Question

21-32 Find the derivative of the function using the definition of the derivative. State the domain of the function and the domain of its derivative.

27. \(f\left( x \right) = \frac{{\bf{1}}}{{{x^{\bf{2}}} - {\bf{4}}}}\)

Short answer

The derivative is \(f'\left( x \right) = - \frac{{2x}}{{{{\left( {{x^2} - 4} \right)}^2}}}\). The domain of \(f\left( x \right)\) and \(f'\left( x \right)\) is \(\left( { - \infty , - 2} \right) \cup \left( { - 2,2} \right) \cup \left( {2,\infty } \right)\).

Step-by-step solution

Find the derivative of the function by using the definition

The derivative of the function \(f\left( x \right)\) can be calculated as:

\(\begin{aligned}f'\left( x \right) & = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{1}{{{{\left( {x + h} \right)}^2} - 4}} - \frac{1}{{{x^2} - 4}}}}{h}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {{x^2} - 4} \right) - \left( {{{\left( {x + h} \right)}^2} - 4} \right)}}{{h\left( {{{\left( {x + h} \right)}^2} - 4} \right)\left( {{x^2} - 4} \right)}}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {{x^2} - 4} \right) - \left( {{x^2} + {h^2} + 2xh - 4} \right)}}{{h\left( {{{\left( {x + h} \right)}^2} - 4} \right)\left( {{x^2} - 4} \right)}}\end{aligned}\)

Simplify further,

\(\begin{aligned}f'\left( x \right) & = \mathop {\lim }\limits_{h \to 0} \frac{{ - {h^2} - 2xh}}{{h\left( {{{\left( {x + h} \right)}^2} - 4} \right)\left( {{x^2} - 4} \right)}}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{ - \left( {h + 2x} \right)}}{{h\left( {{{\left( {x + h} \right)}^2} - 4} \right)\left( {{x^2} - 4} \right)}}\\ & = \frac{{ - 2x}}{{{{\left( {{x^2} - 4} \right)}^2}}}\end{aligned}\)

Thus, the derivative is \(f'\left( x \right) = - \frac{{2x}}{{{{\left( {{x^2} - 4} \right)}^2}}}\).

Find the domain of \(f\left( x \right)\) and its derivative

The functions \(f\left( x \right) = \frac{1}{{{x^2} - 4}}\) and \(f'\left( x \right) = - \frac{{2x}}{{{{\left( {{x^2} - 4} \right)}^2}}}\) are defined, when;

\(\begin{aligned}{x^2} - 4 \ne 0\\x \ne \pm 2\end{aligned}\)

Thus, the domain of the function and its derivative is \(\left( { - \infty , - 2} \right) \cup \left( { - 2,2} \right) \cup \left( {2,\infty } \right)\).