MCQ
Let y =$\sqrt {x\,\, + \,\,\sqrt {x\,\, + \,\,\sqrt {x\,\, + \,\,......\,\,\infty } } }$ then $\frac{{dy}}{{dx}}$ =
  • A
    $\frac{1}{{2\,y\,\, - \,\,1}}$
  • B
    $\frac{y}{{2x\,\, + \,\,y}}$
  • C
    $\frac{1}{{\sqrt {1\,\, + \,\,4x} }}$
  • All of the above

Answer

Correct option: D.
All of the above
d
$y^2 = x + y ==>\frac{{dy}}{{dx}} =$$\frac{1}{{2\,y\,\, - \,\,1}}$

also $y =\frac{x}{y} + 1 $

$==>\frac{{dy}}{{dx}} =\frac{y}{{2\,x\,\, + \,\,y}}$ 

make a quadratic in $y$ to get explicit function $==> C$ 

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