Is the answer complex?

$y = \sqrt{x+\sqrt{12x-36}}\quad \Rightarrow y^2 = x + \sqrt{12x-36}$

$\Rightarrow 2y\dfrac {dy}{dx} = 1 + \dfrac {1}{2} (12x-36)^{-\frac{1}{2}} (12) = 1 + \dfrac {6}{\sqrt{12x-36}}$

When $x = 4$

$\Rightarrow y = \sqrt{4+\sqrt{48-36}} = \sqrt{4+2\sqrt{3}} = \sqrt{(1+\sqrt{3})^2} = 1 + \sqrt{3}$

Therefore,

$\Rightarrow 2y\dfrac {dy}{dx} = 1 + \dfrac {6}{\sqrt{12x-36}}$

$\Rightarrow 2(1+\sqrt{3} )\dfrac {dy}{dx} = 1 + \dfrac {6}{\sqrt{12}} = 1+\sqrt{3}$

$\Rightarrow \dfrac {dy}{dx} = \dfrac {1}{2} \quad \Rightarrow m+n = 1 + 2 = \boxed{3}$

$y = \sqrt{x+\sqrt{12x-36}}$

$\implies$ $y = \sqrt{\sqrt{(x-3)}^2+\sqrt{3}^2+2\sqrt{3x-9}}$

It is in the form of ( $a^2 + b^2 + 2ab$ ) which is equal to $(a+b)^2$ $\implies$ $y = (\sqrt{x-3} + \sqrt{3}$ )

Now it is easy to differentiate $y$ .