# How do you find the derivative of y=sqrt( x+sqrt( x+sqrt( x)))?

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 y} \left(\frac{2 \setminus \sqrt{x} + 1}{4 \setminus \sqrt{x} \left({y}^{2} - x\right)} + 1\right)$

#### Explanation:

Given that

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

${y}^{2} = x + \setminus \sqrt{x + \setminus \sqrt{x}}$

${y}^{2} - x = \setminus \sqrt{x + \setminus \sqrt{x}}$

${\left({y}^{2} - x\right)}^{2} = x + \setminus \sqrt{x}$

Differentiating above equation w.r.t. $x$ using chain rule as follows

$\frac{d}{\mathrm{dx}} {\left({y}^{2} - x\right)}^{2} = \frac{d}{\mathrm{dx}} \left(x + \setminus \sqrt{x}\right)$

$2 \left({y}^{2} - x\right) \frac{d}{\mathrm{dx}} \left({y}^{2} - x\right) = 1 + \frac{1}{2 \setminus \sqrt{x}}$

$2 \left({y}^{2} - x\right) \left(2 y \frac{\mathrm{dy}}{\mathrm{dx}} - 1\right) = 1 + \frac{1}{2 \setminus \sqrt{x}}$

$2 y \frac{\mathrm{dy}}{\mathrm{dx}} - 1 = \frac{2 \setminus \sqrt{x} + 1}{4 \setminus \sqrt{x} \left({y}^{2} - x\right)}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 y} \left(\frac{2 \setminus \sqrt{x} + 1}{4 \setminus \sqrt{x} \left({y}^{2} - x\right)} + 1\right)$