Answer
$$ y' =\frac{-6\sin 6x+2x\cos x^2}{2\sqrt{\cos 6x +\sin x^2}}.$$
Work Step by Step
Recall that $(\sin x)'=\cos x$.
Recall that $(\cos x)'=-\sin x$.
Recall that $(x^n)'=nx^{n-1}$
Since $ y=(\cos 6x +\sin x^2)^{1/2}$, then by using the chain rule, the derivative $ y'$ is given by
$$ y'=\frac{1}{2}(\cos 6x +\sin x^2)^{-1/2}(-6\sin 6x+2x\cos x^2)\\=\frac{-6\sin 6x+2x\cos x^2}{2\sqrt{\cos 6x +\sin x^2}}.$$
