Question

Find the following for the given function f:

1. $f\text{'}\left(a\right),$where ais in the domain of f, and
2. $f\text{'}\left(3\right)$and$f\text{'}\left(4\right)$.

.$f\left(x\right)=-\frac{1}{x^2}$

Short answer

(a) The value of$f\text{'}\left(a\right)=\frac{2}{a^3}$

(b) The value of $f\text{'}\left(3\right)=\frac{2}{27}andf\text{'}\left(4\right)=\frac{1}{32}$.

Step-by-step solution

Step 1. Given information.

The function here given is,

$f\left(x\right)=-\frac{1}{x^2}$

Step 2. Formula used.

The derivative of a function f at a number,denoted by$f\text{'}\left(a\right),$is

.$f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{f(a+h)-f\left(a\right)}{h}$

if this limit exists.

Step 3. Finding the slope of the tangent line at a given point.

According to the definition of a derivative at a, we have

$\begin{array}{c}f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{f(a+h)-f\left(a\right)}{h}Definitionoff\text{'}\left(a\right)\\ =\underset{h\to 0}{\lim }\frac{(-\frac{1}{{(a+h)}^2})-(-\frac{1}{a^2})}{h}f\left(x\right)=-\frac{1}{x^2}\\ =\underset{h\to 0}{\lim }\frac{\frac{-a^2+{(a+h)}^2}{a^2{(a+h)}^2}}{h}\\ =\underset{h\to 0}{\lim }\frac{-a^2+a^2+h^2+2ah}{h{a}^2{(a+h)}^2}Simplify\\ =\underset{h\to 0}{\lim }\frac{h+2a}{a^2{(a+h)}^2}Cancelh\\ =\frac{0+2a}{a^2{(a+0)}^2}\\ =\frac{2}{a^3}\end{array}$

Therefore, the derivative of the function ata is $f\text{'}\left(a\right)=\frac{2}{a^3}$.

Step 4. Finding the value of derivative at given point.

We have.$f\text{'}\left(a\right)=\frac{2}{a^3}$

Let’s find the value of derivative at$a=3anda=4$.

Substitute the value of$a=3anda=4$ into the$f\text{'}\left(a\right)=\frac{2}{a^3}$

$\begin{array}{c}f\text{'}\left(3\right)=\frac{2}{3^3}\\ =\frac{2}{27}\\ f\text{'}\left(4\right)=\frac{2}{4^3}\\ =\frac{2}{64}\\ =\frac{1}{32}\end{array}$

Therefore, the value of $f\text{'}\left(3\right)=\frac{2}{27}andf\text{'}\left(4\right)=\frac{1}{32}$.