To find the derivative of the function [tex]\( f(x) = x^2 + \frac{1}{x} \)[/tex] using the first principles, we use the definition of the derivative. The first principle, also known as the difference quotient, states that the derivative [tex]\( f'(x) \)[/tex] at a point [tex]\( x \)[/tex] is given by:
[tex]\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \][/tex]
Let's follow these steps to find the derivative:
1. Define the function [tex]\( f(x) \)[/tex] and evaluate [tex]\( f(x+h) \)[/tex]:
[tex]\[
f(x) = x^2 + \frac{1}{x}
\][/tex]
We need to find [tex]\( f(x+h) \)[/tex]:
[tex]\[
f(x+h) = (x+h)^2 + \frac{1}{x+h}
\][/tex]
2. Expand [tex]\( f(x+h) \)[/tex]:
[tex]\[
(x+h)^2 = x^2 + 2xh + h^2
\][/tex]
So,
[tex]\[
f(x+h) = x^2 + 2xh + h^2 + \frac{1}{x+h}
\][/tex]
3. Subtract [tex]\( f(x) \)[/tex] from [tex]\( f(x+h) \)[/tex]:
[tex]\[
f(x+h) - f(x) = \left( x^2 + 2xh + h^2 + \frac{1}{x+h} \right) - \left( x^2 + \frac{1}{x} \right)
\][/tex]
Simplifying this, we get:
[tex]\[
f(x+h) - f(x) = 2xh + h^2 + \frac{1}{x+h} - \frac{1}{x}
\][/tex]
4. Form the difference quotient:
[tex]\[
\frac{f(x+h) - f(x)}{h} = \frac{2xh + h^2 + \frac{1}{x+h} - \frac{1}{x}}{h}
\][/tex]
5. Simplify the difference quotient:
[tex]\[
\frac{f(x+h) - f(x)}{h} = \frac{2xh}{h} + \frac{h^2}{h} + \frac{\frac{1}{x+h} - \frac{1}{x}}{h}
\][/tex]
[tex]\[
= 2x + h + \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)
\][/tex]
6. Simplify the term involving the fractions:
[tex]\[
\frac{1}{x+h} - \frac{1}{x} = \frac{x - (x+h)}{x(x+h)} = \frac{-h}{x(x+h)}
\][/tex]
So,
[tex]\[
\frac{1}{h} \left( \frac{-h}{x(x+h)} \right) = \frac{-1}{x(x+h)}
\][/tex]
7. Combine all terms:
[tex]\[
\frac{f(x+h) - f(x)}{h} = 2x + h - \frac{1}{x(x+h)}
\][/tex]
8. Take the limit as [tex]\( h \)[/tex] approaches 0:
[tex]\[
\lim_{h \to 0} \left( 2x + h - \frac{1}{x(x+h)} \right) = 2x + 0 - \frac{1}{x(x+0)}
\][/tex]
[tex]\[
= 2x - \frac{1}{x^2}
\][/tex]
Therefore, the derivative [tex]\( f'(x) \)[/tex] is:
[tex]\[
f'(x) = 2x - \frac{1}{x^2}
\][/tex]
