Chapter 02

Hard

Derivatives — Definition and Rules

00 · Symbol Glossary

$f'(x)$f prime of x — derivative

The derivative of $f$ : a new function giving the instantaneous rate of change of $f$ at each $x$ . Read "f prime of x." The prime mark $'$ denotes "derivative of."

$\dfrac{dy}{dx}$dy dx — Leibniz notation

An alternate notation for the derivative, read "d y d x," treating $y=f(x)$ . It suggests "an infinitesimal change in $y$ divided by an infinitesimal change in $x$ " and is especially convenient once implicit differentiation appears in Chapter 03.

$h$h — the step size

A small nonzero number representing the horizontal distance between two input points, $x$ and $x+h$ . The derivative is defined by shrinking $h$ toward $0$ .

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

The slope of the line through the two points $(x, f(x))$ and $(x+h, f(x+h))$ on the graph of $f$ — the average rate of change over that interval. Taking $h\to0$ turns this average into an instantaneous rate.

$f''(x)$f double prime of x — second derivative

The derivative of $f'(x)$ — the rate of change of the rate of change. Each additional prime mark differentiates one more time.

$\dfrac{d}{dx}\big[\cdot\big]$d dx — differentiation operator

An instruction, not a fraction on its own: "take the derivative, with respect to $x$ , of whatever is inside the brackets." $\dfrac{d}{dx}[x^3]$ means "differentiate $x^3$ ."

$\sin x,\ \cos x$sine, cosine — trigonometric functions

Standard trigonometric functions from precalculus, taken here as already known and evaluatable. Section 09 derives their derivatives directly from the limit definition of Section 01, using two limit facts about them stated as given below.

$\lim_{h\to0}\dfrac{\sin h}{h} = 1$the fundamental trig limit — given fact

A limit fact about $\sin h$ used as a building block in Section 09. Stated here without proof — proving it (typically via a geometric squeeze argument on the unit circle) is outside this chapter's scope, but the fact itself is essential and is used explicitly, not silently.

$\lim_{h\to0}\dfrac{\cos h - 1}{h} = 0$the companion trig limit — given fact

A second limit fact, paired with the one above, also used in Section 09's derivations. Also stated without proof for the same reason.

$e$Euler's number

The unique real number defined by the property $\displaystyle\lim_{h\to0}\frac{e^h-1}{h}=1$ — this is a definition of $e$ , not a derived fact. $e\approx2.71828\ldots$ . Section 10 uses this defining property directly to differentiate $e^x$ .

$e^x$e to the x — natural exponential function

The exponential function with base $e$ . Distinct in kind from $x^n$ (Section 04's power rule): here the base is fixed and the variable sits in the exponent, the reverse of the power-rule setup.

$\ln x$natural logarithm of x

The inverse function of $e^x$ : $\ln x = y \iff e^y = x$ , defined for $x>0$ . Read "el en x" or "natural log of x." Section 11 derives its derivative from this inverse relationship.

01 · The Derivative as a Limit

Chapter 01 built the difference quotient's average version: the slope between two points on a curve. Shrinking the gap between those two points until it vanishes turns that average slope into the instantaneous slope at a single point — the derivative.

Definition — Derivative at a Point

The derivative of $f$ at $x$ is

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

provided the limit exists. If it exists, $f$ is differentiable at $x$ .

Step-by-step — Differentiate $f(x)=x^2$ from the definition

Write the difference quotient:

\dfrac{f(x+h)-f(x)}{h} = \dfrac{(x+h)^2 - x^2}{h}

Expand $(x+h)^2$ :

(x+h)^2 = x^2 + 2xh + h^2

— this comes from multiplying

(x+h)(x+h)

term by term:

x\cdot x=x^2

x\cdot h + h\cdot x=2xh

h\cdot h=h^2

Substitute and cancel $x^2$ :

\dfrac{(x^2+2xh+h^2)-x^2}{h} = \dfrac{2xh+h^2}{h}

Factor $h$ out of the numerator and cancel:

\dfrac{h(2x+h)}{h} = 2x+h

, valid since

h\neq0

throughout the limit process.

Take the limit as $h\to0$ :

\lim_{h\to0}(2x+h) = 2x

. The remaining

h

vanishes because

h\to0

directly (no division by

h

remains, so no indeterminate form). Conclusion:

f'(x)=2x

Why h cannot simply equal 0

Setting $h=0$ directly in $\dfrac{f(x+h)-f(x)}{h}$ gives $\dfrac{0}{0}$ — the same indeterminate form from Chapter 01. The algebra in steps 2–4 exists specifically to cancel the $h$ in the denominator before taking the limit, avoiding that indeterminate form entirely.

02 · Geometric Meaning — the Tangent Line

The difference quotient $\dfrac{f(x+h)-f(x)}{h}$ is the slope of a secant line — a line cutting through two points on the curve. As $h\to0$ , the second point slides along the curve toward the first, and the secant line rotates until it just grazes the curve at a single point: the tangent line.

Example — Tangent line to $y=x^2$ at $x=3$

From Section 01, $f'(x)=2x$ , so the slope at $x=3$ is $f'(3)=2(3)=6$ . The point on the curve is $(3, f(3))=(3,9)$ .

Using point-slope form $y - y_1 = m(x-x_1)$ with $m=6$ , $x_1=3$ , $y_1=9$ :

y - 9 = 6(x-3) \implies y = 6x - 18 + 9 = 6x - 9

The $-18$ comes from $6\times(-3)$ ; the final constant $-9$ from $-18+9$ . This line touches the parabola $y=x^2$ at exactly $(3,9)$ and matches its steepness there.

03 · Differentiability vs. Continuity

Every differentiable function is continuous — a sharp corner or a break cannot have a well-defined slope. But the converse is false: continuity does not guarantee differentiability.

❌ f(x) = |x| is continuous but not differentiable at 0

$f(x)=\lvert x\rvert$ is continuous everywhere, including at $x=0$ (Chapter 01's three conditions all hold: $f(0)=0$ , the limit exists and equals $0$ , they match).

Check differentiability using one-sided difference quotients at $x=0$ : from the right, $h>0$ , so $\lvert 0+h\rvert = h$ , giving $\dfrac{h-0}{h}=1$ for every $h>0$ ; the right-hand limit of the difference quotient is $1$ . From the left, $h<0$ , so $\lvert 0+h\rvert=-h$ , giving $\dfrac{-h-0}{h}=-1$ for every $h<0$ ; the left-hand limit is $-1$ .

Why it breaks: $1 \neq -1$ , so the two-sided limit defining $f'(0)$ does not exist (Chapter 01, Section 03's one-sided-limit rule applies directly to this limit).

Consequence: $f(x)=\lvert x\rvert$ has no derivative at $x=0$ — graphically, the sharp corner has no single well-defined tangent line, only two different one-sided tangent slopes.

04 · The Power Rule

Repeating the limit definition for every function would be unworkable. The power rule gives an instant shortcut for any power of $x$ .

Definition — Power Rule

For any real constant $n$ :

\frac{d}{dx}\big[x^n\big] = n\,x^{n-1}

Bring the exponent $n$ down as a multiplying constant, then reduce the exponent by $1$ .

Step-by-step — Differentiate $f(x)=x^5$, $g(x)=x^{-2}$, and $h(x)=\sqrt{x}$

Apply to $x^5$ : bring down

n=5

, reduce exponent to

5-1=4

f'(x)=5x^4

Apply to $x^{-2}$ : bring down

n=-2

, reduce exponent to

-2-1=-3

g'(x)=-2x^{-3}

Rewrite $\sqrt{x}$ as a power:

\sqrt{x}=x^{1/2}

, so

n=1/2

Apply the power rule: bring down

n=1/2

, reduce exponent to

\frac12-1=-\frac12

h'(x)=\frac12 x^{-1/2} = \dfrac{1}{2\sqrt{x}}

05 · Sum, Difference, and Constant Multiple Rules

Differentiation is linear — it distributes over addition/subtraction and passes through constant multiplication unchanged.

Definition — Sum, Difference, Constant Multiple

\frac{d}{dx}\big[f(x)\pm g(x)\big] = f'(x)\pm g'(x) \qquad \frac{d}{dx}\big[c\cdot f(x)\big] = c\cdot f'(x) \qquad \frac{d}{dx}[c]=0

$c$ — any constant. Its derivative is $0$ because a constant function never changes — zero slope everywhere.

Step-by-step — Differentiate $f(x) = 4x^3 - 7x^2 + 5x - 9$

Differentiate term by term (linearity): handle each of the four terms separately.

$4x^3$ : power rule gives

3x^2

; multiply by the constant

4

12x^2

$-7x^2$ : power rule gives

2x

; multiply by

-7

-14x

$5x$ : power rule on

x^1

gives

1\cdot x^0 = 1

; multiply by

5

5

$-9$ : derivative of a constant is

0

Assemble:

f'(x) = 12x^2 - 14x + 5 + 0 = 12x^2-14x+5

06 · The Product Rule

❌ The derivative of a product is NOT the product of derivatives

For $f(x)=x^2$ and $g(x)=x^3$ , the product is $f(x)g(x)=x^5$ , and by the power rule, $\dfrac{d}{dx}[x^5]=5x^4$ .

A tempting but wrong shortcut: $f'(x)\cdot g'(x) = 2x \cdot 3x^2 = 6x^3$ .

Why it breaks: $6x^3 \neq 5x^4$ (check at $x=1$ : $6\neq5$ ). Multiplying derivatives ignores the cross-interaction between the two changing factors.

Consequence: products require their own rule, derived next, that accounts for both factors changing simultaneously.

Definition — Product Rule

\frac{d}{dx}\big[f(x)g(x)\big] = f'(x)g(x) + f(x)g'(x)

"Derivative of the first times the second, plus the first times the derivative of the second."

Step-by-step — Differentiate $y = x^2(x^3+1)$ using the product rule

Identify the two factors:

f(x)=x^2

g(x)=x^3+1

Differentiate each factor separately:

f'(x)=2x

(power rule);

g'(x)=3x^2

(power rule; the

+1

differentiates to

0

Apply the formula:

y' = f'(x)g(x)+f(x)g'(x) = 2x(x^3+1) + x^2(3x^2)

Expand each term:

2x(x^3+1)=2x^4+2x

;

x^2(3x^2)=3x^4

Combine like terms:

y' = 2x^4+2x+3x^4 = 5x^4+2x

. Check: expanding

y=x^2(x^3+1)=x^5+x^2

directly and differentiating with the sum rule gives

5x^4+2x

— matches. ✓

07 · The Quotient Rule

Definition — Quotient Rule

For $g(x)\neq0$ :

\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{\big(g(x)\big)^2}

"Bottom times derivative of top, minus top times derivative of bottom, all over bottom squared." The order of subtraction matters — the rule is not symmetric in $f$ and $g$ .

Step-by-step — Differentiate $y = \dfrac{x^2-1}{x+3}$

Identify top and bottom:

f(x)=x^2-1

g(x)=x+3

Differentiate each:

f'(x)=2x

;

g'(x)=1

Apply the formula:

y' = \dfrac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2} = \dfrac{2x(x+3) - (x^2-1)(1)}{(x+3)^2}

Expand the numerator:

2x(x+3)=2x^2+6x

;

(x^2-1)(1)=x^2-1

. So the numerator is

2x^2+6x - (x^2-1) = 2x^2+6x-x^2+1

Combine like terms in the numerator:

2x^2-x^2=x^2

, leaving

x^2+6x+1

Final answer:

y' = \dfrac{x^2+6x+1}{(x+3)^2}

❌ Forgetting to square the denominator

A common slip on the same problem: writing $y' = \dfrac{2x(x+3)-(x^2-1)(1)}{x+3}$ — reusing $g(x)=x+3$ instead of $(g(x))^2=(x+3)^2$ in the denominator.

Why it breaks: the quotient rule's denominator always carries an exponent of $2$ ; omitting it changes the units of the result and gives a formula that fails to match the definition-based derivative on any test value.

Consequence: always double-check the denominator has the squared bottom before finalizing a quotient-rule answer.

08 · Higher-Order Derivatives

Differentiating a function once gives the rate of change. Differentiating that result gives the rate of change of the rate of change — and the process can repeat indefinitely.

Definition — Higher-Order Derivatives

f''(x) = \frac{d}{dx}\big[f'(x)\big] \qquad f'''(x) = \frac{d}{dx}\big[f''(x)\big] \qquad f^{(n)}(x) = \frac{d}{dx}\Big[f^{(n-1)}(x)\Big]

For $n\geq4$ , the notation switches from repeated primes to $f^{(n)}(x)$ (a superscript in parentheses, to avoid confusing it with a power).

Step-by-step — Find $f''(x)$ for $f(x) = x^4 - 6x^2$

First derivative (power rule term by term):

f'(x) = 4x^3 - 12x

Second derivative — differentiate $f'(x)$ again: apply the power rule to

4x^3

12x^2

. Apply it to

-12x

-12

Assemble:

f''(x) = 12x^2 - 12

A preview of what higher-order derivatives measure

If $f(x)$ tracks position, $f'(x)$ tracks the rate of change of position, and $f''(x)$ tracks the rate of change of that rate. Chapter 04 uses $f''(x)$ to determine how a curve bends — a purely calculus question, independent of any physical interpretation.

09 · Derivatives of Sine and Cosine

The rules so far only differentiate polynomials — sums of powers of $x$ . $\sin x$ and $\cos x$ are not built from powers of $x$ at all, so the power rule cannot reach them. Their derivatives are found the same way the power rule itself was justified in Section 01: directly from the limit definition of the derivative.

Two prerequisite facts, assumed as given

Two ingredients are needed and are not derived here: the angle-addition identities from trigonometry, $\sin(x+h) = \sin x\cos h + \cos x\sin h$ and $\cos(x+h) = \cos x\cos h - \sin x\sin h$ (standard precalculus facts), and the two limit facts from the Symbol Glossary, $\lim_{h\to0}\frac{\sin h}{h}=1$ and $\lim_{h\to0}\frac{\cos h-1}{h}=0$ . Both limit facts follow from a geometric squeeze argument on the unit circle, which is a full topic in its own right and out of scope here — they are used below exactly as named quantities, not silently.

Step-by-step — Differentiate $f(x)=\sin x$ from the definition

Write the difference quotient (Section 01's definition):

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

Expand $\sin(x+h)$ using the angle-addition identity:

\sin(x+h) = \sin x\cos h + \cos x\sin h

, so the numerator becomes

\sin x\cos h + \cos x\sin h - \sin x

Group the two terms that contain $\sin x$ :

\sin x\cos h - \sin x = \sin x(\cos h - 1)

, leaving the numerator as

\sin x(\cos h-1) + \cos x\sin h

Split into two separate fractions over $h$ :

f'(x) = \displaystyle\lim_{h\to0}\left[\sin x\cdot\frac{\cos h-1}{h} + \cos x\cdot\frac{\sin h}{h}\right]

—

\sin x

and

\cos x

do not depend on

h

, so they pull outside each limit.

Substitute the two given limit facts:

\dfrac{\cos h-1}{h}\to0

and

\dfrac{\sin h}{h}\to1

h\to0

, so

f'(x) = \sin x\cdot0 + \cos x\cdot1

Conclude:

f'(x) = \cos x

Step-by-step — Differentiate $f(x)=\cos x$ from the definition

Write the difference quotient:

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

Expand $\cos(x+h)$ using its angle-addition identity:

\cos(x+h) = \cos x\cos h - \sin x\sin h

, so the numerator becomes

\cos x\cos h - \sin x\sin h - \cos x

Group the two terms that contain $\cos x$ :

\cos x\cos h - \cos x = \cos x(\cos h-1)

, leaving the numerator as

\cos x(\cos h-1) - \sin x\sin h

Split into two fractions over $h$ :

f'(x) = \displaystyle\lim_{h\to0}\left[\cos x\cdot\frac{\cos h-1}{h} - \sin x\cdot\frac{\sin h}{h}\right]

Substitute the given limit facts:

\dfrac{\cos h-1}{h}\to0

\dfrac{\sin h}{h}\to1

, so

f'(x) = \cos x\cdot0 - \sin x\cdot1

Conclude:

f'(x) = -\sin x

. Note the minus sign — it does not cancel or disappear; it is the direct result of the

-\sin x\sin h

term in step 3, which had no

\cos x

factor to group with

(\cos h-1)

Geometric intuition — the derivative of sine traces out cosine

Plot $\sin x$ : it rises steepest at $x=0$ (slope $1=\cos0$ ), flattens to a momentary standstill at the peak $x=\pi/2$ (slope $0=\cos(\pi/2)$ ), then falls steepest at $x=\pi$ (slope $-1=\cos\pi$ ). The height of the cosine curve at each $x$ exactly matches the steepness of the sine curve at that same $x$ — which is precisely what $f'(x)=\cos x$ states. The two curves are the same shape, one a quarter-period shift of the other, which is why differentiating sine "produces" cosine rather than some unrelated function.

❌ Dropping the negative sign on the derivative of cosine

A common slip: writing $\dfrac{d}{dx}[\cos x] = \sin x$ , omitting the minus sign derived in step 6 above.

Why it breaks: near $x=0$ , $\cos x$ is decreasing — check numerically: $\cos(0.1)\approx0.995$ , $\cos(0)=1$ , so the function's value went down as $x$ increased from $0$ to $0.1$ . A positive derivative $f'(0)=\sin(0)=0$ or, worse, an incorrectly-signed positive value elsewhere would claim the function is flat or rising, contradicting the observed decrease.

Consequence: the correct fact $\dfrac{d}{dx}[\cos x]=-\sin x$ gives $f'(0)=-\sin(0)=0$ at $x=0$ specifically (a momentary flat point, correct since $x=0$ is a peak of cosine), but at $x=0.5$ : $f'(0.5)=-\sin(0.5)\approx-0.479<0$ , correctly signaling a decrease — matching $\cos(0.6)<\cos(0.5)$ . Dropping the sign would instead predict an increase there, the opposite of the true behavior.

10 · Derivative of the Natural Exponential Function

Definition — Derivative of e^x

Using the defining property of $e$ from the Symbol Glossary, $\displaystyle\lim_{h\to0}\frac{e^h-1}{h}=1$ :

\frac{d}{dx}\big[e^x\big] = e^x

$e^x$ is its own derivative — a fixed point of differentiation.

Step-by-step — Differentiate $f(x)=e^x$ from the definition

Write the difference quotient:

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

Split the exponent using the exponent law $e^{x+h}=e^x\cdot e^h$ (a standard property of exponents, assumed known): the numerator becomes

e^x e^h - e^x

Factor $e^x$ out of the numerator:

e^x e^h - e^x = e^x(e^h-1)

— this is valid because

e^x

does not depend on

h

, the variable the limit is taken over.

Pull $e^x$ outside the limit (again, it does not depend on $h$ ):

f'(x) = e^x\cdot\displaystyle\lim_{h\to0}\frac{e^h-1}{h}

Substitute the defining limit of $e$ : by the Symbol Glossary's definition of

e

, this limit equals exactly

1

Conclude:

f'(x) = e^x\cdot1 = e^x

Geometric intuition — the curve whose height equals its own slope

For every other function differentiated so far, the derivative is a different-looking function — $x^2$ 's derivative is $2x$ , a straight line, not a parabola. $e^x$ is the unique (up to a constant multiple) function whose graph's steepness at any point equals the graph's height at that same point. Where $e^x$ is small, it grows slowly; where it is large, it grows fast — the rate of growth is proportional to the current value, which is exactly why exponential growth compounds the way it does in Chapter 08's perpetuity example.

❌ Applying the power rule to an exponential function

A common confusion: differentiating $e^x$ as if it were $x^n$ with $n=e$ -like treatment, producing something like $x\cdot e^{x-1}$ by mimicking the power rule's "bring down the exponent, subtract one" pattern.

Why it breaks: the power rule (Section 04) applies to a fixed exponent and a variable base — $x^n$ for constant $n$ . Here the roles are reversed: $e^x$ has a fixed base and a variable exponent. These are structurally different functions, and the power rule's derivation (Section 04, built by expanding $(x+h)^n$ via the binomial theorem) does not apply when the variable is in the exponent rather than the base — there is no binomial expansion of $e^{x+h}$ that behaves like $(x+h)^n$ does.

Consequence: $x\cdot e^{x-1}$ is not even dimensionally close to the correct derivative; at $x=1$ , the (wrong) power-rule-style guess gives $1\cdot e^0=1$ , while the correct derivative from the step-block above gives $f'(1)=e^1\approx2.718$ — the two disagree, and only the exponential rule derived above is valid whenever the variable sits in the exponent.

11 · Derivative of the Natural Logarithm

$\ln x$ is defined (Symbol Glossary) as the inverse of $e^x$ . Rather than returning to the limit definition directly, its derivative is found using a general fact about inverse functions: reciprocal slopes.

Definition — Inverse Function Derivative Rule

If $y=f^{-1}(x)$ is the inverse of a differentiable, invertible function $f$ (so $x=f(y)$ ), then writing the derivative as a limit of $\dfrac{\Delta y}{\Delta x}$ and noting that $\Delta x = f(y+\Delta y)-f(y)$ :

\frac{dy}{dx} = \lim_{\Delta y\to0}\frac{\Delta y}{f(y+\Delta y)-f(y)} = \frac{1}{\displaystyle\lim_{\Delta y\to0}\dfrac{f(y+\Delta y)-f(y)}{\Delta y}} = \frac{1}{f'(y)}

The last step used the fact that as $\Delta x\to0$ , $\Delta y\to0$ as well (the inverse function is continuous), and the reciprocal of a limit is the limit of the reciprocal, provided $f'(y)\neq0$ . This is a standalone derivation from first principles here; Chapter 03 revisits the same identity using the chain rule, once that tool is available.

Step-by-step — Differentiate $f(x)=\ln x$ using the inverse function rule

Set up the inverse relationship: let

y=\ln x

, so by the Symbol Glossary's definition,

x=e^y

Apply the inverse function derivative rule with $f(y)=e^y$ :

\dfrac{dy}{dx} = \dfrac{1}{f'(y)} = \dfrac{1}{\frac{d}{dy}[e^y]}

Substitute the derivative of $e^y$ from Section 10 (the same rule applies regardless of the variable's name):

\dfrac{d}{dy}[e^y] = e^y

, so

\dfrac{dy}{dx} = \dfrac{1}{e^y}

Replace $e^y$ using the original substitution $x=e^y$ from step 1:

\dfrac{1}{e^y} = \dfrac{1}{x}

Conclude:

\dfrac{d}{dx}[\ln x] = \dfrac{1}{x}

, valid for

x>0

(the domain of

\ln x

Geometric intuition — reflecting across the line y = x

The graph of $y=\ln x$ is the graph of $y=e^x$ reflected across the diagonal line $y=x$ (the defining geometric picture of an inverse function). Reflecting a curve across $y=x$ swaps the roles of "rise" and "run" at every corresponding point, which turns a slope into its reciprocal — exactly the algebra in step 2 above. Where $e^x$ grows steeply (large $x$ ), its reflection $\ln x$ grows correspondingly slowly, matching $\dfrac1x\to0$ as $x\to\infty$ .

12 · Exercises

EXERCISE 2.1

Set up the difference quotient $\dfrac{f(x+h)-f(x)}{h}$ , expand $(x+h)^3$ fully, cancel the $x^3$ terms, factor out $h$ , then let $h\to0$ .

$(x+h)^3 = x^3+3x^2h+3xh^2+h^3$ . So $f(x+h)-f(x) = 3x^2h+3xh^2+h^3 = h(3x^2+3xh+h^2)$ . Dividing by $h$ : $3x^2+3xh+h^2$ . Taking $h\to0$ : $f'(x)=3x^2$ — matching the power rule shortcut.

Use the limit definition to find $f'(x)$ for $f(x) = x^3$ . Show the full binomial expansion.

EXERCISE 2.2

Differentiate each term with the power rule; remember the derivative of a lone constant is $0$ .

$f(x) = 5x^4 - 3x^2 + 8x - 12$ . Term by term: $5x^4 \to 20x^3$ ; $-3x^2\to-6x$ ; $8x\to8$ ; $-12\to0$ . So $f'(x)=20x^3-6x+8$ .

Differentiate $f(x) = 5x^4 - 3x^2 + 8x - 12$ .

EXERCISE 2.3

Identify the two factors, differentiate each, and combine with "derivative of first times second, plus first times derivative of second."

$f=x^3$ , $g=2x^2-5$ . $f'=3x^2$ , $g'=4x$ . $y' = 3x^2(2x^2-5) + x^3(4x) = 6x^4-15x^2+4x^4 = 10x^4-15x^2$ .

Differentiate $y = x^3(2x^2-5)$ using the product rule. Verify by first expanding the product and differentiating term by term.

EXERCISE 2.4

Apply the quotient rule directly: bottom times derivative of top, minus top times derivative of bottom, over bottom squared.

$f = 3x+1$ , $g = x-2$ . $f'=3$ , $g'=1$ . $y' = \dfrac{3(x-2)-(3x+1)(1)}{(x-2)^2} = \dfrac{3x-6-3x-1}{(x-2)^2} = \dfrac{-7}{(x-2)^2}$ .

Differentiate $y = \dfrac{3x+1}{x-2}$ .

EXERCISE 2.5

Check the one-sided difference quotients at the corner point separately, as done for $|x|$ in Section 03.

$f(x) = \begin{cases}x^2 & x<1\\2x-1 & x\geq1\end{cases}$ . From the left: $f'(x)=2x$ for $x<1$ , so approaching $x=1$ gives slope $2(1)=2$ . From the right: $f'(x)=2$ for $x>1$ (constant slope of the line $2x-1$ ). Both one-sided slopes equal $2$ , and $f$ is continuous at $x=1$ (both pieces give $f(1)=1$ ), so $f$ is differentiable at $x=1$ , with $f'(1)=2$ .

Determine whether $f(x) = \begin{cases}x^2 & x<1\\2x-1 & x\geq1\end{cases}$ is differentiable at $x=1$ . Show both one-sided difference-quotient limits.

EXERCISE 2.6

An asset's price function's first derivative gives instantaneous price velocity; the second derivative measures how that velocity itself is changing.

$P(t) = -2t^3 + 15t^2 + 100$ (price in dollars, $t$ in days). $P'(t) = -6t^2+30t$ . $P'(2) = -6(4)+30(2) = -24+60=36$ — price is increasing at $36$ dollars/day at $t=2$ . $P''(t) = -12t+30$ . $P''(2) = -24+30=6>0$ — the rate of increase is itself still growing at $t=2$ , i.e. the price is accelerating upward, not just rising.

An asset's price is modeled by $P(t) = -2t^3+15t^2+100$ for $t$ in days. Compute $P'(2)$ and $P''(2)$ , and state what each tells you about the price's behavior at $t=2$ .

EXERCISE 2.7

Expand $\sin(x+h)$ with the angle-addition identity, group the $\sin x$ terms, split into the two given limit facts, and substitute.

$f(x)=\sin x$ at a general point, following Section 09's derivation: $\sin(x+h)-\sin x = \sin x(\cos h-1)+\cos x\sin h$ . Dividing by $h$ and taking the limit: $\sin x\cdot0+\cos x\cdot1=\cos x$ . So $f'(x)=\cos x$ , and at $x=\pi/2$ : $f'(\pi/2)=\cos(\pi/2)=0$ — matching the fact that $\sin x$ has a peak (momentarily flat) at $x=\pi/2$ .

Using the derivation method of Section 09, state $f'(x)$ for $f(x)=\sin x$ and evaluate $f'(\pi/2)$ . Explain in one sentence why a value of $0$ makes sense there.

EXERCISE 2.8

Use the exponential derivative rule from Section 10 directly; no expansion is required once the rule is established.

$f(x) = 3e^x - x^2$ . Differentiate term by term (linearity, Section 05): $\dfrac{d}{dx}[3e^x] = 3e^x$ (Section 10's rule, constant multiple); $\dfrac{d}{dx}[-x^2]=-2x$ (power rule). So $f'(x) = 3e^x-2x$ .

Differentiate $f(x) = 3e^x - x^2$ .

EXERCISE 2.9

Apply the product rule from Section 06, differentiating $\ln x$ using Section 11's rule for one of the two factors.

$y = x\ln x$ . Product rule with $f(x)=x$ , $g(x)=\ln x$ : $f'(x)=1$ , $g'(x)=\dfrac1x$ (Section 11). $y' = 1\cdot\ln x + x\cdot\dfrac1x = \ln x + 1$ . The $x\cdot\frac1x=1$ term comes directly from multiplying $x$ by $\ln x$ 's derivative and cancelling.

Differentiate $y = x\ln x$ using the product rule together with the derivative of $\ln x$ from Section 11.

13 · Chapter Summary

Concept	Formula / Rule
Derivative (definition)	$f'(x) = \lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}$
Tangent line slope	$m = f'(a)$ at the point $(a, f(a))$
Differentiable $\implies$ continuous	Converse false — see $\lvert x\rvert$ at $0$
Power rule	$\dfrac{d}{dx}[x^n] = nx^{n-1}$
Sum/difference/constant multiple	Differentiation is linear; $\dfrac{d}{dx}[c]=0$
Product rule	$(fg)' = f'g+fg'$
Quotient rule	$\left(\dfrac{f}{g}\right)' = \dfrac{f'g-fg'}{g^2}$
Higher-order derivatives	$f''(x) = \dfrac{d}{dx}[f'(x)]$ ; continues to $f^{(n)}(x)$
Derivative of sine	$\dfrac{d}{dx}[\sin x] = \cos x$
Derivative of cosine	$\dfrac{d}{dx}[\cos x] = -\sin x$
Derivative of $e^x$	$\dfrac{d}{dx}[e^x] = e^x$
Derivative of $\ln x$	$\dfrac{d}{dx}[\ln x] = \dfrac{1}{x}$ , $x>0$
Inverse function rule	$\dfrac{dy}{dx} = \dfrac{1}{f'(y)}$ where $x=f(y)$

Next: Chapter 03 — Chain Rule, Implicit Differentiation, where composite and implicitly-defined functions get their own differentiation rules.