Question

Use (a) the $h\to 0$definition of the derivative and then

(b) the $z\to c$definition of the derivative to find $f\text{'}\left(c\right)$ for each function f and value $x=c$ in Exercises 23–38.

24.$f\left(x\right)=x^3,x=1$

Short answer

$f\text{'}\left(1\right)=3$

Step-by-step solution

Part (a) Step 1. Given information.

A function is given as$f\left(x\right)=x^3$and$x=c=1$.

Part (a) Step 2. Find f'(1) using h→0 definition of the derivative.

We have

$$\begin{gathered} f\text{'}\left(c\right)=\underset{h\to 0}{\lim }\frac{f(c+h)-f\left(c\right)}{h} \\ f\text{'}\left(1\right)=\underset{h\to 0}{\lim }\frac{f(1+h)-f\left(1\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{{(1+h)}^3-1^3}{h} \\ =\underset{h\to 0}{\lim }\frac{\left[\right(1+h)-1)\left]\right[{(1+h)}^2+(1+h)1+1\left)\right]}{h} \\ =\underset{h\to 0}{\lim }\frac{h(1+h^2+2h+2+h)}{h} \\ =\underset{h\to 0}{\lim }(h^2+3h+3) \\ =0^2+3\left(0\right)+3 \\ =3 \end{gathered}$$

Part (b) Step 2. Find f'(1) using z→1 definition of the derivative.

We have

$$\begin{gathered} f\text{'}\left(c\right)=\underset{z\to c}{\lim }\frac{f\left(z\right)-f\left(c\right)}{z-c} \\ f\text{'}\left(1\right)=\underset{z\to 1}{\lim }\frac{f\left(z\right)-f\left(1\right)}{z-1} \\ =\underset{z\to 1}{\lim }\frac{{\left(z\right)}^3-1^3}{z-1} \\ =\underset{z\to 1}{\lim }\frac{(z-1)(z^2+z+1)}{z-1} \\ =\underset{z\to 1}{\lim }(z^2+z+1) \\ =1^2+1+1 \\ =1+1+1 \\ =3 \end{gathered}$$