These three errors account for most lost marks on differentiability, drawn from the questions students actually ask on Quora ("why is a function not differentiable, examples on corner points"), Google's People Also Ask ("can a function be differentiable but not continuous"), and JEE-style point-of-non-differentiability problem sets.
Assuming continuity implies differentiability.
Where it slips in:
A student confirms a function is continuous at a point and concludes it must be differentiable there, treating the two words as the same.
Don't do this:
Do not read the theorem backwards. "Differentiable implies continuous" does not give you "continuous implies differentiable."
The correct way:
Use continuity only as a first filter. After it passes, still test the derivative. Remember $f(x) = |x|$: continuous at $0$, yet not differentiable there because the one-sided slopes are $-1$ and $+1$.
Checking only continuity at a piecewise seam.
Where it slips in:
On a piecewise function, a student matches the two pieces' values at the join, sees they agree, and stops, declaring the function differentiable.
Don't do this:
Do not treat "the pieces meet" as the whole test. Matching values is continuity, which is only half of what differentiability needs.
The correct way:
Run both checks at the join: the values must match (continuity) and the one-sided derivatives must match. In Example 4, $a + b = 1$ handled continuity, but only $a = 2$ made the slopes agree.
Forgetting the one-sided derivatives.
Where it slips in:
A student plugs a point into a single derivative formula and reports the answer, without noticing the point is a corner where the left and right derivatives differ.
Don't do this:
Do not compute one derivative and assume it settles the question at a suspicious point (an absolute value, a root, or a piecewise join).
The correct way:
At any suspect point, compute $f'{-}(a)$ and $f'{+}(a)$ separately. They must both exist and be equal, and finite, for $f'(a)$ to exist. A discontinuous function fails this before you even start.
Practice Problems On Differentiability Of A Function
Work each from the definition or the one-sided-derivative test. Answers follow each problem.
Use the limit definition to find $f'(3)$ for $f(x) = x^2$.
(Answer: $\lim_{h \to 0} \frac{(3+h)^2 - 9}{h} = \lim_{h \to 0}(6 + h) = 6$.)Is $f(x) = |x - 2|$ differentiable at $x = 2$?
(Answer: No. The left derivative is $-1$ and the right derivative is $+1$, a corner.)Is $f(x) = x^{1/3}$ differentiable at $x = 0$?
(Answer: No. The slope runs to $+\infty$ from both sides, a vertical tangent.)Find $k$ so that $f(x) = x^2$ for $x \le 2$ and $f(x) = 4x + k$ for $x > 2$ is differentiable at $x = 2$.
(Answer: Continuity needs $4 = 8 + k$, so $k = -4$; the slopes are $2x = 4$ and $4$, which already agree, so $k = -4$ makes it differentiable.)At which point is $f(x) = x^{2/3}$ not differentiable, and what feature sits there?
(Answer: At $x = 0$, a cusp.)True or false: if $f$ is continuous at $a$, then $f$ is differentiable at $a$.
(Answer: False. $f(x) = |x|$ is continuous at $0$ but not differentiable there.)
Where Should You Go Next After Differentiability Of A Function?
Differentiability is the gate into differential calculus, and a few natural doors open from here.
Derivative. Once you know the derivative exists, learn the rules (power, product, quotient, chain) that compute it without returning to the limit each time.
Differentiation of trigonometric functions. Apply differentiability to sine, cosine, and tangent, and see where the tangent function's own breaks appear.
Tangents and normals. Use the derivative as a slope to write the equation of the tangent line, the geometric payoff of differentiability.
If your child is building these calculus foundations, a live Bhanzu trainer teaches differentiability starting from the "why" (the smoothness and tangent-line ideas the definition is built to capture) in the Bhanzu math classes.
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