Differentiate the relation \(\sin y = \tan x\) with respect to \(x\):
\(\cos y \frac{dy}{dx} = \sec^2 x\).
Therefore, \(\frac{dy}{dx} = \frac{\sec^2 x}{\cos y}\).
Since \(y \in (-\tfrac{\pi}{2}, \tfrac{\pi}{2})\), we know \(\cos y > 0\),
so \(\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - \tan^2 x}\).
Express \(1 - \tan^2 x\) in terms of \(\cos 2x\) and \(\cos x\):
\(1 - \tan^2 x = \frac{\cos^2 x - \sin^2 x}{\cos^2 x}
= \frac{\cos 2x}{\cos^2 x}\).
Taking square roots gives
\(\sqrt{1 - \tan^2 x} = \frac{\sqrt{\cos 2x}}{|\cos x|}\).
Hence, \(\frac{dy}{dx} = \frac{\sec^2 x}{\sqrt{1 - \tan^2 x}}
= \frac{1/\cos^2 x}{\sqrt{\cos 2x}/|\cos x|}
= \frac{1}{|\cos x| \sqrt{\cos 2x}}\).
Therefore,
\(\frac{dy}{dx} = \frac{1}{|\cos x| \sqrt{\cos 2x}}\),
valid where \(\cos 2x > 0\).
If we restrict \(x\) so that \(\cos x > 0\)
(for example \(x \in (-\tfrac{\pi}{4}, \tfrac{\pi}{4})\) modulo \(\pi\)),
the absolute value can be omitted:
\(\frac{dy}{dx} = \frac{1}{\cos x \sqrt{\cos 2x}}\).
