Differentiability and First Principles

MATDifferentiability

Differentiability at a point means the derivative exists there, i.e. the limit of the difference quotient exists and the left-hand and right-hand derivatives are equal.

The first-principles (ab initio) definition derives a derivative directly from this limit. ISC tests both deriving standard derivatives from scratch and examining differentiability of modulus and piecewise functions at corners.

Derivative from first principles

f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}

The derivative exists at

x

only if this limit exists and is finite.

Left- and right-hand derivatives

Lf'(a) = \lim_{h \to 0^-} \dfrac{f(a+h)-f(a)}{h}, \quad Rf'(a) = \lim_{h \to 0^+} \dfrac{f(a+h)-f(a)}{h}

f

is differentiable at

a

iff

Lf'(a)=Rf'(a)

(both finite and equal).

Differentiability implies continuity

f \text{ differentiable at } a \;\Rightarrow\; f \text{ continuous at } a

The converse is false:

|x-3|

is continuous at

3

but not differentiable there.

First-principles results

\dfrac{d}{dx}\sqrt{x} = \dfrac{1}{2\sqrt{x}}, \qquad \dfrac{d}{dx}\cos x = -\sin x

Derived using conjugates for

\sqrt{x}

and the identity

\cos(x+h)-\cos x=-2\sin\!\left(x+\tfrac h2\right)\sin\tfrac h2

To test differentiability at a corner point, compute $Lf'(a)$ and $Rf'(a)$ separately; equality is required.
For $f(x)=|x-3|$ , $Lf'(3)=-1$ and $Rf'(3)=+1$ , so it is not differentiable at $3$ though continuous.
Continuity is necessary but not sufficient for differentiability; always check continuity first when a function fails to be differentiable.
For $\sqrt{x}$ , multiply the difference quotient by the conjugate $\sqrt{x+h}+\sqrt{x}$ to clear the surd.
For trig functions, use sum-to-product identities and the limits $\dfrac{\sin h}{h}\to 1$ , $\dfrac{1-\cos h}{h}\to 0$ .
$f(x)=|x|+|x-1|$ has corners at $x=0$ and $x=1$ , so it is non-differentiable at both while continuous everywhere.
A function with a vertical tangent (infinite slope) is also non-differentiable even if continuous.

Where the marks go

Claiming a function is non-differentiable without computing both one-sided derivatives explicitly.
Forgetting that $h\to 0^-$ means $h$ is negative, which flips the sign when simplifying $|h|/h$ .
Assuming continuity guarantees differentiability (the converse error).
Mishandling the sign in $\dfrac{d}{dx}\cos x$ , giving $+\sin x$ instead of $-\sin x$ from first principles.

How the board asks it

Derive / provederivative from first principles
Differentiate $\sqrt{\sin x}$ with respect to $x$ from first principles.
Numericalleft- and right-hand derivatives
Examine the differentiability of $f(x)=|x-2|$ at $x=2$ by computing $Lf'(2)$ and $Rf'(2)$ separately.
Numericaldifferentiability of piecewise functions
Find the values of $a$ and $b$ so that the function $f(x)=\begin{cases} x^2+1 & x\le 1\\ ax+b & x>1 \end{cases}$ is differentiable at $x=1$ .
Give reasonsdifferentiability implies continuity
Show that the function $f(x)=|x|+|x-1|$ is continuous everywhere but is not differentiable at $x=0$ and $x=1$ .
Multiple choicevertical tangent (infinite slope)
The function $f(x)=x^{1/3}$ at $x=0$ is: (a) differentiable, (b) continuous but not differentiable, (c) discontinuous, (d) neither continuous nor differentiable. Choose the correct option.

Practise this topic

PreviousContinuity at a Point and on an Interval NextChain Rule and Standard Derivatives

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.