Question

19-20 Use Definition 4 to find \(f'\left( a \right)\) at the given number \(a\).

20. \(f\left( x \right) = 5{x^4},\,\,x = - 1\)

Short answer

The required answer is \(f'\left( { - 1} \right) = - 20\).

Step-by-step solution

Derivative of a function using limit

For a given function \(f\) we can find the derivative of the function by using limit. Which is given by \(\mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = f'\left( x \right)\).

Finding the derivative of a function at a given point

The function is given by \(f\left( x \right) = 5{x^4}\).

Now the derivative of the function at the point \(x = - 1\) will be:

\(\begin{aligned}f'\left( { - 1} \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{f\left( { - 1 + h} \right) - f\left( { - 1} \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{5{{\left( { - 1 + h} \right)}^4} - 5}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{5{{\left( {1 + {h^2} - 2h} \right)}^2} - 5}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{5\left( {1 + {h^4} + 4{h^2} + 2{h^2} - 4h - 4{h^3}} \right) - 5}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{5\left( {{h^4} + 4{h^2} + 2{h^2} - 4h - 4{h^3}} \right)}}{h}\end{aligned}\)

Further solving we get,

\(\begin{aligned}f'\left( { - 1} \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{5h\left( {{h^3} + 4h + 2h - 4 - 4{h^2}} \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} 5\left( {{h^3} + 4h + 2h - 4 - 4{h^2}} \right)\\ &= 5 \times \left( { - 4} \right)\\ &= - 20\end{aligned}\)

Hence, the required value is \(f'\left( { - 1} \right) = - 20\).