Question

Find dy/dx using first principlex+1/x​

Answer 1

Answer:

The derivative of the function is:

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

Step-by-step explanation:

The first principle is given by:

$f'(x)=\frac{dy}{dx}=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

The function here is:

$f(x)=x+\frac{1}{x}$

Now, using the first principle we have:

$f'(x)=lim_{h\rightarrow 0}\frac{(x+h)+\frac{1}{(x+h)}-x-\frac{1}{x}}{h}$  

$=lim_{h\rightarrow 0}\frac{h+\frac{1}{(x+h)}-\frac{1}{x}}{h}$    

$=lim_{h\rightarrow 0}\frac{h-\frac{h}{x(x+h)}}{h}$    

$=lim_{h\rightarrow 0}\frac{h(1-\frac{1}{x(x+h)})}{h}$

$=lim_{h\rightarrow 0}1-\frac{1}{x(x+h)}$

$=1-\frac{1}{x^{2}}$

 

Therefore, the derivative of the function is:

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

I hope it helps you!