How can I differentiate this function?
$y$ = $\displaystyle\frac{x}{x+\frac{\sqrt{x}}{x+\frac{\sqrt{x}}{x+....}}}$
I think I should break this function into an implicit function:
$\displaystyle\frac{y}{\sqrt{x}}=\frac{\sqrt{x}}{x+\frac{y}{\sqrt{x}}}$
this is the equation I get and was able to differentiate it but all the options are in $x$'s. When I differentiate the above function I got this
$\displaystyle\frac{dy}{dx}=-\frac{y(1+2\sqrt{x})}{\sqrt{x}(\sqrt{x}+x)}$
but since the options are all in $x$, I didn't get any solution.
The solution to the derivative of the above function is $\displaystyle\frac{1}{2x\sqrt{x}}$
You have to first solve for the function and then differentiate. You already know that
$\frac{y}{\sqrt{x}} = \frac{\sqrt{x}}{x + \frac{y}{\sqrt{x}}}$
so let $z = \frac{y}{\sqrt{x}}$, then we have
$z = \frac{\sqrt{x}}{x + z}$
Then $z^2 + xz = \sqrt{x}$. By the quadratic formula, $z = \frac{-x \pm \sqrt{x^2 + 4 \sqrt{x}}}{2}$. We know that $y$ is positive, hence $z$ is positive, hence $z = \frac{-x + \sqrt{x^2 + 4 \sqrt{x}}}{2}$. Then $y = \sqrt{x} \frac{-x + \sqrt{x^2 + 4 \sqrt{x}}}{2}$.
From here, it's a straightforward-ish derivative. But the answer you were provided is incorrect.