${2^{{{\cos }^2}x}}$

Let $y = {2^{{{\cos }^2}x}}$

therefore,$\log y = \log {2^{{{\cos }^2}x}} = {\cos ^2}x \cdot \log 2$

On differentiating w.r.t. $x$, we get

$\frac{d}{{dy}}\log y \cdot \frac{{dy}}{{dx}} = \frac{d}{{dx}}\log 2 \cdot {\cos ^2}x$

$\Rightarrow$ $\frac{1}{y} \cdot \frac{{dy}}{{dx}} = \log 2\frac{d}{{dx}}{(\cos x)^2}$

$\Rightarrow$ $\frac{1}{y} \cdot \frac{{dy}}{{dx}} = \log 2 \cdot [2\cos x] \cdot \frac{d}{{dx}}\cos x$

$= \log 2 \cdot 2\cos x \cdot ( - \sin x)$

$= \log 2 \cdot [ - (\sin 2x)]$

therefore,$\frac{{dy}}{{dx}} = - y \cdot \log 2(\sin 2x)$

$= - {2^{{{\cos }^2}x}} \cdot \log 2(\sin 2x)$[ putting the value of y = $2^{cos^2x}$]