Question

Find \(f'\left( a \right)\).

\(f\left( t \right) = {t^3} - 3t\)

Short answer

The value of \(f'\left( a \right)\) is \(3{a^2} - 3\).

Step-by-step solution

Use the definition 4

The derivative of a function at a number \(a\), denoted by \(f'\left( a \right)\) can be obtained using the formula given below:

\(f'\left( a \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a + h} \right) - f\left( a \right)}}{h}\)

The derivative of the given function

Substitute the values in the above formula and solve it as follows:

\(\begin{aligned}f'\left( a \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a + h} \right) - f\left( a \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {a + h} \right)}^3} - 3\left( {a + h} \right) - \left( {{a^3} - 3a} \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{{a^3} + {h^3} + 3{a^2}h + 3a{h^2} - 3a - 3h - {a^3} + 3a}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{{h^3} + 3{a^2}h + 3a{h^2} - 3h}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} {h^2} + 3{a^2} + 3ah - 3\\ &= 0 + 3{a^2} + 0 - 3\\ &= 3{a^2} - 3\end{aligned}\)

Thus, the value of \(f'\left( a \right)\) is \(3{a^2} - 3\).