# Given y = \frac{\sqrt{x} - 2}{\sqrt{x} + 2}, calculate dy/dx.

## Question:

Given {eq}\displaystyle\, y = \frac{\sqrt{x} - 2}{\sqrt{x} + 2} {/eq},{eq}\, {/eq} calculate {eq}\displaystyle\; \frac{dy}{dx} {/eq}.

## Rules of Differentiation:

Differentiation is a powerful operation in calculus. In fact, because of the rules that we have in finding the derivative, we can essentially differentiate any function that we know and the combinations of those functions: sums, products, and compositions. From the Product Rule, we can come up with a similar rule called a Quotient Rule, which is mainly used for quotients, as the name suggests.

Since the given expression $$y = \frac{\sqrt{x} - 2}{\sqrt{x} + 2}$$ is a quotient, we can use the Quotient Rule to calculate the derivative. Recall that if we take {eq}f(x) = \sqrt{x} - 2 {/eq} and {eq}g(x) = \sqrt{x} + 2 {/eq} then the derivative of {eq}\dfrac{f(x)}{g(x)} {/eq} is given by $$\frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$$

Following this rule, we have \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\left( \frac{1}{2\sqrt{x}} \right)(\sqrt{x} + 2) - (\sqrt{x} - 2)\left( \frac{1}{2\sqrt{x}} \right)}{(\sqrt{x} + 2)^2} \\ &= \frac{\frac{1}{2} + \frac{1}{\sqrt{x}} - \frac{1}{2} + \frac{1}{\sqrt{x}}}{x + 2 \sqrt{x} + 4} \\ &= \frac{2}{\sqrt{x} (x + 2 \sqrt{x} + 4)} \\ &= \boxed{\frac{2}{x^{\frac{3}{2}} + 2x + 4x^{\frac{1}{2}}}} \end{align*}