Math · Calculus I · Worked example

Find the derivative of 1/x from the definition

Use the limit definition to find $f^{'} (2)$ for $f (x) = 1 / x$ . What does the sign of the answer say about the graph?

f^{'} (2) = lim_{h \to 0} \frac{\frac{1}{2 + h} - \frac{1}{2}}{h}

Write the difference quotient

$f (2 + h) = 1 / (2 + h)$ and $f (2) = 1 / 2$ . Substituting $h = 0$ gives $0 / 0$ again, but this time there is nothing to expand. The numerator is a difference of fractions.

\frac{\frac{1}{2 + h} - \frac{1}{2}}{h}

Combine the top over a common denominator

The common denominator of $1 / (2 + h)$ and $1 / 2$ is $2 (2 + h)$ . The numerator becomes $2 - (2 + h)$ over $2 (2 + h)$ , and dividing by $h$ puts $h$ in the denominator too.

\frac{2 - (2 + h)}{2 h (2 + h)}

Simplify the numerator

$2 - (2 + h) = - h$ . The minus sign applies to both terms in the bracket, and the 2s cancel.

\frac{- h}{2 h (2 + h)}

Cancel $h$ , keeping $h \neq 0$

$h$ is not $0$ on the way to the limit, so it cancels from the top and the bottom. What is left can be evaluated at $h = 0$ .

\frac{- 1}{2 (2 + h)} (h \neq 0)

Let $h$ approach $0$

The denominator approaches $2 (2) = 4$ , so the quotient approaches $- 1 / 4$ . That is the slope of the tangent at $(2, 1 / 2)$ .

f^{'} (2) = lim_{h \to 0} \frac{- 1}{2 (2 + h)} = - \frac{1}{4}

\begin{aligned} f^{'} (2) & = lim_{h \to 0} \frac{- 1}{2 (2 + h)} \\ = - \frac{1}{4} \end{aligned}

Compare with nearby secants

With $h = 0.1$ the quotient is $- 1 / 4.2 \approx - 0.238$ , and with $h = - 0.1$ it is $- 1 / 3.8 \approx - 0.263$ . The secant slopes close in on $- 0.25$ from both sides.

Secant slopes of $1 / x$ near $x = 2$
$h$	Secant slope $- 1 / (2 (2 + h))$
$0.1$	$- 0.238$
$0.01$	$- 0.2488$
$- 0.01$	$- 0.2513$
$- 0.1$	$- 0.263$

Read the sign

The slope is negative, so the graph of $1 / x$ is falling at $x = 2$ . Near $x = 2$ , raising $x$ by a small amount lowers $1 / x$ by about a quarter of that amount. With $a$ in place of $2$ , the same steps give $f^{'} (a) = - 1 / a^{2}$ , which is negative at every $a \neq 0$ .

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

Result

$f^{'} (2) = - 1 / 4$ . The graph of $1 / x$ is falling at $x = 2$ , and its tangent there is $y = 1 / 2 - (1 / 4) (x - 2)$ $= - x / 4 + 1$ .

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

Your turn

1. One like the example

Use the limit definition to find $f^{'} (3)$ for $f (x) = 1 / x$ .

Show the answer and explanation

$f^{'} (3) = - 1 / 9$ .

The common denominator of $1 / (3 + h)$ and $1 / 3$ is $3 (3 + h)$ , so the numerator is $(3 - (3 + h)) / (3 (3 + h))$ $= - h / (3 (3 + h))$ . Dividing by $h$ and canceling gives $- 1 / (3 (3 + h))$ , which approaches $- 1 / 9$ . That matches $- 1 / a^{2}$ at $a = 3$ .

\begin{array}{l} lim_{h \to 0} \frac{\frac{1}{3 + h} - \frac{1}{3}}{h} \\ lim_{h \to 0} \frac{- 1}{3 (3 + h)} \\ - \frac{1}{9} \end{array}

2. A variation

Use the limit definition to find $f^{'} (4)$ for $f (x) = \sqrt{x}$ . Here the numerator holds a square root.

Show the answer and explanation

$f^{'} (4) = 1 / 4$ .

Multiply the top and bottom of $(\sqrt{4 + h} - 2) / h$ by $\sqrt{4 + h} + 2$ . The top becomes $(4 + h) - 4 = h$ , which cancels the $h$ underneath and leaves $1 / (\sqrt{4 + h} + 2)$ . As $h$ approaches $0$ this approaches $1 / (2 + 2) = 1 / 4$ . In a Math box the conjugate row reads ✓ “Equivalent.”

\begin{array}{l} \frac{\sqrt{4 + h} - 2}{h} \\ \frac{(4 + h) - 4}{h (\sqrt{4 + h} + 2)} \\ \frac{1}{\sqrt{4 + h} + 2} \end{array}

3. Somewhere new

At a fixed temperature, a gas in a cylinder has pressure $P (V) = 600 / V$ kilopascals at volume $V$ liters. How fast is the pressure changing with volume at $V = 3 L$ ?

Show the answer and explanation

$P^{'} (3)$ $= - 200 / 3$ $\approx - 66.7 kPa$ per liter.

$P (3) = 200$ . Write $200$ as $200 (3 + h) / (3 + h)$ , so the numerator $600 / (3 + h) - 200$ is $(600 - 600 - 200 h) / (3 + h)$ $= - 200 h / (3 + h)$ . Dividing by $h$ gives $- 200 / (3 + h)$ , which approaches $- 200 / 3 \approx - 66.7$ . The pressure falls by about $67 kPa$ for each extra liter, near $3 L$ .

\begin{array}{l} lim_{h \to 0} \frac{\frac{600}{3 + h} - 200}{h} \\ lim_{h \to 0} \frac{- 200}{3 + h} \\ - \frac{200}{3} \end{array}

Keep exploring

Open Secant → Tangent on $1 / x$ at $x = 2$ and shrink $h$ : the secant slope settles at $- 0.25$ from both sides.

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Study the 6 key terms Open Secant → Tangent on 1/x Check the quotient in Math Open worked example on a board Derivative definition in Math Reference

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Write the difference quotient

Combine the top over a common denominator

Simplify the numerator

Cancel hh, keeping h≠0h=0

Let hh approach 00

Compare with nearby secants

Read the sign

Result

Your turn

1. One like the example

f′(3)=−1⁢/9f′(3)=−1/9.

2. A variation

f′(4)=1⁢/4f′(4)=1/4.

3. Somewhere new

P′(3)P′(3) =−200⁢/3=−200/3 ≈−66.7 kPa≈−66.7 kPa per liter.

Keep exploring

Study it, then try it

Cancel $h$ , keeping $h \neq 0$

Let $h$ approach $0$

$f^{'} (3) = - 1 / 9$ .

$f^{'} (4) = 1 / 4$ .

$P^{'} (3)$ $= - 200 / 3$ $\approx - 66.7 kPa$ per liter.