Suppose we wanted to differentiate $$h(x)=\sqrt{\sin(x)}.$$ Here, $h(x)$ is given by the composition $$h(x)=f(g(x)),$$ where $$f(x)=\sqrt{x}\quad\mbox{and}\quad g(x)=\sin(x).$$ Then chain rule says that $$h'(x)=f'(g(x))g'(x).$$

Since $$f'(x)=\frac{1}{2\sqrt{x}},\quad\mbox{and}\quad g'(x)=\cos(x),$$ we have by chain rule $$h'(x) = \left(\frac{1}{2\sqrt{\sin x}}\right)\cos x=\frac{\cos x}{2\sqrt{\sin x}}$$

Using the Leibniz formalism, the above calculation would have the following appearance. First we describe the functional relation as $$z = \sqrt{\sin(x)}.$$ Next, we introduce an auxiliary variable $y$ , and write $$z= \sqrt{y},\qquad y=\sin(x).$$ We then have $$\frac{dz}{dy} = \frac{1}{2\sqrt{y}},\qquad \frac{dy}{dx} = \cos(x),$$ and hence the chain rule gives