Differentiate the functions with respect to x.

Let $f(x)=\cos (\sqrt{x})$

Also, let $u(x)=\sqrt{x}$

And, $v(t)=\cos t$

Then, $(v o u)(x)=v(u(x))$

$=v(\sqrt{x})$

$=\cos \sqrt{x}$

$=f(x)$

Clearly, f is a composite function of two functions, u and v, such that

Then, $\frac{d t}{d x}=\frac{d}{d x}(\sqrt{x})=\frac{d}{d x}\left(x^{\frac{1}{2}}\right)=\frac{1}{2} x^{-\frac{1}{2}}$

$=\frac{1}{2 \sqrt{x}}$

And, $\frac{d v}{d t}=\frac{d}{d t}(\cos t)=-\sin t$

$=-\sin (\sqrt{x})$

By using chain rule, we obtain

$\frac{d t}{d x}=\frac{d v}{d t} \cdot \frac{d t}{d x}$

$=-\sin (\sqrt{x}) \cdot \frac{1}{2 \sqrt{x}}$

$=-\frac{1}{2 \sqrt{x}} \sin (\sqrt{x})$

$=-\frac{\sin (\sqrt{x})}{2 \sqrt{x}}$

Alternate method

$\frac{d}{d x}[\cos (\sqrt{x})]=-\sin (\sqrt{x}) \cdot \frac{d}{d x}(\sqrt{x})$

$=-\sin (\sqrt{x}) \times \frac{d}{d x}\left(x^{\frac{1}{2}}\right)$

$=-\sin \sqrt{x} \times \frac{1}{2} x^{-\frac{1}{2}}$

$=\frac{-\sin \sqrt{x}}{2 \sqrt{x}}$