Find y' for y = (sqrt(a)- sqrt(x))², where a is

Find $y'$ for $y = \left(\sqrt{a}-\sqrt{x}\right)^{2}$, where $a$ is a positive constant. (Do not use the chain rule.)

Step 1: Expand the square

Using $(u-v)^{2} = u^{2} - 2uv + v^{2}$ with $u = \sqrt{a}$ and $v = \sqrt{x}$: \[y = \left(\sqrt{a}-\sqrt{x}\right)^{2} = a - 2\sqrt{a}\,\sqrt{x} + x = a - 2\sqrt{a}\,x^{1/2} + x.\]

Each term is now either a constant or a simple power of $x$.

Step 2: Differentiate term by term

\[ y' = \frac{d}{dx}(a) - 2\sqrt{a}\,\frac{d}{dx}\bigl(x^{1/2}\bigr) + \frac{d}{dx}(x), \]

where

\[ \frac{d}{dx}(a) = 0, \qquad \frac{d}{dx}\bigl(x^{1/2}\bigr) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}, \qquad \frac{d}{dx}(x) = 1. \]

Step 3: Simplify

\[ y' = 0 - 2\sqrt{a}\cdot\frac{1}{2\sqrt{x}} + 1 = 1 - \frac{\sqrt{a}}{\sqrt{x}}. \]

Step 4: Verification

The chain rule, which we were asked to avoid, confirms the answer: \[y' = 2\bigl(\sqrt{a}-\sqrt{x}\bigr)\cdot\left(-\frac{1}{2\sqrt{x}}\right) = \frac{\sqrt{x}-\sqrt{a}}{\sqrt{x}} = 1 - \sqrt{\frac{a}{x}} \;\;\checkmark\]

\[x = a:\quad y' = 1 - 1 = 0, \text{ and indeed } y(a) = 0 \text{ is the minimum of } y \;\;\checkmark\]