Worked examples — Derivative from first principles — difference quotient definition

Before any symbol: recall that $f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$ means "take a buddy point $h$ away, measure the slope of the line between them, then slide the buddy in until $h$ vanishes." The whole difficulty is always the same: at $h=0$ the raw fraction reads $\frac{0}{0}$ (see Indeterminate forms 0 over 0), so we must cancel the $h$ by algebra first, then let $h\to 0$.

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The scenario matrix

Every first-principles problem falls into one of these cells. The examples below are labelled so each cell is covered at least once.

Cell	What makes it different	The algebra tool needed	Example
A. Polynomial (higher power)	$(x+h{)}^n$ must be expanded	Binomial expansion, cancel $x^n$	Ex 1
B. Negative slope region	derivative comes out negative	sign-tracking of the answer	Ex 2
C. Rational (fraction) function	subtracting fractions	common denominator	Ex 3
D. Root / irrational	no $h$ to factor at first	conjugate multiplication	Ex 4
E. Constant function (degenerate)	numerator is exactly $0$	recognise $0/h$	Ex 5
F. Non-differentiable corner	left slope ≠ right slope	one-sided limits	Ex 6
G. Real-world word problem	units + interpretation	apply recipe, read meaning	Ex 7
H. Exam twist (evaluate at a point)	numeric slope + tangent line	plug value after differentiating	Ex 8

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Ex 1 — Cell A: a higher power, $f(x)=x^3$

Forecast: For $x^2$ we got $2x$. What pattern would you guess for $x^3$? (Jot it down before reading.) Many guess $3x^2$ — let us earn it.

Step 1 — write $f(x+h)$. $f(x+h)=(x+h{)}^3=x^3+3x^{2}h+3x{h}^2+h^3$ Why this step? We must know the output at the buddy point. Expanding the cube is the only new work versus $x^2$; every $x$ becomes the block $(x+h)$.

Step 2 — form the numerator. $f(x+h)-f(x)=(x^3+3x^{2}h+3x{h}^2+h^3)-x^3=3x^{2}h+3x{h}^2+h^3$ Why this step? The lone $x^3$ terms cancel. That cancellation is why a finite slope survives — without it we would be dividing something-that-isn't-small by $h$.

Step 3 — factor out $h$ and cancel. $\frac{3x^{2}h+3x{h}^2+h^3}{h}=\frac{h(3x^2+3xh+h^2)}{h}=3x^2+3xh+h^2$ Why this step? Every surviving term already carries an $h$, so $h$ factors cleanly. Cancelling it removes the $\frac{0}{0}$ trap — the expression is now safe to evaluate at $h=0$.

Step 4 — shrink $h\to 0$. $f^{\prime }(x)={\lim }_{h\to 0}(3x^2+3xh+h^2)=3x^2$ Why this step? Both leftover terms $3xh$ and $h^2$ contain $h$, so they vanish. Answer: $f^{\prime }(x)=3x^2$.

Verify: At $x=2$, $f^{\prime }(2)=3(2{)}^2=12$. Sanity: $x^3$ is climbing fast at $x=2$ (from $8$ toward $27$ over the next unit), so a steepness of $12$ is believable. Also $f^{\prime }(0)=0$ — the curve flattens through the origin. ✓ This matches the Power Rule $\frac{d}{dx}{x}^n=n{x}^{n-1}$.

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Ex 2 — Cell B: a region of negative slope, $f(x)=5-x^2$

Forecast: The parabola opens down. To the right of its peak it falls. Do you expect a positive or negative slope at $x=3$?

Step 1. $f(x+h)=5-(x+h{)}^2=5-x^2-2xh-h^2$. Why this step? Careful with the minus sign — the whole $(x+h{)}^2$ block is subtracted.

Step 2. $f(x+h)-f(x)=(5-x^2-2xh-h^2)-(5-x^2)=-2xh-h^2$ Why this step? The $5$ and $-x^2$ cancel, leaving only $h$-carrying terms — exactly the sign we need to survive.

Step 3. $\frac{-2xh-h^2}{h}=\frac{h(-2x-h)}{h}=-2x-h$ Why this step? Factor the $h$, cancel, escape $\frac{0}{0}$.

Step 4. $f^{\prime }(x)={\lim }_{h\to 0}(-2x-h)=-2x$ Answer: $f^{\prime }(x)=-2x$.

Verify: At $x=3$, $f^{\prime }(3)=-6$ — negative, matching the forecast that the curve falls to the right of its peak. At $x=0$ the slope is $0$ (the top of the hill is flat). ✓

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Ex 3 — Cell C: a fraction, $f(x)=\frac{1}{x^2}$

Forecast: For $1/x$ the parent got $-1/x^2$ (always negative). Do you expect $1/x^2$ to also slope negatively for $x>0$?

Step 1. $f(x+h)=\frac{1}{(x+h{)}^2}$.

Step 2 — common denominator. $\frac{1}{(x+h{)}^2}-\frac{1}{x^2}=\frac{x^2-(x+h{)}^2}{x^2(x+h{)}^2}=\frac{x^2-(x^2+2xh+h^2)}{x^2(x+h{)}^2}=\frac{-2xh-h^2}{x^2(x+h{)}^2}$ Why this step? Subtracting two fractions needs a shared denominator; doing so surfaces the hidden $h$ in the numerator so we can eventually cancel it.

Step 3 — divide by $h$, cancel. $\frac{1}{h}\cdot \frac{-2xh-h^2}{x^2(x+h{)}^2}=\frac{h(-2x-h)}{h{x}^2(x+h{)}^2}=\frac{-2x-h}{x^2(x+h{)}^2}$ Why this step? Factor $h$ from the numerator, cancel with the $h$ from the denominator — the $\frac{0}{0}$ is gone.

Step 4. $f^{\prime }(x)={\lim }_{h\to 0}\frac{-2x-h}{x^2(x+h{)}^2}=\frac{-2x}{x^2\cdot x^2}=\frac{-2x}{x^4}=-\frac{2}{x^3}$ Answer: $f^{\prime }(x)=-\frac{2}{x^3}$.

Verify: At $x=1$: $f^{\prime }(1)=-2$. For $x>0$ this is negative — yes, $1/x^2$ falls as $x$ grows, matching the forecast. Power-rule cross-check: $x^{-2}\Rightarrow -2x^{-3}=-2/x^3$. ✓

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Ex 4 — Cell D: a root, $f(x)=\sqrt{2x+1}$

Forecast: The parent showed $\sqrt{x}\to \frac{1}{2\sqrt{x}}$. This root has $2x+1$ inside — will an extra factor of $2$ appear?

Step 1. $f(x+h)=\sqrt{2(x+h)+1}=\sqrt{2x+2h+1}$.

Step 2. Numerator $=\sqrt{2x+2h+1}-\sqrt{2x+1}$ — stuck, no $h$ to factor.

Step 3 — conjugate move. Multiply top and bottom by $\sqrt{2x+2h+1}+\sqrt{2x+1}$: $\frac{\sqrt{2x+2h+1}-\sqrt{2x+1}}{h}\cdot \frac{\sqrt{2x+2h+1}+\sqrt{2x+1}}{\sqrt{2x+2h+1}+\sqrt{2x+1}}=\frac{(2x+2h+1)-(2x+1)}{h(\sqrt{2x+2h+1}+\sqrt{2x+1})}$ Why this step? Using $(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})=a-b$ manufactures a difference in the numerator. The two roots cancel down to a plain subtraction that contains an $h$. $=\frac{2h}{h(\sqrt{2x+2h+1}+\sqrt{2x+1})}=\frac{2}{\sqrt{2x+2h+1}+\sqrt{2x+1}}$ The $h$ cancels — trap escaped.

Step 4. $f^{\prime }(x)={\lim }_{h\to 0}\frac{2}{\sqrt{2x+2h+1}+\sqrt{2x+1}}=\frac{2}{2\sqrt{2x+1}}=\frac{1}{\sqrt{2x+1}}$ Answer: $f^{\prime }(x)=\frac{1}{\sqrt{2x+1}}$. The extra $2$ from inside did appear, then halved back.

Verify: At $x=4$: $2x+1=9$, so $f^{\prime }(4)=1/\sqrt{9}=\frac{1}{3}$. Slope is positive (the root climbs), and gently ($\frac{1}{3}$) since roots flatten out. ✓

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Ex 5 — Cell E: a degenerate case, $f(x)=7$ (constant)

Forecast: A horizontal line never changes height. What must its slope be everywhere?

Step 1. $f(x+h)=7$ — substituting $x+h$ changes nothing, because the formula never uses $x$. Why this step? This is the whole subtlety: there is no $x$ to replace, so the buddy point has the same output.

Step 2. $f(x+h)-f(x)=7-7=0$. Why this step? The rise is exactly zero — the two points sit at the same height.

Step 3. $\frac{0}{h}=0$ for every $h\ne 0$. Why this step? Do not call this $\frac{0}{0}$: the numerator is a genuine $0$ while $h$ is nonzero, so the quotient is honestly $0$, not indeterminate. This is the key distinction from Indeterminate forms 0 over 0.

Step 4. $f^{\prime }(x)=\underset{h\to 0}{\lim }0=0$. Answer: $f^{\prime }(x)=0$.

Verify: A flat line has zero steepness at every point. $f^{\prime }(1)=0$, $f^{\prime }(100)=0$. ✓

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Ex 6 — Cell F: a corner where the derivative FAILS, $f(x)=|x|$

Forecast: The graph of $|x|$ is a sharp "V" with its point at the origin. If you tried to lay a single tangent line on that spike, could one line fit both sides?

We must compute the difference quotient at $x=0$: $\frac{f(0+h)-f(0)}{h}=\frac{|h|-|0|}{h}=\frac{|h|}{h}$

Step 1 — approach from the right ($h\to 0^{+}$, so $h>0$). When $h>0$, $|h|=h$, so $\frac{|h|}{h}=\frac{h}{h}=1$. Why this step? From the right the V rises with slope $+1$; a one-sided limit (see Limits — formal definition and one-sided limits) captures exactly that. So the right-hand slope is $+1$.

Step 2 — approach from the left ($h\to 0^{-}$, so $h<0$). When $h<0$, $|h|=-h$, so $\frac{|h|}{h}=\frac{-h}{h}=-1$. Why this step? From the left the V falls with slope $-1$. So the left-hand slope is $-1$.

Step 3 — compare. ${\lim }_{h\to 0^{+}}\frac{|h|}{h}=+1\ne {\lim }_{h\to 0^{-}}\frac{|h|}{h}=-1$ Why this step? A two-sided limit exists only if both one-sided limits agree. Here they disagree.

Step 4 — conclude. The limit does not exist, so $f$ is not differentiable at $0$. Answer: $f^{\prime }(0)$ does not exist (a corner).

Verify: Right slope $=1$, left slope $=-1$, and $1\ne -1$. ✓

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Ex 7 — Cell G: a real-world word problem (instantaneous speed)

Forecast: Between $t=0$ and $t=3$ the ball fell $45$ m, an average speed of $15$ m/s. Since it keeps accelerating, do you expect the speed at $t=3$ to be more or less than $15$?

Step 1 — buddy point. $s(t+h)=5(t+h{)}^2=5(t^2+2th+h^2)=5t^2+10th+5h^2$. Why this step? This is the average-vs-instantaneous idea: we compare position now with position a tiny $h$ seconds later.

Step 2 — change in position (the rise). $s(t+h)-s(t)=(5t^2+10th+5h^2)-5t^2=10th+5h^2$ Why this step? This is the distance fallen during the tiny interval $h$.

Step 3 — divide by $h$ (the run = elapsed time) and cancel. $\frac{10th+5h^2}{h}=\frac{h(10t+5h)}{h}=10t+5h$ Why this step? $\frac{\text{distance}}{\text{time}}$ is a speed; the units are m/s. Factoring $h$ dodges the $\frac{0}{0}$ that a single instant would give.

Step 4 — shrink the interval. $s^{\prime }(t)={\lim }_{h\to 0}(10t+5h)=10t\text{m/s}$ At $t=3$: $s^{\prime }(3)=10(3)=30$ m/s. Answer: the instantaneous speed at $t=3$ s is $30$ m/s.

Verify (units + sanity): $s^{\prime }(t)=10t$ has units (m)/(s), a speed. At $t=3$ the instantaneous $30$ m/s exceeds the average $15$ m/s over $[0,3]$ — correct, since the ball keeps speeding up. ✓

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Ex 8 — Cell H: exam twist, find the tangent line to $f(x)=x^2-4x$ at $x=1$

Forecast: The tangent touches the curve at $(1,f(1))$. First guess: is the curve rising or falling at $x=1$? (Its vertex is at $x=2$, so $x=1$ is to the left of the bottom.)

Step 1. $f(x+h)=(x+h{)}^2-4(x+h)=x^2+2xh+h^2-4x-4h$. Why this step? Replace every $x$ by the block $(x+h)$ — both in $x^2$ and in $-4x$.

Step 2. $f(x+h)-f(x)=(x^2+2xh+h^2-4x-4h)-(x^2-4x)=2xh+h^2-4h$ Why this step? The $x^2$ and $-4x$ terms cancel, leaving only $h$-carrying terms.

Step 3. $\frac{2xh+h^2-4h}{h}=\frac{h(2x+h-4)}{h}=2x+h-4$ Why this step? Factor and cancel $h$ to escape $\frac{0}{0}$.

Step 4. $f^{\prime }(x)={\lim }_{h\to 0}(2x+h-4)=2x-4$ So $f^{\prime }(1)=2(1)-4=-2$ — the tangent slope at $x=1$.

Step 5 — build the tangent line. The touch point is $(1,f(1))=(1,1-4)=(1,-3)$. Using point-slope $y-y_0=m(x-x_0)$: $y-(-3)=-2(x-1)\Rightarrow y=-2x-1$ Why this step? A line is fixed by one point and one slope; the derivative supplies the slope, the function value supplies the point. Answer: tangent line $y=-2x-1$.

Verify: Slope $f^{\prime }(1)=-2$ is negative — the curve is falling at $x=1$ (left of its vertex at $x=2$), matching the forecast. The line passes through $(1,-3)$: $-2(1)-1=-3$ ✓. And $f^{\prime }(2)=0$ at the vertex, as expected. ✓

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Recall Which tool for which cell?

Higher power ::: expand the binomial, cancel $x^n$ Fraction ::: common denominator to surface the hidden $h$ Root ::: multiply by the conjugate to manufacture an $h$ Constant ::: numerator is genuinely $0$, quotient is $0$ (not $0/0$) Corner (like $|x|$) ::: check left and right one-sided limits — if they differ, no derivative Word problem ::: same recipe; then read units and interpret

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Connections

• Derivative from first principles — difference quotient definition — the recipe these examples drill.
• Limits — formal definition and one-sided limits — powers the corner case (Ex 6) and every "shrink" step.
• Indeterminate forms 0 over 0 — why constants (Ex 5) are not indeterminate while true derivatives are.
• Secant and Tangent lines — the tangent-line construction in Ex 8.
• Power Rule — cross-checks Ex 1 and Ex 3.
• Continuity and Differentiability — the moral of Ex 6.
• Average vs Instantaneous Rate of Change — the heart of the word problem Ex 7.