Unit 03 · Term 1

Derivatives

The derivative is a limit. Once limits are solid, the derivative and all its rules follow — and in MAT137 you prove the rules, not just apply them.

By the end you can

Compute derivatives from the limit definition.
Prove that differentiability implies continuity.
Prove and apply the sum, product, quotient, and chain rules.
Differentiate implicitly.

Why it matters & where it comes from

Why it exists

An average rate of change loses the moment. The derivative captures the instantaneous one — the exact velocity now, the marginal cost of the next unit, the slope right here.

In the real world

Derivatives are everywhere modern: physics (velocity, acceleration), economics (marginal analysis), finance (the option "Greeks"), and above all machine learning, where most modern neural networks are trained with gradient-based optimization.

Story & open questions

Newton and Leibniz discovered it independently, then feuded bitterly over credit and helping isolate parts of British mathematics from Continental developments. Leibniz's $dy/dx$ notation became standard. Today "automatic differentiation" computes these derivatives at enormous scale.

The learning spine

Every derivative calculation in this unit is a disguised limit. Rules like product, quotient, and chain rules are shortcuts only after their hypotheses are met. When learning a new rule, first understand the limit proof, then practice recognizing the pattern quickly.

Derivative-from-definition routine

Write the correct quotient: $\dfrac{f(a+h)-f(a)}{h}$ or $\dfrac{f(x)-f(a)}{x-a}$.

Simplify before taking the limit. Usually this means factoring, expanding, rationalizing, or combining fractions.

Only after the expression is defined for $h\ne0$ and simplified should you let $h\to0$.

Proofs you should own

MAT137 treats the differentiation rules as theorems, not magic formulas. A student who can reproduce the following proof ideas is ready for later analysis.

Differentiable implies continuous. Write the change in $f$ as a difference quotient times the change in $x$.
Product rule. Add and subtract the same middle term so the difference quotient splits into two known limits.
Quotient rule. Derive it from the product rule and the derivative of the reciprocal.
Chain rule. Introduce the inner change $g(x+h)-g(x)$ and handle the case where that change may be zero.

The definition

Definition Derivative

$f$ is differentiable at $a$ if the limit $$f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$ exists. Geometrically it is the slope of the tangent line at $\big(a,f(a)\big)$. Equivalently, $f'(a)=\lim_{x\to a}\dfrac{f(x)-f(a)}{x-a}$.

A secant line becoming the tangent line

For $f(x)=x^2$ at $a=1$, the secant slope is $\frac{(1+h)^2-1}{h}=2+h$. As $h\to0$, it approaches the tangent slope $2$.

Horizontal gap h h = 0.90 secant slope = 2.90

Example From the definition

For $f(x)=\tfrac1x$, $a\ne0$:

$\dfrac{f(a+h)-f(a)}{h}=\dfrac{1}{h}\!\left(\dfrac{1}{a+h}- \dfrac1a\right)=\dfrac{-1}{a(a+h)}.$

As $h\to0$ this is $-\dfrac1{a^2}$, so $f'(a)=-1/a^2$.

Example A corner is not differentiable

For $f(x)=|x|$ at $0$, the difference quotient is $|h|/h$. It tends to $-1$ from the left and $1$ from the right, so the two-sided derivative does not exist.

Differentiability implies continuity

Theorem

If $f$ is differentiable at $a$, then $f$ is continuous at $a$. (The converse fails: $|x|$ is continuous but not differentiable at $0$.)

Proof idea

$f(x)-f(a)=\dfrac{f(x)-f(a)}{x-a}\cdot(x-a)$ for $x\ne a$.

As $x\to a$ the first factor $\to f'(a)$ and the second $\to0$, so $f(x)-f(a)\to0$, i.e. $f(x)\to f(a)$. ∎

Do not reverse the theorem Differentiable $\Rightarrow$ continuous, but continuous $\not\Rightarrow$ differentiable. Corners, cusps, vertical tangents, and sufficiently irregular oscillation can all destroy differentiability.

The differentiation rules

Theorem Rules

For differentiable functions, $(f+g)'=f'+g'$, $(fg)'=f'g+fg'$, $\left(\dfrac fg\right)'=\dfrac{f'g-fg'}{g^2}$ where $g\ne0$, and the chain rule $(f\circ g)'(x)=f'\!\big(g(x)\big)\,g'(x)$ when the composition is defined.

Example Product rule from the definition

$\dfrac{f(x+h)g(x+h)-f(x)g(x)}{h}$

$=\dfrac{f(x+h)-f(x)}{h}\,g(x+h)+f(x)\,\dfrac{g(x+h)-g(x)}{h}.$

Let $h\to0$: $g(x+h)\to g(x)$ (continuity), giving $f'g+fg'$. ∎

Example Chain rule

$\dfrac{d}{dx}\,(x^2+1)^{10}=10(x^2+1)^9\cdot 2x.$

Proof idea Why the chain rule needs care

For $u=g(x+h)-g(x)$, one wants to split the quotient as

$\dfrac{f(g(x+h))-f(g(x))}{g(x+h)-g(x)} \cdot\dfrac{g(x+h)-g(x)}h.$

As $h\to0$, the factors approach $f'(g(x))$ and $g'(x)$. The subtlety is that $u$ may equal $0$ for some $h$. A complete proof extends the first quotient by defining its value at $u=0$ to be $f'(g(x))$; that extended function is continuous at $0$.

Local linear approximation

Differentiability says more than "a slope exists": near $a$, the function is well approximated by its tangent line.

Linearization

$f(a+h)=f(a)+f'(a)h+r(h)$, where $\displaystyle\lim_{h\to0}\frac{r(h)}h=0$. Informally, $f(a+h)\approx f(a)+f'(a)h$ for small $h$.

Example Estimate without a calculator

For $f(x)=\sqrt{x}$ at $a=4$, $f(4)=2$ and $f'(4)=1/4$. Thus $\sqrt{4.1}\approx 2+\frac14(0.1)=2.025$.

Implicit differentiation

When $y$ is defined implicitly by an equation, differentiate both sides in $x$, treating $y=y(x)$ and using the chain rule, then solve for $y'$.

Example

For $x^2+y^2=1$: differentiate to get $2x+2yy'=0$, so $y'=-x/y$ where $y\ne0$. At $(\pm1,0)$ the circle has a vertical tangent, so $y$ is not locally a differentiable function of $x$ there.

Common pitfalls

• Treating $\dfrac{f(a+h)-f(a)}{h}$ as $0/0$ — it is a limit, evaluate it.
• Dropping the inner factor $g'(x)$ in the chain rule.
• Using a rule before checking its hypotheses, especially dividing by a quantity that may be zero.
• Assuming continuous $\Rightarrow$ differentiable. It is the other direction.

Practice

Problems adapted from the MAT137 question bank by Alfonso Gracia-Saz, used under CC BY-SA 4.0. Write full solutions before opening the answer.

Practice 1 · Tangent line

What is the equation of the line tangent to $y=x$ at the point with $x$-coordinate $7$?

Answer

The graph is already a line, so it is its own tangent: $y=x$. (Check: slope $1$, passes through $(7,7)$.)

Practice 2 · Definition

From the definition, compute $f'(a)$ for $f(x)=\sqrt{x}$, $a>0$.

Solution

Rationalize: $\dfrac{\sqrt{a+h}-\sqrt a}{h}=\dfrac{1}{\sqrt{a+h}+\sqrt a}\to \dfrac{1}{2\sqrt a}.$

Practice 3 · Definition again

Compute $g'(4)$ directly from the definition for $g(x)=\dfrac{2}{\sqrt{x}}$.

Solution

$g(4)=1$, and $\dfrac{g(4+h)-g(4)}{h}=\dfrac{2-\sqrt{4+h}}{h\sqrt{4+h}} =\dfrac{-1}{\sqrt{4+h}\,(2+\sqrt{4+h})}\to\dfrac{-1}{2\cdot4}=-\dfrac18.$

Practice 4 · Theory

Which implication is true: continuous $\Rightarrow$ differentiable, or differentiable $\Rightarrow$ continuous? Justify.

Answer

Only differentiable $\Rightarrow$ continuous. The converse fails: $|x|$ is continuous at $0$ but not differentiable there.

Practice 5 · A corner

Let $h(x)=x|x|$. Compute $h'(0)$ from the definition.

Solution

$\dfrac{h(t)-h(0)}{t} =\dfrac{t|t|}{t}=|t|\to0$, so $h'(0)=0$. (Indeed $h'(x)=2|x|$.)

Practice 6 · Chain rule

Differentiate $f(x)=\tan(3x^2+1)$ and justify each step.

Answer

Chain rule: $f'(x)=\sec^2(3x^2+1)\cdot 6x$.

Practice 7 · Quotient rule

Prove the quotient rule from the product and chain rules.

Hint

Write $f/g=f\cdot(1/g)$, use the product rule and $(1/g)'=-g'/g^2$ (apply the chain rule to the reciprocal function $r(u)=1/u$): $(f/g)'=\dfrac{f'g-fg'}{g^2}.$

Practice 8 · Implicit

Find $y'$ if $x^3+y^3=6xy$ (the folium).

Answer

$3x^2+3y^2y'=6y+6xy'$, so $y'=\dfrac{6y-3x^2}{3y^2-6x}=\dfrac{2y-x^2}{y^2-2x}.$

Expanded topic checklist

Course coverage. This unit is part of the complete MAT137 topic map. It should cover the following items before you move on:

Definition. Use the derivative as a limit of difference quotients and know what differentiability says geometrically and analytically.
Rules. Prove and use linearity, product, quotient, and chain rules from the limit definition.
Local linearity. Interpret $f(a+h)=f(a)+f\prime(a)h+o(h)$ as the reason tangent-line approximations work.
Failures. Recognize corners, cusps, vertical tangents, and discontinuities as common causes of non-differentiability.

Textbook/source alignment

Use the Alfonso Gracia-Saz MAT137 materials as the primary local spine, with the current UofT single-variable calculus topic split as the completeness check. The added items are original study prompts, not copied textbook exercises.

Core theorem map

The theorem chain is: differentiability implies continuity, algebra of limits gives derivative rules, and the chain rule turns composition into a reusable engine.

Study routine. When solving derivative problems, separate the proof of the rule from the computation using the rule; both are tested in proof-based calculus.

Deep practice extensions

Use these after the main practice set. They are designed to force definition-level reasoning, theorem selection, and mixed application rather than routine calculation only.

Extension 1 · Definition computation

Compute the derivative of $f(x)=x^2$ at $a$ from the limit definition.

Guide

Use $[(a+h)^2-a^2]/h=2a+h$, whose limit as $h o0$ is $2a$.

Extension 2 · Proof rule

Prove the product rule assuming $f$ and $g$ are differentiable at $a$.

Guide

Add and subtract $f(a+h)g(a)$ in the difference quotient, split into two terms, and use continuity of $f$ or $g$ that follows from differentiability.

Extension 3 · Non-differentiable point

Show $|x|$ is not differentiable at $0$.

Guide

The right difference quotient is $1$ and the left difference quotient is $-1$, so the derivative limit does not exist.

Extension 4 · Linear approximation

Use differentials to estimate $\sqrt{4.04}$.

Guide

For $f(x)=\sqrt{x}$ at $a=4$, $f(4)=2$ and $f\prime(4)=1/4$. The estimate is $2+(1/4)(0.04)=2.01$.

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