Question

If $y=x+\sqrt{x+\sqrt{x}}$, find $\frac{dy}{dx}$.

Answer 1

$y=x+\sqrt{x+\sqrt{x}}$

$y=x+\sqrt{y}$

Differentiating w.r.t. $x$, we get,

$\frac{dy}{dx}=1+\frac{1}{2\sqrt{y}}\frac{dy}{dx}$

$\frac{dy}{dx}(1-\frac{1}{2\sqrt{y}})=1$

$\frac{dy}{dx}\left(\frac{2\sqrt{y}-1}{2\sqrt{y}}\right)=1$

$\frac{dy}{dx}=\frac{2\sqrt{y}}{2\sqrt{y}-1}$