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Let $y = \frac{\sin x}{\sqrt{\cos x}}$. We need to find $\frac{dy}{dx}$.
The quotient rule states that if $y = \frac{u}{v}$, then $\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$. Here, $u = \sin x$ and $v = \sqrt{\cos x} = (\cos x)^{1/2}$.
We have $\frac{du}{dx} = \frac{d}{dx}(\sin x) = \cos x$. Also, $\frac{dv}{dx} = \frac{d}{dx}(\sqrt{\cos x}) = \frac{d}{dx}((\cos x)^{1/2}) = \frac{1}{2}(\cos x)^{-1/2} \cdot (-\sin x) = -\frac{\sin x}{2\sqrt{\cos x}}$.
Now, we substitute these into the quotient rule formula: $$ \frac{dy}{dx} = \frac{\sqrt{\cos x} \cdot \cos x - \sin x \cdot \left(-\frac{\sin x}{2\sqrt{\cos x}}\right)}{(\sqrt{\cos x})^2} $$
$$ \frac{dy}{dx} = \frac{\cos x \sqrt{\cos x} + \frac{\sin^2 x}{2\sqrt{\cos x}}}{\cos x} = \frac{\frac{2\cos^2 x + \sin^2 x}{2\sqrt{\cos x}}}{\cos x} = \frac{2\cos^2 x + \sin^2 x}{2\cos x \sqrt{\cos x}} $$ $$ \frac{dy}{dx} = \frac{2\cos^2 x + \sin^2 x}{2\cos^{3/2} x} = \frac{2\cos^2 x + (1 - \cos^2 x)}{2\cos^{3/2} x} = \frac{\cos^2 x + 1}{2\cos^{3/2} x} $$
Therefore, $\frac{dy}{dx} = \frac{\cos^2 x + 1}{2\cos^{3/2} x}$.
Final Answer: $\frac{\cos^2 x + 1}{2\cos^{3/2} x}$
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