Single-Variable-Calculus · Unit 1 · Video 3 · Interactive Practice

Differentiating 1/x Straight From the Definition

IKey Formulas

Formula	Name	What you need
$f'(x_0) = \lim\limits_{\Delta x \to 0} \dfrac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}$	Definition of the derivative	A function and a base point
$\dfrac{\Delta f}{\Delta x} = \dfrac{\frac{1}{x_0 + \Delta x} - \frac{1}{x_0}}{\Delta x} = \dfrac{-1}{(x_0 + \Delta x)\,x_0}$	Difference quotient for $1/x$ , after cancellation	Common denominator, then cancel $\Delta x$
$f'(x_0) = -\dfrac{1}{x_0^2}$	Derivative of $1/x$	The limit $\Delta x \to 0$

Key Insight: Substituting $\Delta x = 0$ at the start always yields $\tfrac{0}{0}$ — the definition is built so numerator and denominator vanish together. The algebra must cancel the $\Delta x$ first; only then is the limit plain substitution.

IIVisualization 1 — The Secant Becomes the Tangent

As $\Delta x$ shrinks, the secant through $P$ and $Q$ turns into the tangent at $P$ .

Δx 2.50

IIIVisualization 2 — Cancel First, Then Substitute

Lines 1–4 are the same number, yet only the last of them survives $\Delta x = 0$ .

Line 1 Δx 0.50

1 · difference quotient

\frac{\Delta f}{\Delta x} = \frac{\dfrac{1}{x_0 + \Delta x} - \dfrac{1}{x_0}}{\Delta x}

2 · common denominator

= \frac{1}{\Delta x}\left[\frac{x_0 - (x_0 + \Delta x)}{(x_0 + \Delta x)\,x_0}\right]

3 · x-terms cancel

= \frac{1}{\Delta x}\cdot\frac{-\Delta x}{(x_0 + \Delta x)\,x_0}

4 · increments cancel

= \frac{-1}{(x_0 + \Delta x)\,x_0}

5 · pass to the limit

f'(x_0) = \lim_{\Delta x \to 0} \frac{-1}{(x_0 + \Delta x)\,x_0} = -\frac{1}{x_0^2}

IVVisualization 3 — Reading the Slope Off the Hyperbola

One unit to the right, the tangent falls by exactly $1/x_0^2$ — always down, never zero.

x₀ 3.0

💡 The negative branch obeys the same formula: $x_0^2$ is positive for every nonzero $x_0$ , so the slope $-1/x_0^2$ is negative on both branches of the hyperbola.

VQuiz Questions

Problem 1 · Apply the Formula

Given: $f(x) = \dfrac{1}{x}$ — find $f'(4)$ .

✅ Correct!

f'(x_0) = -1/x_0^2

, so at

x_0 = 4

the slope is

-1/16 = -0.0625

: shallow, because the hyperbola has already flattened out by

x = 4

❌ Not quite. That is

-1/x_0

, the negative of the function value. The derivative squares the base point:

-1/x_0^2

❌ Right size, wrong sign. The magnitude

1/16

is correct, but

-1/x_0^2

is negative for every

x_0

— the branch falls as

x

increases.

❌ Not quite. Substitute

x_0 = 4

into

f'(x_0) = -\dfrac{1}{x_0^2}

, not into

x_0^2

alone.

Show solution

The derivative of $1/x$ computed from the definition is

f'(x_0) = -\frac{1}{x_0^2}.

At $x_0 = 4$ :

f'(4) = -\frac{1}{4^2} = -\frac{1}{16} = -0.0625.

Two sanity checks: the sign is negative (the curve falls), and the magnitude is small (far from the axis the curve is nearly flat).

Problem 2 · When Is Substitution Legal?

Given: the four equal expressions produced while differentiating $f(x) = \dfrac{1}{x}$ — find the first one in which setting $\Delta x = 0$ produces a finite number instead of an indeterminate form.

✅ Correct! Only after the

\Delta x

out front cancels against the

-\Delta x

upstairs does nothing vanish:

\dfrac{-1}{(x_0 + 0)x_0} = -\dfrac{1}{x_0^2}

❌ Not quite. Setting

\Delta x = 0

there gives

\dfrac{1/x_0 - 1/x_0}{0} = \dfrac{0}{0}

— the trap the whole derivation exists to avoid.

❌ Close — one cancellation short. The numerator has collapsed to

-\Delta x

, but the factor

\dfrac{1}{\Delta x}

is still out front: at

\Delta x = 0

this reads

\dfrac{1}{0}\cdot 0

, still indeterminate.

❌ Not quite. Any expression that still contains a

\dfrac{1}{\Delta x}

factor blows up at

\Delta x = 0

. Look for the line where that factor is gone.

Show solution

Track what $\Delta x = 0$ does to each line.

(a) $\dfrac{1/x_0 - 1/x_0}{0} = \dfrac{0}{0}$ — indeterminate.

(b) $\dfrac{1}{0}\left[\dfrac{x_0 - x_0}{x_0 \cdot x_0}\right] = \dfrac{1}{0}\cdot 0$ — indeterminate.

(c) $\dfrac{1}{0}\cdot\dfrac{-0}{x_0 \cdot x_0} = \dfrac{1}{0}\cdot 0$ — still indeterminate; the $\Delta x$ 's have not been cancelled against each other yet.

(d) $\dfrac{-1}{(x_0 + 0)\,x_0} = -\dfrac{1}{x_0^2}$ — finite. Nothing vanishes and nothing blows up.

The template: form the difference quotient, do algebra until the $\Delta x$ cancels, then take the limit as plain substitution.

Problem 3 · Tangent Line to the Hyperbola

Given: the curve $y = \dfrac{1}{x}$ at the point where $x_0 = 2$ — find the slope of the tangent line there and its $y$ -intercept in $y = mx + b$ .

What is the slope?

What is the y-intercept?

✅ Correct! The tangent is

y = -\tfrac{1}{4}x + 1

, and it meets the curve at

\left(2, \tfrac{1}{2}\right)

❌ Check the slope. The slope of the tangent is

f'(2) = -1/2^2

, not the height

f(2) = 1/2

and not

-1/x_0

❌ Right size, wrong sign. The magnitude

\tfrac{1}{4}

is correct, but

f'(x_0) = -1/x_0^2

is negative for every

x_0

— the branch falls as

x

increases.

❌ That is $-x_0$ . You negated the base point instead of using it. The derivative squares it and inverts:

f'(2) = -1/2^2 = -\tfrac{1}{4}

❌ Check the intercept. Start from

y - \tfrac{1}{2} = -\tfrac{1}{4}(x - 2)

and distribute before collecting constants.

❌ That tangent would pass through the origin.

y = -\tfrac{1}{4}x

gives height

-\tfrac{1}{2}

x = 2

, but the tangent must touch the curve at height

+\tfrac{1}{2}

❌ Sign slip collecting the constants. Distributing gives

y = -\tfrac{1}{4}x + \tfrac{1}{2} + \tfrac{1}{2}

; the two halves add to

+1

, not

-1

Show solution

Step 1 — the point. $f(2) = \dfrac{1}{2}$ , so the tangent touches at $\left(2, \tfrac{1}{2}\right)$ .

Step 2 — the slope.

m = f'(2) = -\frac{1}{2^2} = -\frac{1}{4}

Step 3 — point-slope, then solve for $y$ .

y - \frac{1}{2} = -\frac{1}{4}(x - 2)

y - \frac{1}{2} = -\frac{1}{4}x + \frac{1}{2}

y = -\frac{1}{4}x + 1

So $m = -\tfrac{1}{4}$ and $b = 1$ .

Verify: at $x = 2$ : $y = -\tfrac{1}{4}(2) + 1 = -\tfrac{1}{2} + 1 = \tfrac{1}{2}$ ✅ — the tangent passes through the point of tangency.

Problem 4 · Run the Formula Backwards

Given: $f(x) = \dfrac{1}{x}$ and a point with $x_0 > 0$ where the tangent has slope $-4$ — find $x_0$ .

✅ Correct!

-1/x_0^2 = -4

forces

x_0^2 = \tfrac{1}{4}

, so

x_0 = \tfrac{1}{2}

— close to the axis, exactly where the formula predicts a steep plunge.

❌ Not quite. That solves

1/x_0 = 4

. The derivative is

-1/x_0^2

, so the equation to solve is

x_0^2 = \tfrac{1}{4}

❌ Reciprocal flipped. From

1/x_0^2 = 4

you get

x_0^2 = \tfrac{1}{4}

, not

x_0^2 = 4

. A slope of magnitude

4

needs a point inside

x_0 = 1

, not outside it.

❌ Not quite. Set

-\dfrac{1}{x_0^2} = -4

, clear the fraction, and take the positive square root.

Show solution

Set the derivative equal to the required slope:

-\frac{1}{x_0^2} = -4 \quad\Longrightarrow\quad \frac{1}{x_0^2} = 4 \quad\Longrightarrow\quad x_0^2 = \frac{1}{4}

With $x_0 > 0$ this gives

x_0 = \frac{1}{2}.

Check: $f'\!\left(\tfrac{1}{2}\right) = -\dfrac{1}{(1/2)^2} = -\dfrac{1}{1/4} = -4$ ✅

Since $1/x_0^2$ grows without bound as $x_0 \to 0$ , every steep negative slope is achieved once on this branch — and every shallow one is achieved far out to the right.

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