y=sqrt(3-sqrt(x)) y'=?

$y\ =\ \sqrt{3-\sqrt{x}}\\~\\ y\ =\ (3-x^\frac12)^\frac12\\~\\ y'\ =\ \frac{d}{dx}(3-x^\frac12)^\frac12$

Using the power rule and chain rule,

$y'\ =\ \frac12(3-x^\frac12)^{-\frac12}(\ \frac{d}{dx}(3-x^\frac12)\ )\\~\\ y'\ =\ \frac12(3-x^\frac12)^{-\frac12}(\ \frac{d}{dx}3-\frac{d}{dx}x^\frac12\ )\\~\\ y'\ =\ \frac12(3-x^\frac12)^{-\frac12}(\ 0-\frac12x^{-\frac12}\ )$

And we can rewrite the right side of the equation like this:

$y'\ =\ \dfrac{-1}{4\,\cdot\,\sqrt{3-\sqrt{x}}\,\cdot\,\sqrt{x}}$_

y=$\sqrt{3-\sqrt{x}}$

$y'=\ ?$

$\begin{array}{|rcll|} \hline y &=& \sqrt{3-\sqrt{x}} \\ y^2 &=& 3-\sqrt{x} \quad | \quad \text{derivate both sides} \\\\ 2yy' &=& -\dfrac{d\ ( \sqrt{x})}{dx} \\\\ 2yy' &=& -\dfrac{d\ (x^{\frac12})}{dx} \\\\ 2yy' &=& -\dfrac{1}{2} x^{-\frac{1}{2}} \\\\ 2yy' &=& -\dfrac{1}{2x^{\frac{1}{2}}} \\\\ 2yy' &=& -\dfrac{1}{2\sqrt{x}} \\\\ y' &=& -\dfrac{1}{2\sqrt{x}2y} \\\\ y' &=& -\dfrac{1}{4\sqrt{x} y} \\\\ \mathbf{y'} &=& \mathbf{-\dfrac{1}{4\sqrt{x} \sqrt{3-\sqrt{x}}}} \\ \hline \end{array}$