Find the derivative of $\dfrac{x^2-6}{x}$ with respect to x from first principle of derivatives.
Let $f\left(x\right)=\dfrac{x^2-6}{x}$
$\begin{aligned}
f'\left(x\right) & = \lim\limits_{h \to 0} \; \dfrac{f\left(x+h\right)-f\left(x\right)}{h} \\
& = \lim\limits_{h \to 0} \; \dfrac{\dfrac{\left(x+h\right)^2 -6}{x+h}-\left(\dfrac{x^2-6}{x}\right)}{h} \\
& = \lim\limits_{h \to 0} \; \dfrac{\dfrac{x^2+2hx+h^2-6}{x+h}-\dfrac{x^2-6}{x}}{h} \\
& = \lim\limits_{h \to 0} \; \dfrac{x^3 + 2hx^2 + h^2 x -6x -x^3 -hx^2 +6x +6h}{hx\left(x+h\right)} \\
& = \lim\limits_{h \to 0} \; \dfrac{hx^2 + h^2 x +6h}{hx\left(x+h\right)} \\
& = \lim\limits_{h \to 0} \; \dfrac{x^2+hx+6}{x\left(x+h\right)} \\
& = \dfrac{x^2+0+6}{x\left(x+0\right)} \\
& = \dfrac{x^2+6}{x^2}
\end{aligned}$