Derivative from first principles — difference quotient definition: Exercises

Remember the recipe from the parent note: Plug, Subtract, Cancel, Shrink.

Figure — Derivative from first principles — difference quotient definition

Every problem is the same story as this picture: a secant line through two nearby points slowly rotating into the tangent as the second point slides in. Read Secant and Tangent lines if that picture is fuzzy.

Level 1 — Recognition

(Can you spot the pieces and run the machine on the simplest inputs?)

Exercise 1.1

State the difference quotient for $f (x) = 5 x - 2$ , then find $f^{'} (x)$ from first principles.

Recall Solution 1.1

Step 1 — Plug. $f (x + h) = 5 (x + h) - 2 = 5 x + 5 h - 2$ . Step 2 — Subtract. $f (x + h) - f (x) = (5 x + 5 h - 2) - (5 x - 2) = 5 h$ Step 3 — Cancel. $\frac{5 h}{h} = 5$ . Step 4 — Shrink. $f^{'} (x) = lim_{h \to 0} 5 = 5$ . Answer: $f^{'} (x) = 5$ . A straight line has one constant slope everywhere — exactly its coefficient. ✓

Exercise 1.2

Find $f^{'} (x)$ from first principles for the constant $f (x) = 7$ .

Recall Solution 1.2

Plug. $f (x + h) = 7$ (no $x$ to replace). Subtract. $7 - 7 = 0$ . Cancel. $\frac{0}{h} = 0$ for every $h \neq = 0$ . Shrink. $f^{'} (x) = lim_{h \to 0} 0 = 0$ . Answer: $f^{'} (x) = 0$ . A flat horizontal line never rises — zero steepness everywhere. ✓

Exercise 1.3

For $f (x) = x^{2}$ (derived in the parent note as $f^{'} (x) = 2 x$ ), what is the slope of the tangent line at $x = - 3$ ?

Recall Solution 1.3

We already have $f^{'} (x) = 2 x$ . Substitute the single point: $f^{'} (- 3) = 2 (- 3) = - 6$ Answer: $- 6$ . On the left arm of the parabola the curve slopes downhill, so a negative slope is exactly what we expect. ✓

Level 2 — Application

(Run the full four-step machine on genuinely curved functions.)

Exercise 2.1

Find $f^{'} (x)$ from first principles for $f (x) = x^{2} + 3 x$ .

Recall Solution 2.1

Plug. $f (x + h) = (x + h)^{2} + 3 (x + h) = x^{2} + 2 x h + h^{2} + 3 x + 3 h$ . Subtract. $f (x + h) - f (x) = (x^{2} + 2 x h + h^{2} + 3 x + 3 h) - (x^{2} + 3 x) = 2 x h + h^{2} + 3 h$ The $x^{2}$ and $3 x$ terms cancel — this cancellation is why a finite slope survives. Cancel. Factor $h$ : $\frac{h ( 2 x + h + 3 )}{h} = 2 x + h + 3$ Shrink. $f^{'} (x) = lim_{h \to 0} (2 x + h + 3) = 2 x + 3$ . Answer: $f^{'} (x) = 2 x + 3$ . ✓

Exercise 2.2

Find $f^{'} (x)$ from first principles for $f (x) = x^{3}$ .

Recall Solution 2.2

Plug. Use $(x + h)^{3} = x^{3} + 3 x^{2} h + 3 x h^{2} + h^{3}$ . Subtract. $f (x + h) - f (x) = 3 x^{2} h + 3 x h^{2} + h^{3}$ Cancel. Every term has an $h$ : $\frac{h ( 3 x ^{2} + 3 x h + h ^{2} )}{h} = 3 x^{2} + 3 x h + h^{2}$ Shrink. As $h \to 0$ both $3 x h$ and $h^{2}$ vanish: $f^{'} (x) = 3 x^{2}$ Answer: $f^{'} (x) = 3 x^{2}$ . This matches the Power Rule pattern $\frac{d}{d x} x^{n} = n x^{n - 1}$ with $n = 3$ . ✓

Exercise 2.3

Find $f^{'} (x)$ from first principles for $f (x) = \frac{1}{x ^{2}}$ .

Recall Solution 2.3

Plug. $f (x + h) = \frac{1}{( x + h ) ^{2}}$ . Subtract (common denominator). $\frac{1}{( x + h ) ^{2}} - \frac{1}{x ^{2}} = \frac{x ^{2} - ( x + h ) ^{2}}{x ^{2} ( x + h ) ^{2}} = \frac{x ^{2} - ( x ^{2} + 2 x h + h ^{2} )}{x ^{2} ( x + h ) ^{2}} = \frac{- 2 x h - h ^{2}}{x ^{2} ( x + h ) ^{2}}$ Cancel. Divide by $h$ ; factor $h$ from the top: $\frac{1}{h} \cdot \frac{- h ( 2 x + h )}{x ^{2} ( x + h ) ^{2}} = \frac{- ( 2 x + h )}{x ^{2} ( x + h ) ^{2}}$ Shrink. Let $h \to 0$ : $f^{'} (x) = \frac{- 2 x}{x ^{2} \cdot x ^{2}} = \frac{- 2 x}{x ^{4}} = - \frac{2}{x ^{3}}$ Answer: $f^{'} (x) = - \frac{2}{x ^{3}}$ . ✓

Level 3 — Analysis

(The algebra hides the $h$ — you must engineer it into view.)

Exercise 3.1

Find $f^{'} (x)$ from first principles for $f (x) = x + 1$ (a shifted square root).

Recall Solution 3.1

Plug. $f (x + h) = x + h + 1$ . Subtract. $x + h + 1 - x + 1$ — stuck, no $h$ to factor. Use the conjugate trick (multiply top and bottom by $x + h + 1 + x + 1$ , which turns $a - b$ into $a - b$ and manufactures an $h$ ): $\frac{x + h + 1 - x + 1}{h} \cdot \frac{x + h + 1 + x + 1}{x + h + 1 + x + 1} = \frac{( x + h + 1 ) - ( x + 1 )}{h ( x + h + 1 + x + 1 )} = \frac{h}{h ( x + h + 1 + x + 1 )}$ Cancel. $= \frac{1}{x + h + 1 + x + 1}$ . Shrink. $h \to 0$ makes both roots equal: $f^{'} (x) = \frac{1}{2 x + 1}$ Answer: $f^{'} (x) = \frac{1}{2 x + 1}$ . ✓

Exercise 3.2

Find $f^{'} (x)$ from first principles for $f (x) = \frac{x}{x + 1}$ .

Recall Solution 3.2

Plug. $f (x + h) = \frac{x + h}{x + h + 1}$ . Subtract (common denominator). $\frac{x + h}{x + h + 1} - \frac{x}{x + 1} = \frac{( x + h ) ( x + 1 ) - x ( x + h + 1 )}{( x + h + 1 ) ( x + 1 )}$ Expand the top: $(x + h) (x + 1) = x^{2} + x + h x + h$ and $x (x + h + 1) = x^{2} + h x + x$ . Subtract: $(x^{2} + x + h x + h) - (x^{2} + h x + x) = h$ So the numerator is just $h$ : $= \frac{h}{( x + h + 1 ) ( x + 1 )}$ Cancel. Divide by $h$ : $\frac{1}{( x + h + 1 ) ( x + 1 )}$ . Shrink. $h \to 0$ : $f^{'} (x) = \frac{1}{( x + 1 ) ^{2}}$ Answer: $f^{'} (x) = \frac{1}{( x + 1 ) ^{2}}$ . Always positive — this function only ever climbs where it is defined. ✓

Exercise 3.3

Show, from first principles, that $f (x) = ∣ x ∣$ is not differentiable at $x = 0$ , and identify what goes wrong. (Uses Limits — formal definition and one-sided limits.)

Recall Solution 3.3

At $x = 0$ the difference quotient is $\frac{f ( 0 + h ) - f ( 0 )}{h} = \frac{∣ h ∣ - 0}{h} = \frac{∣ h ∣}{h}$ Now split by the sign of $h$ (this is a one-sided-limit question — the two sides need not agree):

From the right ( $h \to 0^{+}$ ): $∣ h ∣ = h$ , so $\frac{h}{h} = + 1$ . Limit $= + 1$ .
From the left ( $h \to 0^{-}$ ): $∣ h ∣ = - h$ , so $\frac{- h}{h} = - 1$ . Limit $= - 1$ .

The right-hand slope $(+ 1)$ and left-hand slope $(- 1)$ disagree, so the two-sided limit does not exist. $f$ is not differentiable at $0$ . Geometrically the graph has a sharp corner — the secant approaches a different tangent from each side, so no single tangent slope exists. (Note $∣ x ∣$ is continuous at $0$ ; see Continuity and Differentiability — continuous does not force differentiable.) ✓

Level 4 — Synthesis

(Combine the machine with unknown constants and general structure.)

Exercise 4.1

For the general quadratic $f (x) = a x^{2} + b x + c$ , prove from first principles that $f^{'} (x) = 2 a x + b$ .

Recall Solution 4.1

Plug. $f (x + h) = a (x + h)^{2} + b (x + h) + c = a x^{2} + 2 a x h + a h^{2} + b x + bh + c$ . Subtract. $f (x + h) - f (x) = (a x^{2} + 2 a x h + a h^{2} + b x + bh + c) - (a x^{2} + b x + c) = 2 a x h + a h^{2} + bh$ Every "position" term ( $a x^{2}, b x, c$ ) cancels; only "change" terms — each carrying an $h$ — remain. Cancel. $\frac{h ( 2 a x + ah + b )}{h} = 2 a x + ah + b$ Shrink. $h \to 0$ kills the $ah$ : $f^{'} (x) = 2 a x + b$ Answer: $f^{'} (x) = 2 a x + b$ . Check against Exercise 2.1 ( $a = 1, b = 3, c = 0$ ): $2 x + 3$ . ✓

Exercise 4.2

Prove from first principles the sum rule: if $f (x) = u (x) + v (x)$ and both $u, v$ are differentiable, then $f^{'} (x) = u^{'} (x) + v^{'} (x)$ .

Recall Solution 4.2

Plug & Subtract. Write the difference quotient and group the $u$ -parts and $v$ -parts: $\frac{f ( x + h ) - f ( x )}{h} = \frac{[ u ( x + h ) + v ( x + h ) ] - [ u ( x ) + v ( x ) ]}{h} = \frac{u ( x + h ) - u ( x )}{h} + \frac{v ( x + h ) - v ( x )}{h}$ Shrink. A limit of a sum is the sum of the limits (a limit law from Limits — formal definition and one-sided limits), and each piece is a derivative by definition: $f^{'} (x) = lim_{h \to 0} \frac{u ( x + h ) - u ( x )}{h} + lim_{h \to 0} \frac{v ( x + h ) - v ( x )}{h} = u^{'} (x) + v^{'} (x)$ Answer: $f^{'} (x) = u^{'} (x) + v^{'} (x)$ . ✓ No algebra tricks — the whole proof is regrouping plus one limit law.

Exercise 4.3

Find the equation of the tangent line to $f (x) = x^{2}$ at $x = 2$ , using the first-principles derivative. (Ties to Secant and Tangent lines.)

Recall Solution 4.3

From the parent note $f^{'} (x) = 2 x$ , so the tangent slope at $x = 2$ is $m = f^{'} (2) = 2 (2) = 4$ The point of tangency is $(2, f (2)) = (2, 4)$ . Point–slope form $y - y_{0} = m (x - x_{0})$ : $y - 4 = 4 (x - 2) ⟹ y = 4 x - 4$ Answer: $y = 4 x - 4$ . As a check, this line passes through $(2, 4)$ and has slope $4$ ; a nearby secant to $x = 2.5$ has slope $\frac{f ( 2.5 ) - f ( 2 )}{0.5} = \frac{6.25 - 4}{0.5} = 4.5$ , already close to $4$ and shrinking toward it as $h \to 0$ . ✓

Level 5 — Mastery

(Subtle limits, degenerate points, and forecasting.)

Exercise 5.1

Find $f^{'} (0)$ from first principles for the piecewise function $f (x) = {x^{2} sin (\frac{1}{x}) 0 x \neq = 0 x = 0$ and note why the ordinary "plug into a formula" instinct fails here.

Recall Solution 5.1

We must use the definition at $x = 0$ because $f$ is defined by a special rule there. $\frac{f ( 0 + h ) - f ( 0 )}{h} = \frac{h ^{2} s i n ( \frac{1}{h} ) - 0}{h} = h sin (\frac{1}{h})$ Now bound it. Since $- 1 \leq sin (\cdot) \leq 1$ for every input, $- ∣ h ∣ \leq h sin (\frac{1}{h}) \leq ∣ h ∣$ As $h \to 0$ both outer bounds go to $0$ , so the middle is squeezed to $0$ : $f^{'} (0) = lim_{h \to 0} h sin (\frac{1}{h}) = 0$ Answer: $f^{'} (0) = 0$ . The derivative exists even though $sin (1/ h)$ oscillates wildly, because the $h$ out front crushes the oscillation. You could not get this by "differentiating the formula and plugging in $0$ " — the formula $x^{2} sin (1/ x)$ isn't even defined at $0$ . Only the difference quotient reaches the answer. ✓

Exercise 5.2

Forecast, then verify. For $f (x) = x^{4}$ , forecast $f^{'} (x)$ using the pattern you've seen ( $x^{2} \to 2 x$ , $x^{3} \to 3 x^{2}$ ). Then verify from first principles.

Recall Solution 5.2

Forecast. The pattern $x^{n} \to n x^{n - 1}$ predicts $f^{'} (x) = 4 x^{3}$ . Verify. Use $(x + h)^{4} = x^{4} + 4 x^{3} h + 6 x^{2} h^{2} + 4 x h^{3} + h^{4}$ . Subtract. $f (x + h) - f (x) = 4 x^{3} h + 6 x^{2} h^{2} + 4 x h^{3} + h^{4}$ . Cancel. Factor $h$ : $\frac{h ( 4 x ^{3} + 6 x ^{2} h + 4 x h ^{2} + h ^{3} )}{h} = 4 x^{3} + 6 x^{2} h + 4 x h^{2} + h^{3}$ Shrink. Every term with an $h$ vanishes: $f^{'} (x) = 4 x^{3}$ Answer confirmed: $f^{'} (x) = 4 x^{3}$ . ✓ The forecast held — this is exactly the Power Rule being built from first principles.

Exercise 5.3

The difference quotient $\frac{f ( x + h ) - f ( x )}{h}$ at $h = 0$ is the indeterminate form $0/0$. Using $f (x) = x^{2}$ at $x = 3$ , evaluate the quotient at $h = 1, 0.1, 0.01$ and state the limit it approaches. Explain why the value is not $0/0$ even though the fraction looks like it becomes $0/0$ .

Recall Solution 5.3

With $f (x) = x^{2}$ , at $x = 3$ the quotient simplifies (Step 3) to $2 x + h = 6 + h$ . Tabulate:

h & \frac{f(3+h)-f(3)}{h}=6+h\\ \hline 1 & 7\\ 0.1 & 6.1\\ 0.01 & 6.01 \end{array}$$ The values march toward $\mathbf{6}$, which equals $f'(3)=2(3)=6$. ✓ **Why not $0/0$?** At $h=0$ *exactly*, both rise and run are $0$ — genuinely undefined. But a limit never asks for the value *at* $h=0$; it asks what the fraction *approaches near* $h=0$. Cancelling the $h$ (Step 3) rewrites the same expression in a form that *is* defined there, and that form settles on $6$. The $0/0$ was only a disguise the algebra removed. ✓

Self-test recap

Constant

f (x) = c

has derivative

0

everywhere.

Slope of

f (x) = x^{2}

x = - 3

- 6

f^{'} (x)

for

f (x) = x^{3}

from first principles

3 x^{2}

f^{'} (x)

for

f (x) = 1/ x^{2}

- 2/ x^{3}

Why is

∣ x ∣

not differentiable at

0

The right slope is

+ 1

and the left slope is

- 1

; they disagree, so the limit fails — a corner.

Tangent to

y = x^{2}

x = 2

y = 4 x - 4

General quadratic

a x^{2} + b x + c

has derivative

2 a x + b

f^{'} (0)

for

x^{2} sin (1/ x)

(with

f (0) = 0

)

0

, by the squeeze

- ∣ h ∣ \leq h sin (1/ h) \leq ∣ h ∣

Connections

Limits — formal definition and one-sided limits — powers Exercises 3.3, 4.2, 5.1.
Indeterminate forms 0 over 0 — the heart of Exercise 5.3.
Secant and Tangent lines — the picture behind 4.3.
Power Rule — built by 2.2, 4.1, 5.2.
Continuity and Differentiability — Exercise 3.3 (continuous but not differentiable).
Average vs Instantaneous Rate of Change — the shrinking-secant view of 5.3.
Parent: Derivative from first principles — difference quotient definition.