Derivative Proof Workshop

Organizing a Derivative Proof

The derivative is defined by a limit of difference quotients. Many derivative proofs become manageable once that quotient is written down and simplified before any limit law is used. The proof of Cauchy’s Mean Value Theorem in the previous tutorial illustrated a broader habit: choose a construction that makes the desired conclusion emerge from a simple calculation. Here, the construction is usually local. We examine $f(a+h)-f(a)$, factor or estimate it, and then divide by $h$.

For a function $f$ defined near an interior point $a$, differentiability with derivative $L$ means

\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}=L.

A proof should make clear that $h\ne 0$ while the quotient is formed, that $a+h$ lies in the domain, and why the simplified expression tends to the claimed value. The derivative rules established earlier in this course can shorten calculations when their hypotheses apply. In a proof workshop, however, it is useful to return to the definition whenever a function is given by different formulas, has a removable-looking expression, or contains a term whose behavior is not captured by routine algebra.

Write the quotient.
Start with $\bigl(f(a+h)-f(a)\bigr)/h$, not an assumed derivative formula.

Use the function’s actual definition.
If $f$ is piecewise or has a special value at $a$, substitute that definition on each relevant side.

Expose the decisive limit.
Factor, cancel only a nonzero quantity, or bound the quotient by an expression tending to zero.

Check both sides and the hypotheses.
A two-sided derivative requires a common finite limit from positive and negative $h$, with the function defined on a neighborhood of $a$.

When Two Formula Branches Meet

Piecewise functions make the two-sided requirement especially visible. At a joining point, the formulas on either side may have separate one-sided derivative limits. If those limits agree and the function values fit together, the combined function is differentiable. The next theorem records this fact in a form that can be checked directly from the definition.

Theorem (Differentiability at a Joined Branch): Let $\delta>0$, and let $u:[a-\delta,a]\to\mathbb{R}$ and $v:[a,a+\delta]\to\mathbb{R}$. Suppose $u(a)=v(a)$. Define $F:[a-\delta,a+\delta]\to\mathbb{R}$ by $F(x)=u(x)$ for $x\leq a$ and $F(x)=v(x)$ for $x\geq a$. Suppose the one-sided limits $$ \lim_{h\to0^-}\frac{u(a+h)-u(a)}{h}=L_-,\qquad \lim_{h\to0^+}\frac{v(a+h)-v(a)}{h}=L_+ $$ exist as finite real numbers and satisfy $L_-=L_+$. Then $F$ is differentiable at $a$, and $F'(a)=L_-=L_+$.

Proof. Since $u(a)=v(a)$, the definition of $F$ gives a single value $F(a)$ that is equal to both $u(a)$ and $v(a)$. For every sufficiently small $h<0$, $a+h$ is on the left branch, so

\frac{F(a+h)-F(a)}{h} =\frac{u(a+h)-u(a)}{h}.

Consequently, as $h\to0^-$, the quotient for $F$ tends to $L_-$. For every sufficiently small $h>0$, $a+h$ is on the right branch, and

\frac{F(a+h)-F(a)}{h} =\frac{v(a+h)-v(a)}{h}.

As $h\to0^+$, this quotient tends to $L_+$. By hypothesis, both one-sided limits equal the same finite number $L$. The two-sided derivative criterion therefore gives

\lim_{h\to0}\frac{F(a+h)-F(a)}{h}=L.

Thus $F$ is differentiable at $a$ and $F'(a)=L$, as claimed. The matching value hypothesis is essential to these quotient identities: it ensures that the $F(a)$ subtracted in the derivative definition is the endpoint value of each branch. $\square$

Worked Example: Matching Branches with Equal Slopes

Define $F:\mathbb{R}\to\mathbb{R}$ by

F(x)= \begin{cases} x^2,&x\leq 1,\\ 2x-1,&x>1. \end{cases}

At the joining point, the branch values agree: the left formula gives $1^2=1$, and the right formula would give $2(1)-1=1$. For $h<0$, the difference quotient at $1$ is

\frac{F(1+h)-F(1)}{h} =\frac{(1+h)^2-1}{h} =\frac{2h+h^2}{h} =2+h.

Its limit as $h\to0^-$ is $2$. For $h>0$, the quotient is

\frac{F(1+h)-F(1)}{h} =\frac{2(1+h)-1-1}{h} =\frac{2h}{h} =2.

Its limit as $h\to0^+$ is also $2$. The joined-branch theorem applies, so $F'(1)=2$. Notice that the proof checks the two sides separately; a calculation from only one branch would not establish the two-sided derivative.

Worked Example: Matching Values but Unequal Slopes

Now define $G:\mathbb{R}\to\mathbb{R}$ by

G(x)= \begin{cases} x^2,&x\leq 0,\\ 3x,&x>0. \end{cases}

Both branches give the value $0$ at the joining point. For $h<0$, however,

\frac{G(h)-G(0)}{h} =\frac{h^2}{h} =h,

which tends to $0$ from the left. For $h>0$,

\frac{G(h)-G(0)}{h} =\frac{3h}{h} =3,

which tends to $3$ from the right. Since the one-sided limits differ, the two-sided derivative does not exist. This example shows why matching function values alone is not enough: the slopes of the branches must also agree for differentiability at the join.

Bounding a Difficult Quotient

Not every difference quotient simplifies to a polynomial or a constant. Sometimes an oscillating factor prevents the numerator from having a simple limit after division by $h$. In such cases, estimate the whole quotient rather than trying to determine the oscillating factor’s limit. A bounded factor multiplied by a term tending to zero can still produce a limit, provided the bound is explicit.

Proposition: Define $f:\mathbb{R}\to\mathbb{R}$ by $f(0)=0$ and $f(x)=x^2\sin(1/x)$ for $x\ne0$. Then $f$ is differentiable at $0$, and $f'(0)=0$.

Proof. For $h\ne0$, the difference quotient at $0$ is

\frac{f(0+h)-f(0)}{h} =\frac{h^2\sin(1/h)}{h} =h\sin(1/h).

The sine factor need not approach a limit as $h\to0$, so evaluating its limit is not the right strategy. Instead, use $|\sin t|\leq1$ for every real $t$. Then

\left|\frac{f(h)-f(0)}{h}\right| =|h|\,|\sin(1/h)| \leq |h|.

Given $\varepsilon>0$, if $0<|h|<\varepsilon$, this inequality gives $\left|(f(h)-f(0))/h\right|<\varepsilon$. Thus the difference quotient tends to $0$, which proves $f'(0)=0$. $\square$

Worked Example: A Removable Expression at the Differentiation Point

Define $q:\mathbb{R}\to\mathbb{R}$ by

q(x)= \begin{cases} \dfrac{x^2-9}{x-3},&x\ne3,\\ 6,&x=3. \end{cases}

To find the derivative at $3$, use the value assigned at $3$ and the quotient definition. For $h\ne0$, substitution gives

\frac{q(3+h)-q(3)}{h} =\frac{\dfrac{(3+h)^2-9}{(3+h)-3}-6}{h}.

Because $h\ne0$, the denominator in the inner fraction is nonzero. Expanding and simplifying,

\frac{\dfrac{9+6h+h^2-9}{h}-6}{h} =\frac{6+h-6}{h} =1.

This quotient equals $1$ for every nonzero $h$ for which it is defined, so its limit is $1$. Therefore $q'(3)=1$. The key detail is that the special value $q(3)=6$ participates in the quotient; it cannot be inferred merely by substituting $x=3$ into the original fraction.

Proof Checks and Common Pitfalls

A derivative proof can fail even when its algebra looks familiar. In the examples above, the calculation was guided by the definition and each cancellation had a stated condition. The following checks help identify gaps before they become hidden inside a limit argument.

Use the correct value at the point. For a piecewise function, $f(a)$ is determined by its definition at $a$. The left- and right-hand formulas do not automatically define that value.
Do not cancel at a forbidden input. In a difference quotient, $h\ne0$. A cancellation by $h$ is valid for the quotient’s domain, but does not mean the original expression was defined at $h=0$.
Check both directions. A limit calculated for $h>0$ proves only a right-hand limit. Differentiability requires the left and right limits to exist and agree.
Bound the entire quotient. In the oscillatory example, $\sin(1/h)$ itself does not need a limit. The estimate $|h\sin(1/h)|\leq|h|$ settles the derivative directly.
Separate a theorem’s hypotheses from a useful shortcut. A formula that simplifies a quotient away from $a$ may help evaluate the derivative, but the value at $a$ and the domain near $a$ still need checking.

These methods complement the derivative rules rather than replace them. The Sum Rule, Product Rule, Quotient Rule, and Chain Rule are efficient when their hypotheses fit. Direct proofs are especially useful when the definition itself carries the main difficulty: at a branch point, at a specially assigned value, or in the presence of a bounded but nonconvergent factor.

Check Your Understanding

Use the definition and the results proved above to answer the following questions.

In the joined-branch theorem, where is the hypothesis $u(a)=v(a)$ used when forming the difference quotients?
For the function in the first worked example, what are the left and right difference quotients at $x=1$, and what derivative do they establish?
Why does matching the values of two branches fail to guarantee differentiability, as the second worked example demonstrates?
For $f(h)=h^2\sin(1/h)$ when $h\ne0$ and $f(0)=0$, give an inequality that proves the difference quotient tends to zero.
In the removable-expression example, why must $h\ne0$ before cancelling the factor $h$?

What You'll Learn