<s>[INST] Differentiating w.r.t x Please help me understand this question. 
\begin{gathered}
  {(\sqrt x  - \frac{1}{{\sqrt x }})^2} = y \hfill \\
  {\text{Differentiating w}}{\text{.r}}{\text{.t x}} \hfill \\
   \hfill \\
  \frac{{dy}}{{dx}} = \frac{d}{{dx}}{\left[ {(\sqrt x ) - \frac{1}{{\sqrt x }}} \right]^2} \hfill \\
   \hfill \\
   = 2\left( {\sqrt x  - \frac{1}{{\sqrt x }}} \right)\frac{d}{{dx}}{\left[ {(\sqrt x ) - \frac{1}{{\sqrt x }}} \right]^{}}{\text{From Where 2 Come?}} \hfill \\
   \hfill \\
  \frac{{dy}}{{dx}} = 2\left( {\sqrt x  - \frac{1}{{\sqrt x }}} \right){\left[ {\frac{d}{{dx}}({x^{\frac{1}{2}}}) - \frac{d}{{dx}}({x^{ - \frac{1}{2}}})} \right]^{}} \hfill \\
   \hfill \\
   = 2\left( {\sqrt x  - \frac{1}{{\sqrt x }}} \right){\left[ {\frac{1}{2}({x^{\frac{1}{2} - 1}}) - \left( { - \frac{1}{2}} \right)({x^{ - \frac{1}{2} - 1}})} \right]^{}} \hfill \\
   \hfill \\
   = 2\left( {\sqrt x  - \frac{1}{{\sqrt x }}} \right){\left[ {\frac{1}{2}{x^{ - \frac{1}{2}}}) + \frac{1}{2}{x^{ - \frac{3}{2}}}} \right]^{}} \hfill \\
   \hfill \\
   = 2\left( {\sqrt x  - \frac{1}{{\sqrt x }}} \right){\text{x}}\frac{1}{2}{\left[ {\frac{1}{{\sqrt x }} + {{\frac{1}{{x\sqrt x }}}^{}}} \right]^{}} \hfill \\
   \hfill \\
   = \left( {\sqrt x  - \frac{1}{{\sqrt x }}} \right)\left[ {\frac{{x + 1}}{{x\sqrt x }}} \right]{\text{How this equation come?}} \hfill \\
   \hfill \\
   = \left[ {\frac{{x - 1}}{{\sqrt x }}} \right]\left[ {\frac{{x + 1}}{{x\sqrt x }}} \right]{\text{How this equation come?}} \hfill \\
   = \frac{{{x^2} - 1}}{{{x^2}}}Answer \hfill \\ 
\end{gathered} ]
 [/INST] it is by the power and Quotient rule
$$2\left(\frac{x-1}{\sqrt{x}}\right)\left(\frac{\sqrt{x}-(x-1)\cdot \frac{1}{2}x^{-1/2}}{x}\right)$$
it can be simplified ton $$1-\frac{1}{x^2}$$
</s>