Chapter 3.1, Problem 4E

To determine

To calculate: The derivative of the given function at the given value of a using the definition $f^{\prime }\left(a\right)=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$ .

Answer to Problem 4E

The required value is 1.

Explanation of Solution

Given information:

The function: $f\left(x\right)=x^3+x$ , $a=0$

Calculation:

The given function is $f\left(x\right)=x^3+x$ .

Write the given definition using the above function as:

  $f^{\prime }\left(a\right)=\underset{h\to 0}{\lim }\left[\frac{{\left(a+h\right)}^3+\left(a+h\right)}{h}\right]$

Substitute 0 for a in the above limit and simplify.

  $\begin{array}{c}{f}^{\prime }\left(0\right)=\underset{h\to 0}{\lim }\left[\frac{{\left(0+h\right)}^3+\left(0+h\right)}{h}\right]\\ =\underset{h\to 0}{\lim }\left[\frac{h^3+h}{h}\right]\\ =\underset{h\to 0}{\lim }\left[\frac{\cancel{h}\left(h^2+1\right)}{\cancel{h}}\right]\\ =\underset{h\to 0}{\lim }\left(h^2+1\right)\end{array}$

Substitute the 0 for h in the limit and simplify.

  $\begin{array}{c}{f}^{\prime }\left(0\right)=\underset{h\to 0}{\lim }\left(h^2+1\right)\\ ={\left(0\right)}^2+1\\ =1\end{array}$

Hence, the required value is 1.