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2.1 Defining Average and Instantaneous Rates of Change at a Point

Average rate of change measures output change per unit of input change across an interval. Geometrically, it is the slope of the secant line through the two endpoint points on the graph.

\text{Average rate on }[a,b]=\frac{f(b)-f(a)}{b-a}=\frac{f(a+h)-f(a)}{h},\qquad h=b-a\ne0

For $f(x)=x^2$ on $[2,5]$ , the output change is $f(5)-f(2)=25-4=21$ and the input change is $5-2=3$ . The average rate of change is $21/3=7$ , so the secant line through $(2,4)$ and $(5,25)$ has slope $7$ .

This is an average over the whole interval, not necessarily the rate at either endpoint. Its units are output units per input unit, and the order of subtraction must match in numerator and denominator.

The instantaneous rate of change at $x=a$ is the limit of average rates over intervals whose second endpoint approaches $a$ . If this limit exists, it is the derivative $f'(a)$ and the slope of the tangent line at $(a,f(a))$ .

f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}

For $f(x)=x^2$ at $a=3$ , $\dfrac{f(3+h)-f(3)}{h}=\dfrac{(3+h)^2-9}{h}=\dfrac{6h+h^2}{h}=6+h$ for $h\ne0$ . Taking $h\to0$ gives $f'(3)=6$ . Thus the instantaneous rate and tangent slope at $x=3$ are $6$ .

Do not substitute $h=0$ into the original quotient; it is undefined there. Simplify for nonzero $h$ , then take the limit. If the two-sided limit does not exist, the function is not differentiable at $a$ .

2.2 Defining the Derivative of a Function and Using Derivative Notation

The derivative $f'$ is a new function: at each input $x$ where the defining limit exists, $f'(x)$ gives the instantaneous rate of change of $f$ and the tangent slope at $(x,f(x))$ .

f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}

For $y=f(x)$ , the symbols $f'(x)$ , $y'$ , and $\dfrac{dy}{dx}$ name the same derivative function. The expression $f'(a)$ is one value of that function: the rate and tangent slope specifically at $x=a$ .

If $f(x)=x^2$ , then $\dfrac{(x+h)^2-x^2}{h}=2x+h$ for $h\ne0$ . Letting $h\to0$ gives $f'(x)=2x$ . Thus the derivative is a whole function; for example, $f'(3)=6$ is one of its values.

Representation	Meaning of $f'(a)$
Graphical	Slope of the tangent to $f$ at $x=a$
Numerical	Limiting value of nearby secant slopes
Analytical	Value obtained from a derivative formula or defining limit
Verbal	Instantaneous output change per input unit at $a$

A derivative formula applies only at inputs where the defining limit exists. Do not confuse $f'(x)$ , which varies with $x$ , with the single number $f'(a)$ .

At $x=a$ , the tangent line uses the point $(a,f(a))$ on the curve and the slope $f'(a)$ . Put those two pieces into point-slope form.

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

Evaluate $f(a)$ to find the point on the curve.
Evaluate $f'(a)$ to find the tangent slope.
Substitute both values into point-slope form.
Simplify only if another line form is requested.

For $f(x)=x^2$ at $x=3$ , the curve point is $(3,9)$ and $f'(x)=2x$ gives slope $f'(3)=6$ . Therefore $y-9=6(x-3)$ , or equivalently $y=6x-9$ . Substituting $x=3$ returns $y=9$ , confirming that the line passes through the required point.

Use $f(a)$ as the point's $y$ -coordinate and $f'(a)$ only as the slope; interchanging them gives the wrong line. A tangent line describes the curve's local linear behavior and may cross the curve elsewhere.

2.3 Estimating Derivatives of a Function at a Point

To estimate $f'(a)$ from a table, use values close to $a$ on opposite sides when available. Their secant slope approximates the tangent slope because the interval is centered on the target input.

f'(a)\approx\frac{f(a+h)-f(a-h)}{(a+h)-(a-h)}=\frac{f(a+h)-f(a-h)}{2h}

For the explicitly defined function $f(x)=x^3$ :

$x$	$f(x)$
$1.9$	$6.859$
$2.0$	$8.000$
$2.1$	$9.261$

Using the values equally spaced around $2$ , $f'(2)\approx\dfrac{9.261-6.859}{2.1-1.9}=\dfrac{2.402}{0.2}=12.01$ . The symbol $\approx$ is appropriate because a finite secant interval is being used in place of the limiting tangent slope.

From a graph, estimate the slope of the tangent at the target point using two readable points on the tangent line. With technology, evaluate a numerical derivative at the target input and report reasonable precision; the displayed result is still an estimate unless exact analysis is supplied.

Do not use the slope of a visibly distant secant when closer data are available, and do not read two arbitrary points on the curve as if they lay on the tangent. Table spacing, graph scale, and rounded values all affect the estimate.

2.4 Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist

If $f$ is differentiable at $x=a$ , then $f$ is continuous at $x=a$ . Differentiability is therefore the stronger condition. The converse is false: a function can be continuous at a point without having a derivative there.

When $x\ne a$ , write $f(x)-f(a)=\frac{f(x)-f(a)}{x-a}(x-a).$ If $f'(a)$ exists as a finite two-sided limit, the first factor approaches $f'(a)$ while the second approaches $0$ . Their product approaches $0$ , so $f(x)\to f(a)$ : this is continuity at $a$ .

For $f(x)=|x|$ at $x=0$ , the graph is continuous, but the difference quotient approaches $-1$ from the left and $1$ from the right. The two-sided derivative does not exist because those limits disagree.

For $f(x)=\sqrt[3]{x}$ at $x=0$ , the graph is continuous, but its tangent is vertical. The difference quotient grows without a finite limit, so $f'(0)$ does not exist as a finite derivative.

Use the implication in the correct direction: differentiable $\Rightarrow$ continuous. A discontinuity guarantees non-differentiability, but continuity alone never guarantees differentiability. Also, if $a$ is not in the domain of $f$ , neither continuity nor differentiability at $a$ is defined.

2.5 Applying the Power Rule

A power function has the form $f(x)=x^r$ . Its derivative is found by moving the exponent $r$ in front as a coefficient, then subtracting $1$ from the exponent.

\frac{d}{dx}(x^r)=r x^{r-1}

Keep the base $x$ unchanged. Use the original exponent as the coefficient, calculate $r-1$ carefully, and then simplify only if the new form is clearer.

For $f(x)=x^{5/2}$ , $f'(x)=\frac{5}{2}x^{5/2-1}=\frac{5}{2}x^{3/2}.$ For $g(x)=x^{-3}$ , $g'(x)=-3x^{-4}=-\frac{3}{x^4}.$ The same rule handles positive, negative, and fractional powers; only the exponent arithmetic changes.

Do not multiply the exponent by $x$ and leave the exponent unchanged: the new exponent is always $r-1$ . Also preserve real-domain restrictions. For example, $x^{-3}$ and its derivative are not defined at $x=0$ .

2.6 Derivative Rules: Constant, Sum, Difference, and Constant Multiple

Derivatives distribute across addition and subtraction, and constant factors stay attached. This lets you differentiate a polynomial one term at a time instead of returning to the limit definition.

\frac{d}{dx}(c)=0,\qquad (f+g)'=f'+g',\qquad (f-g)'=f'-g',\qquad (cf)'=cf'

For each nonconstant term $ax^n$ , keep the coefficient $a$ and apply the power rule: $ax^n\mapsto an x^{n-1}$ . Preserve the operation sign between terms, and replace every constant term by $0$ .

If $p(x)=4x^5-3x^2+7$ , then $p'(x)=4(5x^4)-3(2x)+0=20x^4-6x.$ The coefficient of each power is multiplied by its old exponent, the exponent decreases by one, and the constant disappears.

The constant-multiple rule applies when the multiplier is constant with respect to $x$ . Do not treat a product of two variable functions as a constant multiple, and do not lose a negative sign when differentiating a difference.

2.7 Derivatives of cos x, sin x, eˣ, and ln x

The functions $\sin x$ , $\cos x$ , $e^x$ , and $\ln x$ have standard derivatives that can be applied directly. For the trigonometric rules, $x$ is measured in radians.

Function $f(x)$	Derivative $f'(x)$	Condition
$\sin x$	$\cos x$	$x$ in radians
$\cos x$	$-\sin x$	$x$ in radians
$e^x$	$e^x$	all real $x$
$\ln x$	$\dfrac{1}{x}$	$x>0$

If $f(x)=3\sin x-2e^x+\ln x$ , differentiate term by term: $f'(x)=3\cos x-2e^x+\frac{1}{x},\qquad x>0.$ The domain condition comes from the original $\ln x$ term as well as its derivative.

The cosine rule is the one with a negative sign: $(\cos x)'=-\sin x$ . These formulas apply directly only when the input is exactly $x$ ; differentiating a composite input such as $\sin(x^2)$ requires a later rule.

A difference-quotient limit can often be evaluated by identifying it as $f'(a)$ . Match the whole numerator to the change in one function and the denominator to the corresponding change in input.

f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}

Identify the underlying function $f$ , identify the point $a$ , verify that the numerator and denominator follow the same derivative-definition form, and then replace the limit by the known value $f'(a)$ .

Consider $\lim_{h\to0}\frac{e^{2+h}-e^2}{h}.$ This is the derivative definition for $f(x)=e^x$ at $a=2$ . Since $f'(x)=e^x$ , the limit equals $f'(2)=e^2$ .

Do not match only the denominator. Both function values in the numerator must correspond to the same point change. If the pattern does not match a derivative definition exactly, another limit method may be needed.

2.8 The Product Rule

For a product $y=u(x)v(x)$ , the derivative has two contributions: first let $u$ change while $v$ stays as it is, then let $v$ change while $u$ stays as it is.

\frac{d}{dx}[u(x)v(x)]=u'(x)v(x)+u(x)v'(x)

Label the factors $u$ and $v$ .
Find $u'$ and $v'$ separately.
Form $u'v+uv'$ , keeping one original factor in each term.
Simplify or factor only after both terms are present.

For $y=x^2e^x$ , take $u=x^2$ and $v=e^x$ , so $u'=2x$ and $v'=e^x$ . Then $y'=(2x)e^x+x^2(e^x)=e^x(x^2+2x).$ Both terms are needed because both factors vary with $x$ .

In general, $(uv)'\ne u'v'$ . If one factor is a true constant, its derivative contribution is zero and the product rule reduces to the constant-multiple rule.

2.9 The Quotient Rule

For $y=\dfrac{u(x)}{v(x)}$ , both $u$ and $v$ may change. The quotient rule combines those changes in a fixed order and applies wherever $u$ and $v$ are differentiable and $v(x)\ne0$ .

\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{u'v-uv'}{v^2}

Label the numerator $u$ and denominator $v$ .
Find $u'$ and $v'$ separately.
Build the numerator as $u'v-uv'$ in that order.
Divide by $v^2$ , then simplify without changing the original domain.

For $y=\dfrac{\sin x}{x}$ , let $u=\sin x$ and $v=x$ , so $u'=\cos x$ and $v'=1$ . Then $y'=\frac{x\cos x-(\sin x)(1)}{x^2}=\frac{x\cos x-\sin x}{x^2},\qquad x\ne0.$

In general, $\left(\dfrac{u}{v}\right)'\ne\dfrac{u'}{v'}$ . Do not reverse the subtraction, forget the square on $v$ , or include inputs where the original denominator equals zero.

2.10 Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions

Rewrite tangent, cotangent, secant, and cosecant in terms of sine and cosine; then the quotient rule and known sine/cosine derivatives produce their formulas. Angles are measured in radians.

Function identity	Derivative
$\tan x=\dfrac{\sin x}{\cos x}$	$\sec^2 x$
$\cot x=\dfrac{\cos x}{\sin x}$	$-\csc^2 x$
$\sec x=\dfrac{1}{\cos x}$	$\sec x\tan x$
$\csc x=\dfrac{1}{\sin x}$	$-\csc x\cot x$

For tangent, the quotient rule gives $\frac{d}{dx}(\tan x)=\frac{(\cos x)(\cos x)-(\sin x)(-\sin x)}{\cos^2 x}=\frac{\cos^2 x+\sin^2 x}{\cos^2 x}=\sec^2 x.$ The Pythagorean identity turns the numerator into $1$ .

The tangent/secant pair has positive derivatives. The cotangent/cosecant pair has negative derivatives. Identities explain these results; they are not four unrelated facts.

Apply each formula only where the original function is defined: $\tan x$ and $\sec x$ require $\cos x\ne0$ , while $\cot x$ and $\csc x$ require $\sin x\ne0$ . Composite inputs require the chain rule, which belongs to the next Topic.

Unit 2: Differentiation: Definition and Fundamental Properties

Exam analysis

Most tested topics

2.1 Defining Average and Instantaneous Rates of Change at a Point

Average Rate of Change as a Difference Quotient

Instantaneous Rate from the Derivative Limit

2.2 Defining the Derivative of a Function and Using Derivative Notation

The Derivative as a Function and Across Representations

Writing the Tangent Line at a Point

2.3 Estimating Derivatives of a Function at a Point

Estimating a Derivative from a Table or Graph

2.4 Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist

Differentiability Is Stronger Than Continuity

2.5 Applying the Power Rule

Applying the Power Rule to $x^r$

2.6 Derivative Rules: Constant, Sum, Difference, and Constant Multiple

Differentiate Polynomials Term by Term

2.7 Derivatives of cos x, sin x, eˣ, and ln x

Four Familiar Derivative Rules

Recognize a Limit as a Derivative

2.8 The Product Rule

Differentiate a Product of Two Functions

2.9 The Quotient Rule

Differentiate a Quotient of Two Functions

2.10 Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions

Derivatives of $\tan x$, $\cot x$, $\sec x$, and $\csc x$