Find y' for y = (1 - 1/(sqrt(x)))². (Do not use the chain
Find $y'$ for $y = \left(1 - \dfrac{1}{\sqrt{x}}\right)^{2}$. (Do not use the chain rule.)
Explanation of Method
Expand a square containing a negative power, then differentiate.
Step 1: Expand the square
Write $\dfrac{1}{\sqrt{x}} = x^{-1/2}$ and expand: $y = \left(1 - x^{-1/2}\right)^{2} = 1 - 2x^{-1/2} + x^{-1}$.
Note that $\bigl(x^{-1/2}\bigr)^{2} = x^{-1}$.
Step 2: Differentiate term by term
\[ y' = \frac{d}{dx}(1) - 2\,\frac{d}{dx}\bigl(x^{-1/2}\bigr) + \frac{d}{dx}\bigl(x^{-1}\bigr), \]
where by the power rule
\[ \frac{d}{dx}(1) = 0, \qquad \frac{d}{dx}\bigl(x^{-1/2}\bigr) = -\frac{1}{2}x^{-3/2}, \qquad \frac{d}{dx}\bigl(x^{-1}\bigr) = -x^{-2}. \]
Step 3: Simplify
\[ y' = 0 - 2\left(-\frac{1}{2}\right)x^{-3/2} - x^{-2} = x^{-3/2} - x^{-2}. \]
In explicit form, using $x^{3/2} = x\sqrt{x}$: $y' = \frac{1}{x^{3/2}} - \frac{1}{x^{2}} = \frac{1}{x\sqrt{x}} - \frac{1}{x^{2}}$.
Step 4: Verification
\[\text{Chain rule:}\quad y' = 2\left(1-x^{-1/2}\right)\cdot\frac{1}{2}x^{-3/2} = x^{-3/2} - x^{-2} \;\;\checkmark\]
\[x = 1:\quad y' = 1 - 1 = 0, \text{ and } y(1) = 0 \text{ is the minimum of } y \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \dfrac{1}{x\sqrt{x}} - \dfrac{1}{x^{2}}, \qquad x > 0\,}\]
Remark
Remark. A common form of the answer is $y' = \dfrac{\sqrt{x}-1}{x^{2}}$, obtained by putting both terms over the denominator $x^{2}$. Written this way the sign is immediate: the function decreases on $(0,1)$ and increases on $(1,\infty)$.