Worked examples · Sheet 1

Derivatives from the definition and tangent lines

Theory used: Derivative at a point Tangent line

Exercise 1

Compute from the definition the derivative of $f(x) = \dfrac{1}{x}$ at the point $x_0 = 2$ and write the tangent line to the graph at that point.

\begin{aligned} \frac{f(2 + h) - f(2)}{h} &= \frac{1}{h} \left( \frac{1}{2 + h} - \frac{1}{2} \right) \\[6pt] &= \frac{1}{h} \cdot \frac{2 - (2 + h)}{2\,(2 + h)} \\[6pt] &= -\frac{1}{2\,(2 + h)} \end{aligned}

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

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

y = \dfrac{1}{x}

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

The tangent to the graph of

\frac{1}{x}

at the point

(2, \frac{1}{2})

has slope

- \frac{1}{4}

The difference quotient compares the value of $f$ at $2 + h$ with the value at $2$ , divided by $h$ . Bringing the difference of the two fractions to a common denominator, the numerator that remains is $-h$ , which cancels with the $\dfrac{1}{h}$ in front. Before the simplification the limit was of the form $\dfrac{0}{0}$ ; after it, the limit is computed by substituting $h = 0$ and gives $-\dfrac{1}{4}$ . The tangent line passes through the point $\left( 2, \dfrac{1}{2} \right)$ with slope $-\dfrac{1}{4}$ , and is $y = -\dfrac{x}{4} + 1$ . The slope is negative, consistent with the fact that $\dfrac{1}{x}$ decreases for $x > 0$ .

Exercise 2

Compute from the definition the derivative of $f(x) = \sqrt{x}$ at the point $x_0 = 9$ and write the tangent line. Use it to approximate $\sqrt{9.6}$ .

\begin{aligned} \frac{\sqrt{9 + h} - 3}{h} &= \frac{(\sqrt{9 + h} - 3)(\sqrt{9 + h} + 3)}{h\,(\sqrt{9 + h} + 3)} \\[6pt] &= \frac{9 + h - 9}{h\,(\sqrt{9 + h} + 3)} \\[6pt] &= \frac{1}{\sqrt{9 + h} + 3} \end{aligned}

f'(9) = \lim_{h \to 0} \frac{1}{\sqrt{9 + h} + 3} = \frac{1}{6}

y = 3 + \frac{1}{6}\,(x - 9), \qquad \sqrt{9.6} \approx 3 + \frac{0.6}{6} = 3.1

The difference quotient contains a difference between a root and a number, and as $h \to 0$ it is of the form $\dfrac{0}{0}$ . It is rationalized by multiplying numerator and denominator by $\sqrt{9 + h} + 3$ ; in this way the numerator becomes $h$ and cancels. The limit equals $\dfrac{1}{6}$ , in agreement with the formula $\left( \sqrt{x} \right)' = \dfrac{1}{2\sqrt{x}}$ . The tangent line at $(9, 3)$ has slope $\dfrac{1}{6}$ . Near $9$ the graph and the tangent almost overlap, so the value of the tangent at $9.6$ , that is, $3.1$ , approximates $\sqrt{9.6} = 3.098\ldots$