$y=x+x^2+\frac{1}{x}+\frac{1}{x^3}\text{ Find }\frac{dy}{dx}$

Solution:

$\begin{array}{l}y=x+x^2+\frac{1}{x}+\frac{1}{x^3}\\ \frac{dy}{dx}=\frac{dx}{dx}+\frac{d{x}^2}{dx}+\frac{d{x}^{-1}}{dx}+\frac{d{x}^{-3}}{dx}\\ =1+2x+\left(-1x^{-1-1}\right)+\left(-3x^{-3-1}\right)\left(-\frac{dx}{dx}=1,\frac{d{x}^n}{dx}=n{x}^{n-1}\right)\\ =1+2x-\frac{1}{x^2}-\frac{3}{x^4}\end{array}$