Find y' for y = (sqrt(2x-1))/x
Find $y'$ for $y = \dfrac{\sqrt{2x-1}}{x}$.
Explanation of Method
The chain rule inside the quotient rule, with a tidy cancellation.
Step 1: Differentiate the numerator by the chain rule
With $u = (2x-1)^{1/2}$, the inner derivative is $2$, so
\[ u' = \frac{1}{2}(2x-1)^{-1/2}\cdot 2 = \frac{1}{\sqrt{2x-1}}. \]
Step 2: Apply the quotient rule
With $v = x$, $v' = 1$: $y' = \frac{u'v - uv'}{v^{2}} = \frac{\dfrac{x}{\sqrt{2x-1}} - \sqrt{2x-1}}{x^{2}}$.
Step 3: Simplify the numerator
Put the two terms over the common denominator $\sqrt{2x-1}$, using $\sqrt{2x-1}\cdot\sqrt{2x-1} = 2x-1$: $\frac{x}{\sqrt{2x-1}} - \sqrt{2x-1} = \frac{x-(2x-1)}{\sqrt{2x-1}} = \frac{1-x}{\sqrt{2x-1}}$.
Step 4: Verification
\[x = 1:\quad y' = \frac{0}{1\cdot 1} = 0, \text{ a critical point; } y(1) = 1 \;\;\checkmark\]
\[x = 2:\quad y' = \frac{-1}{4\sqrt{3}} \approx -0.14434; \qquad \frac{y(2+10^{-5})-y(2-10^{-5})}{2\times 10^{-5}} \approx -0.14434 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \dfrac{1-x}{x^{2}\sqrt{2x-1}}, \qquad x > \tfrac12\,}\]
Remark
Remark. The domain is $x \gt \tfrac12$: the radicand must be non-negative and $x \neq 0$, and on $x \geq \tfrac12$ the second condition is automatic. The sign of $y'$ is the sign of $1-x$, so $y$ rises to the maximum $y(1) = 1$ and then decays to zero.