Differentiation: Definition and Fundamental Properties

Average vs. Instantaneous Rate of Change (and the Tangent Line Problem)

What “rate of change” means in calculus

A rate of change describes how one quantity changes as another quantity changes. If you have a function $f$, you can think of $f(x)$ as an output that depends on an input $x$. The rate of change tells you how sensitive the output is to small changes in the input.

In Algebra, you may have seen slope as “rise over run.” In calculus, slope becomes more powerful because we can talk about the slope at a single point on a curve, not just between two points on a line.

There are two closely related ideas:

• Average rate of change: change over an interval.
• Instantaneous rate of change: change at an instant (at a point).

The derivative will formalize instantaneous rate of change.

Average rate of change and secant lines

The average rate of change of $f$ from $x=a$ to $x=b$ is

$\frac{f(b)-f(a)}{b-a}$

A common “algebra slope” way to say the same thing is

$\frac{y_2-y_1}{x_2-x_1}$

Geometrically, this average rate of change is the slope of the secant line through the points $\bigl(a,f(a)\bigr)$ and $\bigl(b,f(b)\bigr)$.

Why it matters: average rate of change is how you estimate speed from “distance traveled over time,” and it is the stepping stone to instantaneous rate of change because you can “zoom in” by shrinking the interval.

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From secant slope to tangent slope (the key limiting idea)

Suppose you want the slope of the curve at $x=a$. You cannot compute slope “at a point” using just one point; you need two points. The calculus idea is to use a second point very close to $a$.

Let the second point be at $a+h$, where $h$ is a small number (positive or negative). The slope of the secant line through $\bigl(a,f(a)\bigr)$ and $\bigl(a+h,f(a+h)\bigr)$ is

$\frac{f(a+h)-f(a)}{h}$

As $h$ approaches $0$, the second point approaches the first point. If the slopes of these secant lines approach a single value, that limiting value is the slope of the **tangent line** at $x=a$. The closer the points are, the more accurate the secant-line approximation becomes.

This is the heart of differentiation: instantaneous rate of change is a limit of average rates of change.

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Worked example: average rate of change

Let $f(x)=x^2$. Find the average rate of change from $x=1$ to $x=3$.

Compute outputs: $f(3)=9$ and $f(1)=1$. Then

$\frac{f(3)-f(1)}{3-1}=\frac{9-1}{2}=4$

Interpretation: over the interval from input 1 to input 3, the output increases on average by 4 units per 1 unit of input.

Worked example: a secant slope expression that leads to a tangent slope

For $f(x)=x^2$, write the secant slope from $x=a$ to $x=a+h$.

Start with

$\frac{f(a+h)-f(a)}{h}$

Compute

$f(a+h)=(a+h)^2=a^2+2ah+h^2$

So

$\frac{f(a+h)-f(a)}{h}=\frac{(a^2+2ah+h^2)-a^2}{h}=\frac{2ah+h^2}{h}=2a+h$

You can already see the limit as $h\to 0$ will be $2a$, which becomes the derivative at $a$.

Exam Focus

• Typical question patterns:
  • Compute an average rate of change from a table, graph, or formula.
  • Find the slope of a secant line and then use a limit idea to describe a tangent slope.
  • Interpret a rate of change in context (include units).
• Common mistakes:
  • Swapping $a$ and $b$ inconsistently (this flips the sign). Keep numerator and denominator in matching order.
  • Using $b-a$ in the denominator but $f(a)-f(b)$ in the numerator.
  • Treating “average rate of change” as “instantaneous” without a limiting process.

The Derivative Defined as a Limit

What the derivative is

The derivative of a function at a point measures the function’s instantaneous rate of change at that point. Geometrically, it is the slope of the tangent line to the graph at that point (when the tangent line exists in the usual sense).

The derivative is defined using a limit because you approximate the tangent slope by secant slopes and then let the interval shrink to zero.

The (difference quotient) definition at a point

The derivative of $f$ at $x=a$ is

$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$

Key pieces to understand: $h$ represents a small change in the input; $f(a+h)-f(a)$ is the resulting change in output; dividing by $h$ gives an average rate of change over the small interval; and the limit as $h\to 0$ captures the instantaneous rate of change.

A common equivalent form uses a second input value $x$ approaching $a$:

$f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$

Both definitions describe the same idea; which one you use often depends on convenience.

When the derivative exists

The derivative at $a$ exists if the limit exists and is finite.

Graph connections you should recognize:

• At a smooth point, slopes of secant lines from both sides approach the same number.
• At a corner or cusp, the left-hand and right-hand slopes do not match (or blow up differently).
• At a vertical tangent, the slope tends to infinity (the derivative is not a finite real number).
• At a discontinuity, the derivative cannot exist.

Worked example: derivative from the limit definition

Find $f'(a)$ for $f(x)=x^2$ using the definition.

$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}=\lim_{h\to 0}\frac{(a+h)^2-a^2}{h}$

Since

$(a+h)^2=a^2+2ah+h^2$

we get

$\frac{(a+h)^2-a^2}{h}=\frac{a^2+2ah+h^2-a^2}{h}=\frac{2ah+h^2}{h}=2a+h$

So

$f'(a)=\lim_{h\to 0}(2a+h)=2a$

and the derivative function is $f'(x)=2x$.

Worked example: a function with a non-smooth point

Consider $f(x)=|x|$ at $x=0$.

$f'(0)=\lim_{h\to 0}\frac{|h|-|0|}{h}=\lim_{h\to 0}\frac{|h|}{h}$

If $h>0$, then $\frac{|h|}{h}=1$. If $h<0$, then $\frac{|h|}{h}=-1$. The one-sided limits differ, so the overall limit does not exist. Therefore, $f'(0)$ does not exist. Graphically, $|x|$ has a corner at the origin.

This example is important because it shows that “continuous” does not automatically mean “differentiable.”

Exam Focus

• Typical question patterns:
  • Compute $f'(a)$ using the limit definition for a given formula.
  • Use a graph to reason whether $f'(a)$ exists and estimate its value.
  • Decide whether a derivative exists at a point for a piecewise-defined function.
• Common mistakes:
  • Forgetting to simplify before taking the limit (often you must cancel a factor of $h$).
  • Plugging in $h=0$ too early, causing a false “division by zero” dead end.
  • Assuming the derivative exists just because the function is defined or continuous at the point.

Derivative Notation and the Derivative as a Function

Derivative notation (and what it’s actually saying)

Because derivatives show up in many contexts, calculus uses multiple notations.

When you write $f'(a)$, you are naming **a number**: the derivative evaluated at the specific input $a$.

When you write $f'(x)$, you are describing a new function: the derivative function, which takes an input and outputs the slope of the original function at that input.

This distinction matters a lot on exams. A common error is to treat $f'(a)$ (a value) and $f'(x)$ (a function) as interchangeable.

Notation reference table

The Leibniz notation $\frac{dy}{dx}$ is especially useful in applied contexts because it looks like “change in output over change in input,” matching rate-of-change thinking. Later, it will also help you remember rules like the chain rule.

The derivative as a function: from a single slope to a whole slope map

When you compute $f'(x)$, you create a function that tells you the slope of the original curve at every input value where it is differentiable.

Why it matters: the sign of $f'(x)$ tells you whether $f$ is increasing or decreasing (a major idea in later units); the magnitude of $f'(x)$ tells you how steep the curve is; and you can use $f'(x)$ to build tangent line equations quickly.

Interpreting a derivative in context (units and meaning)

If $f(t)$ represents position (meters) as a function of time (seconds), then $f'(t)$ represents velocity in meters per second.

A simple but heavily tested rule is:

• Units of $f'(x)$ are “units of output per unit of input.”

So if output is dollars and input is hours, $f'(x)$ is dollars per hour.

Worked example: turning derivative-at-a-point into a tangent line

If $f'(a)$ is the slope of the tangent line at $x=a$, then the tangent line at $x=a$ is the line through $\bigl(a,f(a)\bigr)$ with slope $f'(a)$. In point-slope form:

$y-f(a)=f'(a)(x-a)$

Example: let $f(x)=x^2$. Find the tangent line at $x=3$.

$f(3)=9$

$f'(x)=2x$

$f'(3)=6$

So

$y-9=6(x-3)$

and equivalently

$y=6x-9$

Worked example: derivative from a table (conceptual)

Suppose a table gives values of $f(x)$ near $x=2$. A common way to estimate $f'(2)$ is to compute slopes of secant lines using points close to $2$:

$\frac{f(2+h)-f(2)}{h}$

for small positive and negative $h$. If the left and right estimates agree, that common value is a good estimate of $f'(2)$.

A key habit is to use both sides when possible. If the left-side and right-side estimates differ substantially, the function might not be differentiable there (or the table spacing is too coarse).

Exam Focus

• Typical question patterns:
  • Given $f$ and a point $a$, find the equation of the tangent line using $f(a)$ and $f'(a)$.
  • Interpret $f'(a)$ in words and with correct units in an applied scenario.
  • Estimate $f'(a)$ from a table or graph using secant slopes near $a$.
• Common mistakes:
  • Confusing $f(a)$ with $f'(a)$ (height vs. slope).
  • Writing the tangent line with slope $f(a)$ instead of $f'(a)$.
  • Ignoring units when interpreting derivatives in context.

Differentiability and Continuity

Differentiable vs. continuous: what’s the relationship?

A function is continuous at $x=a$ if its graph has no break there, meaning the limit equals the function value:

$\lim_{x\to a}f(x)=f(a)$

A function is differentiable at $x=a$ if the derivative limit exists:

$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$

The key theorem you must know is: if $f$ is differentiable at $a$, then $f$ is continuous at $a$. Differentiability is a stronger condition than continuity.

But the converse is false: a function can be continuous but not differentiable.

What non-differentiability looks like on a graph

A function can fail to be differentiable at a point for several common reasons:

1. Discontinuity: if the function is not continuous, it cannot be differentiable.
2. Corner: the left-hand slope and right-hand slope are finite but unequal.
3. Cusp: the slopes become infinite in magnitude with opposite behavior (like a sharp point).
4. Vertical tangent: the slope becomes infinite in the same direction; the tangent line is vertical.

A helpful way to think about it is that differentiability requires the function to be “smooth enough” that the tangent line slope is well-defined and finite.

Why differentiability implies continuity (intuition, not just memorization)

If $f'(a)$ exists, then for small $h$ the expression

$\frac{f(a+h)-f(a)}{h}$

must stay close to some finite number. That can only happen if $f(a+h)-f(a)$ becomes small when $h$ becomes small. In other words, the outputs must approach $f(a)$ as inputs approach $a$, which is exactly continuity.

One-sided derivatives

Sometimes a function behaves differently from the left and right. You can define:

Left-hand derivative:

$\lim_{h\to 0^-}\frac{f(a+h)-f(a)}{h}$

Right-hand derivative:

$\lim_{h\to 0^+}\frac{f(a+h)-f(a)}{h}$

The derivative $f'(a)$ exists only if both one-sided derivatives exist and are equal.

This is especially important for piecewise functions, endpoints of domains, and absolute value style graphs.

Worked example: continuity but not differentiability

Consider

$f(x)=|x-2|$

At $x=2$, the graph has a corner. The function is continuous there because the left and right values meet at the point, but the slopes differ: left side slope is $-1$ and right side slope is $1$. So $f'(2)$ does not exist.

Worked example: vertical tangent

Consider

$f(x)=\sqrt[3]{x}$

You can show (graphically or with algebra) that the tangent line at $x=0$ is vertical, meaning the slope is unbounded. In AP Calculus terms, the derivative at that point does not exist as a finite real number.

The lesson is that “a tangent line exists” in a visual sense does not always mean “the derivative exists as a finite number.”

Exam Focus

• Typical question patterns:
  • From a graph, identify where $f$ is differentiable and where it is not, and explain why (corner, cusp, discontinuity, vertical tangent).
  • Given a piecewise function, find parameter values that make it continuous and differentiable at a junction point.
  • Decide whether statements like “continuous implies differentiable” are true or false and justify with a counterexample.
• Common mistakes:
  • Claiming a function is differentiable at a corner because it is continuous there.
  • Forgetting that discontinuity automatically prevents differentiability.
  • Treating a vertical tangent as having derivative $0$ instead of recognizing the slope is unbounded.

Fundamental Differentiation Rules (Constant, Power, Sum, Difference, Constant Multiple)

Why rules matter when you already have a definition

The limit definition is the foundation, but using it repeatedly for every function would be slow. Differentiation rules are shortcuts justified by limit properties.

A useful way to think about these rules is: derivatives come from limits, and limits have algebraic properties, so derivatives inherit algebraic properties.

Constant function rule

If $f(x)=c$ where $c$ is a constant, then

$f'(x)=0$

Reasoning: the output never changes as the input changes, so the rate of change is zero.

Example: if $f(x)=10$ then $f'(x)=0$.

Power rule

For $f(x)=x^n$ where $n$ is a real number for which the expression makes sense on the domain you are working with,

$\frac{d}{dx}\bigl(x^n\bigr)=nx^{n-1}$

A good memory phrase is “multiply down and decrease the power.” For example, $x^4$ becomes $4x^3$, and $2x^2$ becomes $4x$.

Common misconception: the exponent must become $n-1$, not remain $n$.

Constant multiple rule

If $f(x)=c\,g(x)$, then

$f'(x)=c\,g'(x)$

This is often described as “pull the constant out.”

Sum and difference rules

If $f(x)=g(x)+h(x)$, then

$f'(x)=g'(x)+h'(x)$

If $f(x)=g(x)-h(x)$, then

$f'(x)=g'(x)-h'(x)$

Building polynomial derivatives

A polynomial is a sum of constant multiples of powers of $x$, so the power rule plus the constant multiple rule and sum or difference rules let you differentiate polynomials quickly.

Worked example: differentiating a polynomial

Differentiate

$f(x)=3x^5-7x^2+4x-9$

Differentiate term-by-term:

$\frac{d}{dx}(3x^5)=15x^4$

$\frac{d}{dx}(-7x^2)=-14x$

$\frac{d}{dx}(4x)=4$

$\frac{d}{dx}(-9)=0$

So

$f'(x)=15x^4-14x+4$

Worked example: a non-integer power

Differentiate

$g(x)=x^{1/2}$

Using the power rule:

$g'(x)=\frac{1}{2}x^{-1/2}$

Rewrite to avoid negative exponents:

$g'(x)=\frac{1}{2\sqrt{x}}$

Domain reminder: $\sqrt{x}$ requires $x\ge 0$ in the real numbers, and the derivative expression requires $x>0$ because of the denominator.

Worked example: using the derivative to approximate local behavior

If $f'(2)=5$, then near $x=2$ the function increases about 5 units in output for each 1 unit increase in input. For a tiny change like $\Delta x=0.01$, you expect approximately $5(0.01)=0.05$ change in $f$. This “local linearity” viewpoint is one of the most important meanings of the derivative.

Exam Focus

• Typical question patterns:
  • Differentiate polynomials and simple power functions quickly and accurately.
  • Use given derivative values (like $f'(a)$) to interpret local change or build tangent lines.
  • Combine derivative rules (constant multiple plus sum or difference plus power rule) in one expression.
• Common mistakes:
  • Applying the power rule incorrectly (especially forgetting to subtract 1 from the exponent).
  • Differentiating a constant term as itself instead of 0.
  • Dropping negative signs when differentiating term-by-term.

Product and Quotient Rules

Why these rules exist

If you have two expressions multiplied or divided, you can sometimes expand or simplify first and then use earlier rules. For instance, with polynomials like

$(2x+7)(9x+8)$

you could multiply it out and then use the power rule, but that takes time. The product and quotient rules give direct shortcuts.

Product rule

If

$f(x)=u\,v$

then

$f'(x)=u\frac{dv}{dx}+v\frac{du}{dx}$

A common memory phrase is “1d2 + 2d1” (first times derivative of second, plus second times derivative of first).

Quotient rule

If

$f(x)=\frac{u}{v}$

then

$f'(x)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$

A common memory phrase is “low d high - high d low over low squared.”

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Exam Focus

• Typical question patterns:
  • Differentiate a product directly without expanding, especially when expanding would be messy.
  • Differentiate a quotient and simplify the final expression.
  • Use mnemonics correctly while still keeping track of signs.
• Common mistakes:
  • Forgetting that the product rule is not $\frac{du}{dx}\frac{dv}{dx}$.
  • Dropping parentheses and losing a negative sign in the quotient rule numerator.
  • Squaring the wrong part in the quotient rule (only the denominator function becomes $v^2$).

The Derivatives of Sine and Cosine

Why trig derivatives belong in “fundamental properties”

Sine and cosine are basic building blocks for modeling periodic behavior: waves, oscillations, circular motion, and seasonal patterns. Calculus needs derivatives of these functions because rates of change for periodic phenomena are just as important as rates of change for polynomials.

A big conceptual idea is that the derivative of a trig function is another trig function, which keeps periodic behavior “within the family.”

The fundamental trig derivative facts

These are core results (angles must be in radians):

$\frac{d}{dx}(\sin x)=\cos x$

$\frac{d}{dx}(\cos x)=-\sin x$

Radians matter because they tie angle measure directly to arc length on the unit circle, which makes the limiting process behind the derivative work out cleanly.

Interpreting these derivatives graphically

Sine and cosine are phase-shifted versions of each other. Where $\sin x$ is increasing most rapidly, $\cos x$ is largest; where $\cos x$ is decreasing most rapidly, $-\sin x$ is most negative.

Quick “reasonableness checks” at known angles:

• At $x=0$, $\sin 0=0$ and $\cos 0=1$, so the slope of $\sin x$ at 0 should be 1.
• At $x=0$, $\cos 0=1$ and $-\sin 0=0$, so the slope of $\cos x$ at 0 should be 0.

Worked example: differentiating a trig expression using basic rules

Differentiate

$f(x)=3\sin x-2\cos x$

Using constant multiple and sum or difference rules:

$f'(x)=3\cos x-2(-\sin x)$

So

$f'(x)=3\cos x+2\sin x$

Worked example: tangent line to a trig function

Find the tangent line to $f(x)=\sin x$ at $x=\pi$.

$f(\pi)=\sin \pi=0$

$f'(x)=\cos x$

$f'(\pi)=\cos \pi=-1$

Tangent line:

$y-0=-1(x-\pi)$

So

$y=-x+\pi$

Exam Focus

• Typical question patterns:
  • Differentiate expressions involving $\sin x$ and $\cos x$ using constant multiple and sum or difference rules.
  • Find slopes or tangent lines at special angles like $0$, $\pi/2$, $\pi$.
  • Interpret derivative sign (increasing or decreasing) using trig values.
• Common mistakes:
  • Forgetting the negative sign in $\frac{d}{dx}(\cos x)=-\sin x$.
  • Using degrees instead of radians when interpreting derivatives conceptually.
  • Mixing up values of $\sin$ and $\cos$ at special angles.

Memory Derivatives: Exponential and Logarithmic Basics

Why these are often memorized

Some derivative results are so common that it is usually faster to memorize them than to re-derive them each time. Common “memory derivatives” include those for $\sin x$, $\cos x$, $e^x$, and $\ln x$.

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Core formulas

For the natural exponential function:

$\frac{d}{dx}(e^x)=e^x$

For the natural logarithm (with domain $x>0$):

$\frac{d}{dx}(\ln x)=\frac{1}{x}$

Exam Focus

• Typical question patterns:
  • Differentiate expressions that include $e^x$ or $\ln x$ directly using these formulas.
  • Interpret domain restrictions when a derivative creates a denominator (for example, $\frac{1}{x}$ requires $x\ne 0$, and $\ln x$ requires $x>0$).
• Common mistakes:
  • Treating $\frac{d}{dx}(e^x)$ as $xe^{x-1}$ (incorrectly applying the power rule).
  • Forgetting that $\ln x$ is only defined for $x>0$ in real-valued calculus.

Using Derivatives to Describe Motion and Other Applied Rates of Change

The derivative as “instantaneous rate” in the real world

Many applications of calculus begin with a quantity changing over time. The derivative turns a position function into a velocity function, or a cost function into a marginal cost, or a volume function into a rate of filling.

What makes calculus different from “average rate” thinking is that the derivative describes the rate at a specific moment.

Motion: position, velocity, and acceleration (conceptual)

Let $s(t)$ be position as a function of time.

Velocity is the derivative of position:

$v(t)=s'(t)$

Speed is the magnitude of velocity:

$|v(t)|=|s'(t)|$

Units matter. If $s$ is in meters and $t$ is in seconds, then $v$ is in meters per second.

“Rate of change of a quantity” beyond motion

You should be comfortable translating contexts into derivative language:

• If $P(t)$ is population, $P'(t)$ is population growth rate (people per year).
• If $C(x)$ is cost to produce $x$ items, $C'(x)$ is marginal cost (dollars per item).
• If $T(t)$ is temperature, $T'(t)$ is heating or cooling rate (degrees per minute).

A key interpretation skill is that if $f'(a)=0$, the quantity is momentarily not changing at that input. That does not necessarily mean it is at a maximum or minimum yet, but it does mean the instantaneous rate is zero.

Connecting average and instantaneous rates in context

In applications, you often measure average rates (because data is collected at discrete times) and use them to estimate instantaneous rate.

If you know $s(2)=10$ and $s(2.1)=10.8$, the average velocity from $t=2$ to $t=2.1$ is

$\frac{10.8-10}{0.1}=8$

If you compute similar averages for intervals closer to $t=2$ (like 2.01, 2.001, etc.) and the values stabilize, that stable value is the best estimate of $v(2)$.

Worked example: interpreting a derivative value with units

Suppose $V(t)$ is the volume of water in a tank (liters) at time $t$ (minutes). If $V'(12)=-3$, then at $t=12$ minutes the volume is decreasing at 3 liters per minute. The negative sign indicates direction of change.

Worked example: estimating an instantaneous rate from nearby average rates

Suppose you have position values: $s(5)=100$, $s(5.1)=104$, and $s(4.9)=96.2$.

Right-side average:

$\frac{s(5.1)-s(5)}{5.1-5}=\frac{104-100}{0.1}=40$

Left-side average:

$\frac{s(5)-s(4.9)}{5-4.9}=\frac{100-96.2}{0.1}=38$

Since these are close, a reasonable estimate is around $39$ (units of position per time). On an exam, you would typically state the estimate and include units.

Exam Focus

• Typical question patterns:
  • Interpret statements like $f'(a)=k$ in context, including units and meaning of sign.
  • Estimate an instantaneous rate from tables or graphs by computing nearby secant slopes.
  • Distinguish between velocity and speed (signed vs. absolute value).
• Common mistakes:
  • Omitting units or mixing them up (a very common scoring loss in free-response).
  • Interpreting a negative derivative as “impossible” rather than “decreasing.”
  • Confusing average rate over an interval with instantaneous rate at an endpoint.

Working with Graphs and Tables: Estimating and Reasoning About Derivatives

Derivative information you can read from a graph

Even without an explicit formula, you can learn a lot about $f'(x)$ from the graph of $f$.

At any point where $f$ is smooth:

• $f'(x)$ is the slope of the tangent line.
• If $f$ is increasing, tangent slopes tend to be positive.
• If $f$ is decreasing, tangent slopes tend to be negative.
• If $f$ is flat (horizontal tangent), $f'(x)$ is near zero.

Estimating a tangent slope visually

To estimate $f'(a)$ from a graph:

1. Locate the point at $x=a$.
2. Sketch the tangent line (a line that just “kisses” the curve there and matches its local direction).
3. Pick two convenient points on your tangent line (not necessarily points on the curve) to compute its slope.

This method is approximate, but it matches what many graph-based exam questions expect.

Estimating from a table (difference quotients)

If you have a table of values, you can approximate the derivative using secant slopes.

To estimate $f'(a)$ well:

• Use points on both sides of $a$ if possible.
• Use the closest available inputs to $a$.
• Compare left and right slopes.

With step size $h$, typical estimates are

$\frac{f(a+h)-f(a)}{h}$

and

$\frac{f(a)-f(a-h)}{h}$

If those are similar, their common value is a good estimate of $f'(a)$.

Derivatives of piecewise functions (graph + algebra thinking)

Piecewise functions test whether you understand that derivatives depend on local behavior.

At a point where a piecewise function switches formulas, checking differentiability often requires:

1. Continuity at the switching point.
2. Matching slopes (left derivative equals right derivative).

A “smooth join” needs both the function values and the slopes to match.

Worked example: compare derivative values from a graph (conceptual)

If you see three points $A$, $B$, and $C$ on a curve and the tangent at $B$ is steepest upward, then $f'(B)$ is the largest among $f'(A)$, $f'(B)$, and $f'(C)$.

Similarly, if the curve is decreasing at $C$, then $f'(C)$ is negative.

Exam Focus

• Typical question patterns:
  • Estimate $f'(a)$ from a graph by approximating the tangent slope.
  • Estimate $f'(a)$ from a table using secant slopes on either side.
  • Determine where $f'(x)$ is positive, negative, or zero by analyzing where $f$ increases, decreases, or has horizontal tangents.
• Common mistakes:
  • Using points on the curve (instead of the tangent line) when estimating a tangent slope, which gives a secant slope instead.
  • Using too large an interval from a table, producing a poor “instantaneous” estimate.
  • Confusing the value $f(a)$ (height) with $f'(a)$ (slope) when reading graphs.

Putting It Together: Tangent Lines, Linearization Mindset, and Conceptual Consistency

Tangent lines as the best local linear model

A deep idea in early calculus is that differentiable functions are “locally linear.” If you zoom in far enough near a point of differentiability, the curve looks almost like a straight line.

If you know $f(a)$ and $f'(a)$, the tangent line at $x=a$ is

$y=f(a)+f'(a)(x-a)$

This connects the derivative (a slope) to a usable approximation tool.

Consistency checks you should always do

When differentiating or interpreting derivatives, use quick self-checks:

• Sign check: if the function is increasing at $a$, $f'(a)$ should be positive.
• Magnitude check: if the graph is very steep, $|f'(a)|$ should be large.
• Units check: derivative units must be “output units per input unit.”
• Special-value check (trig): at key angles like $0$ and $\pi$, verify slopes match the graph’s behavior.

These checks do not replace correct computation, but they catch many common errors.

Worked example: full tangent line workflow

Let

$f(x)=2x^3-x$

Find the tangent line at $x=-1$.

First compute the derivative:

$f'(x)=6x^2-1$

Evaluate the slope:

$f'(-1)=6(-1)^2-1=5$

Find the point:

$f(-1)=2(-1)^3-(-1)=-1$

So the point is $(-1,-1)$. Point-slope form:

$y-(-1)=5(x-(-1))$

So

$y+1=5(x+1)$

and equivalently

$y=5x+4$

A quick reasonableness check: the slope is positive, so the function should be increasing at $x=-1$.

Worked example: interpreting derivative values as local change

If you are told $f(10)=200$ and $f'(10)=-3$, then near $x=10$ a small increase of $\Delta x=0.2$ suggests

$\Delta f\approx f'(10)\Delta x=-3(0.2)=-0.6$

So you would expect $f(10.2)$ to be about $199.4$.

Exam Focus

• Typical question patterns:
  • Compute a tangent line equation from a formula (differentiate, evaluate, apply point-slope).
  • Interpret $f'(a)$ as a local rate of change and use it to predict small changes.
  • Combine graph or table interpretation with tangent line meaning (slope as derivative).
• Common mistakes:
  • Using the wrong point in the tangent line equation (mixing up $a$ with $f(a)$).
  • Treating a derivative value as an average rate over a large interval.
  • Ignoring the sign of $f'(a)$ when making a “small change” prediction.