Worked examples · Sheet 1

Derivatives using the definition

Theory used: Derivative and difference quotient One-sided derivatives

Exercise 1

Compute, using the definition, the derivative of $f (x) = x^{2} - 3 x$ at a generic point $x \in R$ .

\frac{f ( x + h ) - f ( x )}{h} = \frac{( x + h ) ^{2} - 3 ( x + h ) - x ^{2} + 3 x}{h} = \frac{2 x h + h ^{2} - 3 h}{h} = 2 x + h - 3

f^{'} (x) = h \to 0 lim (2 x + h - 3) = 2 x - 3

Write the difference quotient and expand the square $(x + h)^{2} = x^{2} + 2 x h + h^{2}$ . The terms $x^{2}$ and $- 3 x$ cancel with those subtracted, therefore only the terms containing $h$ remain in the numerator. After cancelling $h$ the quotient equals $2 x + h - 3$ , a continuous function of $h$ , hence the limit as $h \to 0$ is computed by substitution and equals $2 x - 3$ . The limit exists and is finite for every $x$ , therefore $f$ is differentiable on all of $R$ with $f^{'} (x) = 2 x - 3$ .

Exercise 2

Compute, using the definition, the derivative of $f (x) = \frac{1}{x}$ at the points $x \neq = 0$ .

\frac{f ( x + h ) - f ( x )}{h} = \frac{1}{h} (\frac{1}{x + h} - \frac{1}{x}) = \frac{1}{h} \cdot \frac{x - ( x + h )}{x ( x + h )} = \frac{- h}{h x ( x + h )} = - \frac{1}{x ( x + h )}

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

The difference of the two fractions is brought to the common denominator $x (x + h)$ , and the numerator becomes $x - (x + h) = - h$ . The factor $h$ cancels with the denominator of the difference quotient, therefore for every $h \neq = 0$ small enough the quotient equals $- \frac{1}{x ( x + h )}$ . As $h \to 0$ the denominator tends to $x^{2}$ , hence the derivative is $- \frac{1}{x ^{2}}$ at every point $x \neq = 0$ , in agreement with the power rule applied to $x^{- 1}$ .

Exercise 3

Compute, using the definition, the derivative of $f (x) = x$ at the points $x > 0$ and study differentiability at $x = 0$ .

\frac{x + h - x}{h} = \frac{( x + h - x ) ( x + h + x )}{h ( x + h + x )} = \frac{h}{h ( x + h + x )} = \frac{1}{x + h + x}

f^{'} (x) = h \to 0 lim \frac{1}{x + h + x} = \frac{1}{2 x} (x > 0)

At $x = 0$ :

h \to 0^{+} lim \frac{h - 0}{h} = h \to 0^{+} lim \frac{1}{h} = + \infty,

therefore $f$ is not differentiable at $0$ .

As $h \to 0$ the difference quotient is an indeterminate form $\frac{0}{0}$ and one rationalizes by multiplying numerator and denominator by $x + h + x$ : in the numerator $(x + h) - x = h$ appears, which cancels with the factor $h$ in the denominator. The remaining denominator tends to $2 x$ , hence $f^{'} (x) = \frac{1}{2 x}$ at every point $x > 0$ .

At $x = 0$ the domain extends only to the right, therefore one studies the right derivative. The difference quotient equals $\frac{1}{h}$ and tends to $+ \infty$ , therefore the right derivative does not exist as a finite number and $f$ is not differentiable at $0$ ; the graph leaves the origin with a vertical tangent.