Q. 42 - Calculus

Question

Use the definition of the derivative to find $f\text{'}$ for each function f in Exercises 39-54

$f\left(x\right)=-x^3-x+1$

Step-by-Step Solution

Answer

The value of $f\text{'}\left(x\right)=-3x^2+1$

Step 1. Given information

The given function $f\left(x\right)=-x^3-x+1$

Step 2. Finding the value of f ' ( x )

We know that $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h} ......... \left(1\right)$

Given $f\left(x\right)=-x^3-x+1$ then

$f(x+h)=-{(x+h)}^3-(x+h)+1$

Putting the values in (1)

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

Putting $h=0$

        $$\begin{gathered} =-{\left(0\right)}^2-3x^2-3x\left(0\right)+1 \\ =-3x^2+1 \end{gathered}$$

Hence, $f\text{'}\left(x\right)=-3x^2+1$