Question
If $$y = \frac{{\left( {a - x} \right)\sqrt {a - x} - \left( {b - x} \right)\sqrt {x - b} }}{{\sqrt {a - x} + \sqrt {x - b} }},$$ then $$\frac{{dy}}{{dx}}$$ wherever it is defined is :
A.
$$\frac{{x + \left( {a + b} \right)}}{{\sqrt {\left( {a - x} \right)\left( {x - b} \right)} }}$$
B.
$$\frac{{2x - a - b}}{{2\sqrt {a - x} \sqrt {x - b} }}$$
C.
$$ - \frac{{\left( {a + b} \right)}}{{2\sqrt {\left( {a - x} \right)\left( {x - b} \right)} }}$$
D.
$$\frac{{2x + \left( {a + b} \right)}}{{2\sqrt {\left( {a - x} \right)\left( {x - b} \right)} }}$$
Answer :
$$\frac{{2x - a - b}}{{2\sqrt {a - x} \sqrt {x - b} }}$$
Solution :
$$\eqalign{
& y = \frac{{{{\left( {a - x} \right)}^{\frac{3}{2}}} + {{\left( {x - b} \right)}^{\frac{3}{2}}}}}{{\sqrt {a - x} + \sqrt {x - b} }} \cr
& = \frac{{\left( {\sqrt {a - x} + \sqrt {x - b} } \right)\left( {a - x - \sqrt {a - x} \sqrt {x - b} + x - b} \right)}}{{\sqrt {a - x} + \sqrt {x - b} }} \cr
& = a - b - \sqrt {a - x} \sqrt {x - b} \cr
& {\text{or }}\frac{{dy}}{{dx}} = \frac{1}{{2\sqrt {a - x} }}\sqrt {x - b} - \frac{1}{{2\sqrt {x - b} }}\sqrt {a - x} \cr
& = \frac{{2x - a - b}}{{2\sqrt {a - x} \sqrt {x - b} }} \cr} $$