Find y' for y = 1/(sqrt(x))
Find $y'$ for $y = \dfrac{1}{\sqrt{x}}$.
Explanation of Method
Differentiate a reciprocal square root using a negative fractional exponent.
Step 1: Rewrite as a power function
\[ y = \frac{1}{\sqrt{x}} = \frac{1}{x^{1/2}} = x^{-1/2}. \]
Step 2: Apply the power rule
Using $\bigl(x^{\alpha}\bigr)' = \alpha x^{\alpha-1}$ with $\alpha = -\dfrac{1}{2}$: $y' = -\frac{1}{2}\,x^{-\frac{1}{2}-1} = -\frac{1}{2}\,x^{-3/2}$.
Step 3: Rewrite in radical form
Since $x^{3/2} = x\sqrt{x}$: $y' = -\frac{1}{2}\cdot\frac{1}{x^{3/2}} = -\frac{1}{2x\sqrt{x}}$.
Step 4: Verification
By the quotient rule applied to $y = \dfrac{1}{\sqrt{x}}$, with $\bigl(\sqrt{x}\bigr)' = \dfrac{1}{2\sqrt{x}}$: \[y' = \frac{0\cdot\sqrt{x} - 1\cdot\dfrac{1}{2\sqrt{x}}}{\bigl(\sqrt{x}\bigr)^{2}} = \frac{-\dfrac{1}{2\sqrt{x}}}{x} = -\frac{1}{2x\sqrt{x}} \;\;\checkmark\]
\[x = 1:\quad y'(1) = -\frac{1}{2}; \qquad \frac{y(1.001)-y(1)}{0.001} \approx -0.4996 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = -\dfrac{1}{2x\sqrt{x}}, \qquad x > 0\,}\]
Remark
Remark. The derivative is negative throughout $x \gt 0$, so the curve falls steadily; it also tends to $-\infty$ as $x \to 0^{+}$, reflecting the vertical asymptote of $y = 1/\sqrt{x}$ at the origin.