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**Punch** Given that $\displaystyle \frac{dy}{dx}=\frac{\sqrt{x}-4}{2(\sqrt{x}-\color{red}2)^2}$ for the curve $\displaystyle y=\frac{x}{\sqrt{x}-2}$, obtain the equation of the normal to the curve $\displaystyle y=\frac{x}{\sqrt{x}-2}$ at the point on the curve where x=1.

I have attempted the question by finding the value of y when x=1. Then using the formula $\displaystyle (y-y_1)=\frac{dy}{dx}(x-x_1)$. However, i am unable to simplify it...

Hence, i require a detailed working to examine the problem, thanks!