Question

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

\(F\left( v \right) = \frac{v}{{v + {\bf{2}}}}\)

Short answer

The derivative is \(F'\left( v \right) = \frac{2}{{{{\left( {v + 2} \right)}^2}}}\).

The domain of \(F\left( v \right)\) and \(F'\left( v \right)\) is \(\left( { - \infty , - 2} \right) \cup \left( { - 2,\infty } \right)\).

Step-by-step solution

Find the derivative of the function by using the definition

The derivative of the function \(F\left( v \right)\) can be calculated as:

\(\begin{aligned}F'\left( v \right) & = \mathop {\lim }\limits_{h \to 0} \frac{{F\left( {v + h} \right) - F\left( v \right)}}{h}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {\frac{{v + h}}{{\left( {v + h} \right) + 2}} - \frac{v}{{v + 2}}} \right)}}{h}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {v + 2} \right)\left( {v + h} \right) - v\left( {\left( {v + h} \right) + 2} \right)}}{{h\left( {\left( {v + h} \right) + 2} \right)\left( {v + 2} \right)}}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{{v^2} + 2v + vh + 2h - {v^2} - vh - 2v}}{{h\left( {\left( {v + h} \right) + 2} \right)\left( {v + 2} \right)}}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{2h}}{{h\left( {\left( {v + h} \right) + 2} \right)\left( {v + 2} \right)}}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{2}{{\left( {\left( {v + h} \right) + 2} \right)\left( {v + 2} \right)}}\\ & = \frac{2}{{{{\left( {v + 2} \right)}^2}}}\end{aligned}\)

Thus, the derivative is \(F'\left( v \right) = \frac{2}{{{{\left( {v + 2} \right)}^2}}}\).

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

The functions \(F\left( v \right) = \frac{v}{{v + 2}}\) and \(F'\left( v \right) = \frac{2}{{{{\left( {v + 2} \right)}^2}}}\) is defined, when;

\(\begin{aligned}v + 2 \ne 0\\v \ne - 2\end{aligned}\)

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