Answer
$y'=\dfrac{1}{2\sqrt{x}}$
Work Step by Step
$y=\dfrac{x-a}{\sqrt{x}-\sqrt{a}}$
Factor the numerator as a difference of squares:
$y=\dfrac{x-a}{\sqrt{x}-\sqrt{a}}=\dfrac{(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})}{\sqrt{x}-\sqrt{a}}=...$
Simplify:
$...=\sqrt{x}+\sqrt{a}=...$
Rewrite the roots using rational exponents:
$...=x^{1/2}+a^{1/2}$
Evaluate the derivative:
$y'=\dfrac{1}{2}x^{-1/2}+0=\dfrac{1}{2x^{1/2}}=...$
$...=\dfrac{1}{2\sqrt{x}}$