Differentiability — Theory and Solved Exercises
Theoretical Recall
A function f is differentiable at x₀ if the limit
\[f'(x_0) = \lim_{h\to0} \frac{f(x_0+h)-f(x_0)}{h}\]exists and is finite.
Equivalently, when the corresponding one-sided limits exist, differentiability requires:
\[f'_-(x_0)=f'_+(x_0).\]Differentiability and Continuity
If f is differentiable at x₀, then f is continuous at x₀.
Therefore:
\[f\text{ differentiable at }x_0 \quad\Longrightarrow\quad f\text{ continuous at }x_0.\]The converse is false: a function can be continuous without being differentiable.
Corners and Cusps
A corner occurs when the left and right derivatives are finite but different.
For example:
\[f(x)=|x|\]is continuous at zero, but:
\[f'_-(0)=-1,\]while:
\[f'_+(0)=1.\]A cusp occurs when the one-sided derivatives become infinite with opposite signs.
Vertical Tangents
A vertical tangent may occur when the difference quotient tends to +∞ or −∞.
In that case the ordinary derivative is not finite, so the function is not differentiable at that point in the usual sense.
Piecewise Functions
For a piecewise function to be differentiable at a junction x₀, two conditions are required.
First, continuity:
\[\lim_{x\to x_0^-}f(x) = f(x_0) = \lim_{x\to x_0^+}f(x).\]Second, equality of the one-sided derivatives:
\[f'_-(x_0)=f'_+(x_0).\]Higher-Order Smoothness
The notation:
\[f\in C^k(\mathbb{R})\]means that f has derivatives up to order k and that all these derivatives are continuous on ℝ.
The notation:
\[f\in C^\infty(\mathbb{R})\]means that f is infinitely differentiable.
Exercises
Exercise 1 — Continuous but Not Differentiable
Consider:
\[f(x)= \begin{cases} x^2, & x\ge0,\\ -x, & x<0. \end{cases}\]Verify continuity and differentiability at x = 0.
Solution.
For continuity, consider the left-hand limit:
\[\lim_{x\to0^-}f(x) = \lim_{x\to0^-}(-x) = 0.\]The right-hand limit is:
\[\lim_{x\to0^+}f(x) = \lim_{x\to0^+}x^2 = 0.\]Moreover:
\[f(0)=0.\]Therefore:
\[\lim_{x\to0}f(x)=f(0).\]The function is continuous at zero.
Now compute the left derivative:
\[f'_-(0) = \lim_{h\to0^-} \frac{f(h)-f(0)}{h}.\]For h < 0:
\[f(h)=-h.\]Thus:
\[f'_-(0) = \lim_{h\to0^-} \frac{-h}{h} = -1.\]For the right derivative:
\[f'_+(0) = \lim_{h\to0^+} \frac{h^2}{h}.\]Therefore:
\[f'_+(0) = \lim_{h\to0^+}h = 0.\]Since:
\[-1\ne0,\]the function is not differentiable at zero.
Final Result
\[f\in C^0(\mathbb{R}), \qquad f\notin C^1(\mathbb{R})\]Exercise 2 — Absolute Value and Radical
Study the differentiability of:
\[f(x)=x\sqrt{|x|}.\]Solution.
For x > 0:
\[f(x)=x^{3/2}.\]Therefore:
\[f'(x)=\frac32\sqrt{x}.\]For x < 0:
\[f(x) = x\sqrt{-x} = -(-x)^{3/2}.\]Differentiating:
\[f'(x) = \frac32\sqrt{-x}.\]At zero:
\[f(0)=0.\]The difference quotient is:
\[\frac{f(h)-f(0)}{h} = \frac{h\sqrt{|h|}}{h}.\]For h ≠ 0:
\[\frac{f(h)}{h} = \sqrt{|h|}.\]Hence:
\[f'(0) = \lim_{h\to0}\sqrt{|h|} = 0.\]Moreover:
\[\lim_{x\to0^-}f'(x)=0,\]and:
\[\lim_{x\to0^+}f'(x)=0.\]Thus the derivative is continuous at zero.
Final Result
\[f\in C^1(\mathbb{R})\]Exercise 3 — A Smooth Absolute-Value Power
Consider:
\[f(x)=|x|\sqrt{|x|}.\]Study differentiability at x = 0.
Solution.
We can write:
\[f(x)=|x|^{3/2}.\]For x > 0:
\[f(x)=x^{3/2},\]so:
\[f'(x)=\frac32\sqrt{x}.\]For x < 0:
\[f(x)=(-x)^{3/2}.\]Therefore:
\[f'(x) = -\frac32\sqrt{-x}.\]At zero:
\[\frac{f(h)-f(0)}{h} = \frac{|h|^{3/2}}{h}.\]For h > 0:
\[\frac{|h|^{3/2}}{h} = \sqrt{h}.\]For h < 0:
\[\frac{|h|^{3/2}}{h} = -\sqrt{-h}.\]Both expressions tend to zero.
Therefore:
\[f'(0)=0.\]Furthermore:
\[\lim_{x\to0}f'(x)=0=f'(0).\]Author’s note: Although the function involves an absolute value and a radical, the exponent 3/2 is greater than 1. This is sufficient for first-order differentiability at the origin, but not for arbitrary smoothness.
Final Result
\[f\in C^1(\mathbb{R})\]Exercise 4 — Endpoint and Vertical Tangent
Study:
\[f(x)=|x|\sqrt{1-x}\]at x = 1.
Solution.
The square root requires:
\[1-x\ge0.\]Thus the domain is:
\[(-\infty,1].\]At x = 1:
\[f(1)=0.\]Since 1 is an endpoint of the domain, we consider the derivative from within the domain:
\[\lim_{h\to0^-} \frac{f(1+h)-f(1)}{h}.\]For h sufficiently close to zero from the left:
\[|1+h|=1+h.\]Therefore:
\[\frac{f(1+h)-f(1)}{h} = \frac{(1+h)\sqrt{-h}}{h}.\]Since h < 0:
\[h=-|h|.\]Hence:
\[\frac{(1+h)\sqrt{-h}}{h} = -\frac{1+h}{\sqrt{-h}}.\]As h → 0⁻:
\[-\frac{1+h}{\sqrt{-h}} \to-\infty.\]Thus there is a vertical tangent at the endpoint.
The ordinary finite derivative does not exist.
Final Result
\[f\text{ is not differentiable at }x=1\]Exercise 5 — Failure of Continuity
Consider:
\[f(x)=x^x\log x, \qquad x>0,\]and define:
\[f(0)=0.\]Check continuity and differentiability at x = 0.
Solution.
For x > 0:
\[x^x=e^{x\log x}.\]We know that:
\[\lim_{x\to0^+}x\log x=0.\]Therefore:
\[\lim_{x\to0^+}x^x=1.\]On the other hand:
\[\lim_{x\to0^+}\log x=-\infty.\]Hence:
\[x^x\log x\to-\infty.\]Therefore:
\[\lim_{x\to0^+}f(x)\ne f(0).\]The extension f(0) = 0 is not continuous at zero.
Since differentiability implies continuity, the function cannot be differentiable there.
Final Result
\[f\text{ is neither continuous nor differentiable at }x=0\]Exercise 6 — Differentiability of a Piecewise Function
Consider:
\[f(x)= \begin{cases} ax+b, & x<0,\\ x^2, & x\ge0. \end{cases}\]Find a and b so that f is differentiable at x = 0.
Solution.
Differentiability first requires continuity.
The left-hand limit is:
\[\lim_{x\to0^-}(ax+b)=b.\]Since:
\[f(0)=0,\]continuity requires:
\[b=0.\]Now compute the one-sided derivatives.
From the left:
\[f'_-(0)=a.\]From the right:
\[f'_+(0) = \lim_{h\to0^+} \frac{h^2}{h} = 0.\]Differentiability requires:
\[a=0.\]Thus:
\[a=0, \qquad b=0.\]With these values, the resulting function is x² for x ≥ 0 and 0 for x < 0, and its derivative is continuous at zero.
Final Result
\[a=0, \qquad b=0\]Exercise 7 — Smoothness of Absolute Powers
For which real values of a is:
\[f(x)=|x|^a\]of class Cᵏ on ℝ?
Solution.
First, if a < 0, the function is not defined at zero.
If:
\[a=0,\]then:
\[|x|^0=1,\]so the function is infinitely differentiable.
If a is a positive even integer:
\[a=2m,\]then:
\[|x|^{2m}=x^{2m}.\]Thus the function is a polynomial and belongs to C∞(ℝ).
Now suppose a > 0 and a is not an even integer.
For x ≠ 0, repeated differentiation produces terms whose magnitude behaves like:
\[|x|^{a-j}\]after j derivatives.
If:
\[a>k,\]all derivatives up to order k extend continuously to zero.
If a is an odd positive integer, say:
\[a=2m+1,\]| then | x | ᵃ is of class C²ᵐ but not C²ᵐ⁺¹. |
For non-integer positive a, the same general threshold applies: the function belongs to Cᵏ whenever k < a, while smoothness fails once the differentiation order reaches or exceeds the relevant singular exponent.
Therefore, for an integer k ≥ 0, the complete criterion is:
\[|x|^a\in C^k(\mathbb{R})\]if a = 0, or a is a positive even integer, or a > k.
| Author’s note: The even-integer case is exceptional because the absolute value disappears algebraically: | x | ²ᵐ = x²ᵐ. |
Final Result
\[|x|^a\in C^k(\mathbb{R}) \iff a=0 \ \text{or}\ a\in2\mathbb{N} \ \text{or}\ a>k\]Exercise 8 — Matching Derivatives
Consider:
\[f(x)= \begin{cases} ax^2+bx, & x\ge0,\\ \sin x, & x<0. \end{cases}\]Find a and b so that f is continuous and differentiable at x = 0.
Solution.
For continuity:
\[\lim_{x\to0^-}\sin x=0.\]For the right branch:
\[f(0)=0.\]Therefore continuity is automatic.
Now consider the derivatives.
From the left:
\[f'_-(0) = \cos0 = 1.\]For x > 0:
\[f'(x)=2ax+b.\]Hence:
\[f'_+(0)=b.\]Differentiability requires:
\[b=1.\]The parameter a does not affect differentiability at zero.
Final Result
\[b=1, \qquad a\in\mathbb{R}\]Exercise 9 — The Absolute Value Function
Study continuity and differentiability of:
\[f(x)=|x|\]at x = 0.
Solution.
Since:
\[\lim_{x\to0}|x|=0=f(0),\]the function is continuous at zero.
Now consider the difference quotient:
\[\frac{|h|-0}{h}.\]For h > 0:
\[\frac{|h|}{h}=1.\]Thus:
\[f'_+(0)=1.\]For h < 0:
\[\frac{|h|}{h}=-1.\]Thus:
\[f'_-(0)=-1.\]Since the one-sided derivatives are different:
\[f'_-(0)\ne f'_+(0).\]Therefore f is not differentiable at zero.
Geometrically, the graph has a corner at the origin.
Final Result
\[f\in C^0(\mathbb{R}), \qquad f\notin C^1(\mathbb{R})\]Exercise 10 — A C² Piecewise Function
Find a, b and c so that:
\[f(x)= \begin{cases} ax^2+bx+c, & x<0,\\ \cos x, & x\ge0 \end{cases}\]belongs to C²(ℝ).
Solution.
Both branches are infinitely differentiable away from zero. Therefore we only need to match the function and its first two derivatives at x = 0.
For continuity:
\[\lim_{x\to0^-}f(x)=c.\]Since:
\[f(0)=\cos0=1,\]we require:
\[c=1.\]For x < 0:
\[f'(x)=2ax+b.\]Thus:
\[f'_-(0)=b.\]For x > 0:
\[f'(x)=-\sin x.\]Therefore:
\[f'_+(0)=0.\]Hence:
\[b=0.\]Now compute the second derivatives.
For x < 0:
\[f''(x)=2a.\]Thus:
\[f''_-(0)=2a.\]For x > 0:
\[f''(x)=-\cos x.\]Therefore:
\[f''_+(0)=-1.\]For the second derivative to be continuous:
\[2a=-1.\]Hence:
\[a=-\frac12.\]Final Result
\[a=-\frac12, \qquad b=0, \qquad c=1\]