Why a Sharp Corner Stops a Function From Having a Derivative

A derivative measures slope, so a function needs a single, well-defined tangent at a point to be differentiable there. Smooth curves pass this test, but sharp corners and jumps fail it — and knowing which functions are differentiable, and what their derivatives are, underpins the rest of calculus.

This lesson covers the modulus and greatest integer functions, derivatives of exponential and logarithmic functions, derivatives of inverse trigonometric functions, and the link between continuity and differentiability.

**A function is differentiable at a when its left-hand derivative ${\lim }_{h\to 0^{-}}\frac{f(a+h)-f(a)}{h}$ equals its right-hand derivative ${\lim }_{h\to 0^{+}}\frac{f(a+h)-f(a)}{h}$; $|x|$ is continuous at 0 but not differentiable there, and $[x]$ is neither continuous nor differentiable at any integer.

Modulus function at 0:**

- Continuous, since both one-sided limits and $f(0)$ equal 0
- Left-hand derivative ${\lim }_{h\to 0^{-}}\frac{|h|}{h}=-1$
- Right-hand derivative ${\lim }_{h\to 0^{+}}\frac{|h|}{h}=1$
- Since $-1\ne 1$, $|x|$ is not differentiable at 0

Greatest integer function at 2:

- The left-hand limit is 1 and the right-hand limit is 2, so $[x]$ is discontinuous at 2
- A function that is not continuous at a point cannot be differentiable there
- Between integers $[x]$ is constant, so its derivative there is 0

Worked example. For $f(x)=|x-3|$, the one-sided derivatives at 3 are $-1$ and 1, so f is continuous but not differentiable at 3; at $x=5$, $f(x)=x-3$ and $f^{\prime }(5)=1$.

An everyday example. A cricket ball bouncing off the pitch changes direction abruptly — its height graph has a sharp corner at the bounce.

The substance. **$x|x|$ is differentiable at 0** — both its one-sided derivatives are 0, even though $|x|$ alone is not differentiable there.

**The derivative of $e^x$ is $e^x$, of $a^x$ is $a^x\log a$, of $\log x$ is $\frac{1}{x}$, and of ${\log }_{a}x$ is $\frac{1}{x\log a}$, all derived from first principles using ${\lim }_{h\to 0}\frac{e^h-1}{h}=1$ and ${\lim }_{h\to 0}\frac{\log (1+h)}{h}=1$.

Derivation for $e^x$:**

$$\frac{e^{x+h}-e^x}{h}=e^x\times \frac{e^h-1}{h}\to e^x$$

**Derivation for $\log x$:**

$$\frac{\log (x+h)-\log x}{h}=\frac{1}{x}\times \frac{\log \left(1+\frac{h}{x}\right)}{\frac{h}{x}}\to \frac{1}{x}$$

Worked examples:

- $\frac{d}{dx}(e^{3x})=3e^{3x}$
- $\frac{d}{dx}(2^x)=2^x\log 2$, so at $x=3$ the slope is $8\log 2\approx 5.55$
- $\frac{d}{dx}({\log }_{10}x)=\frac{1}{x\log 10}$, which is about 0.0869 at $x=5$
- $\frac{d}{dx}(x^2{e}^x)=e^x(x^2+2x)$

An everyday example. **A deposit growing continuously as $P{e}^{rt}$ increases at a rate proportional to the amount already there — the defining property of the exponential function.

The substance. Here log means the natural logarithm, to base e** — for base 10 the extra factor $\frac{1}{\log 10}$ appears.

**Writing $y={\sin }^{-1}x$ as $x=\sin y$ and differentiating gives $\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$, and the same method gives $\frac{d}{dx}({\tan }^{-1}x)=\frac{1}{1+x^2}$ and the other inverse trigonometric derivatives.

Derivation for ${\sin }^{-1}x$.** From $x=\sin y$, $\frac{dx}{dy}=\cos y=\sqrt{1-x^2}$, taking the positive root because $\cos y\ge 0$ on $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$. So $\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$.

Standard results:

- $\frac{d}{dx}({\sin }^{-1}x)=\frac{1}{\sqrt{1-x^2}}$ and $\frac{d}{dx}({\cos }^{-1}x)=-\frac{1}{\sqrt{1-x^2}}$
- $\frac{d}{dx}({\tan }^{-1}x)=\frac{1}{1+x^2}$ and $\frac{d}{dx}({\cot }^{-1}x)=-\frac{1}{1+x^2}$
- $\frac{d}{dx}({\sec }^{-1}x)=\frac{1}{|x|\sqrt{x^2-1}}$

Worked example (substitution). Differentiate $y={\tan }^{-1}\frac{2x}{1-x^2}$ for $|x|<1$. Put $x=\tan \theta$: then $y={\tan }^{-1}(\tan 2\theta )=2\theta =2{\tan }^{-1}x$, so $\frac{dy}{dx}=\frac{2}{1+x^2}$, which is 1.6 at $x=0.5$.

An everyday example. A lighthouse lamp turning to follow a boat sailing past the coast changes its angle at a rate given by the derivative of an inverse tangent.

The substance. Substitution turns a messy derivative into a one-liner — the chain rule applied directly to ${\tan }^{-1}\frac{2x}{1-x^2}$ gives the same answer only after far more algebra.

Every function that is differentiable at a point is continuous there, but the converse fails: a function can be continuous yet not differentiable, typically at a sharp corner, a cusp or a vertical tangent.

Proof that differentiable implies continuous.

$$f(a+h)-f(a)=\frac{f(a+h)-f(a)}{h}\times h\to f^{\prime }(a)\times 0=0$$

so $f(a+h)\to f(a)$, which is continuity.

Continuous but not differentiable:

- Corner — $|x|$ at 0
- Cusp — $x^{2/3}$ at 0
- Vertical tangent — $x^{1/3}$ at 0

Worked example. Where is $f(x)=|x-1|+|x-4|$ not differentiable? For $x<1$, $f(x)=5-2x$ with slope $-2$; for $1<x<4$, $f(x)=3$ with slope 0; for $x>4$, $f(x)=2x-5$ with slope 2. The slope changes suddenly at 1 and 4, so f is not differentiable there, though it is continuous everywhere.

An everyday example. A taxi fare graph that switches from one per-kilometre rate to another is continuous at the switch but has a corner there.

The substance. Discontinuity always rules out differentiability — so checking continuity first can save a derivative calculation.