Answer:
The derivative of the function is:
[tex]f'(x)=1-\frac{1}{x^{2}}[/tex]
Step-by-step explanation:
The first principle is given by:
[tex]f'(x)=\frac{dy}{dx}=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}[/tex]
The function here is:
[tex]f(x)=x+\frac{1}{x}[/tex]
Now, using the first principle we have:
[tex]f'(x)=lim_{h\rightarrow 0}\frac{(x+h)+\frac{1}{(x+h)}-x-\frac{1}{x}}{h}[/tex]
[tex]=lim_{h\rightarrow 0}\frac{h+\frac{1}{(x+h)}-\frac{1}{x}}{h}[/tex]
[tex]=lim_{h\rightarrow 0}\frac{h-\frac{h}{x(x+h)}}{h}[/tex]
[tex]=lim_{h\rightarrow 0}\frac{h(1-\frac{1}{x(x+h)})}{h}[/tex]
[tex]=lim_{h\rightarrow 0}1-\frac{1}{x(x+h)}[/tex]
[tex]=1-\frac{1}{x^{2}}[/tex]
Therefore, the derivative of the function is:
[tex]f'(x)=1-\frac{1}{x^{2}}[/tex]
I hope it helps you!
