Math · Calculus I · Worked example

Find a derivative from the limit definition

Use the limit definition to find the slope of $f (x) = x^{2}$ at $x = 2$ , then write the tangent line there.

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

Write the difference quotient at $a = 2$

Replace $f (2 + h)$ by ${(2 + h)}^{2}$ and $f (2)$ by $2^{2} = 4$ . This quotient is the slope of the secant between $x = 2$ and $x = 2 + h$ .

\frac{f (2 + h) - f (2)}{h} = \frac{{(2 + h)}^{2} - 4}{h}

Expand the numerator

${(2 + h)}^{2} = 4 + 4 h + h^{2}$ . Subtracting $4$ cancels the constant term, as it does for every polynomial.

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

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

Factor $h$ out of the numerator and cancel it. The cancellation is allowed because the quotient is only ever used at nonzero $h$ . The simplified form $4 + h$ is the slope of every secant near $x = 2$ .

\frac{h (4 + h)}{h} = 4 + h (h \neq 0)

Take the limit

As $h$ approaches $0$ from either side, $4 + h$ approaches $4$ . The tangent slope at $x = 2$ is $f^{'} (2) = 4$ .

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

Compare with nearby secants

The secant slopes $4 + h$ sit just above $4$ for small positive $h$ and just below it for small negative $h$ . The table supports the limit, and the algebra is what establishes it.

Secant slopes near $x = 2$
$h$	Secant slope $4 + h$
$0.1$	$4.1$
$0.01$	$4.01$
$- 0.01$	$3.99$
$- 0.1$	$3.9$

Write the tangent line

The tangent passes through $(2, f (2)) = (2, 4)$ with slope $4$ . Point-slope form, $y = f (a) + f^{'} (a) (x - a)$ , gives the line. Expanding writes it as $y = 4 x - 4$ .

\begin{array}{l} y = 4 + 4 (x - 2) \\ y = 4 x - 4 \end{array}

Result

$f^{'} (2) = 4$ , and the tangent line at $x = 2$ is $y = 4 x - 4$ .

f^{'} (2) = 4, y = 4 x - 4

Your turn

1. One like the example

Use the limit definition to find the tangent line to $f (x) = x^{2}$ at $x = - 1$ .

Show the answer and explanation

$y = - 2 x - 1$ .

The quotient $({(- 1 + h)}^{2} - 1) / h$ $= (- 2 h + h^{2}) / h$ $= - 2 + h$ for $h \neq 0$ , which approaches $- 2$ , so $f^{'} (- 1) = - 2$ . The point is $(- 1, 1)$ , so the tangent is $y = 1 - 2 (x + 1)$ $= - 2 x - 1$ .

\begin{array}{l} y = 1 - 2 (x + 1) \\ y = - 2 x - 1 \end{array}

2. A variation

Use the limit definition to find $f^{'} (1)$ for $f (x) = 3 x^{2} - x$ , then write the tangent line at $x = 1$ . The function now has two terms.

Show the answer and explanation

$f^{'} (1) = 5$ , and the tangent line is $y = 5 x - 3$ .

$f (1) = 3 - 1 = 2$ , and $f (1 + h)$ $= 3 {(1 + h)}^{2} - (1 + h)$ $= 2 + 5 h + 3 h^{2}$ . The quotient is $(5 h + 3 h^{2}) / h = 5 + 3 h$ for $h \neq 0$ , which approaches $5$ . The tangent is $y = 2 + 5 (x - 1)$ $= 5 x - 3$ .

\begin{array}{l} \frac{3 {(1 + h)}^{2} - (1 + h) - 2}{h} \\ \frac{5 h + 3 h^{2}}{h} \\ 5 + 3 h \end{array}

3. Somewhere new

The stone falls $s (t) = 5 t^{2}$ meters in $t$ seconds. Use the limit definition to find how fast it is falling at $t = 3$ .

Show the answer and explanation

$s^{'} (3) = 30$ meters per second.

$s (3) = 45$ and $s (3 + h)$ $= 5 (9 + 6 h + h^{2})$ $= 45 + 30 h + 5 h^{2}$ . The quotient is $(30 h + 5 h^{2}) / h = 30 + 5 h$ for $h \neq 0$ , which approaches $30$ . The units are meters per second, the units of $s$ over the units of $t$ .

\begin{array}{l} lim_{h \to 0} \frac{5 {(3 + h)}^{2} - 45}{h} \\ lim_{h \to 0} (30 + 5 h) \\ 30 \end{array}

Keep exploring

Open Secant → Tangent at $x = 2$ and drag $h$ toward zero: the secant slope $4 + h$ settles at $4$ .

Return to the concept →

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After reading

Study it, then try it

Practice the key terms as flashcards, then open the example in its tool, change a value and keep a useful result on your board.

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

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Write the difference quotient at a=2a=2

Expand the numerator

Cancel hh, keeping h≠0h=0

Take the limit

Compare with nearby secants

Write the tangent line

Result

Your turn

1. One like the example

y=−2⁢x−1y=−2x−1.

2. A variation

f′(1)=5f′(1)=5, and the tangent line is y=5⁢x−3y=5x−3.

3. Somewhere new

s′(3)=30s′(3)=30 meters per second.

Keep exploring

Study it, then try it

Write the difference quotient at $a = 2$

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

$y = - 2 x - 1$ .

$f^{'} (1) = 5$ , and the tangent line is $y = 5 x - 3$ .

$s^{'} (3) = 30$ meters per second.