Question Detail

To differentiate $y = \frac{e^{\sin 2x}}{\cos 2x}$, we use the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$.Let $u = e^{\sin 2x}$ and $v = \cos 2x$.First, find the derivatives $u'$ and $v'$:$u' = \frac{d}{dx}(e^{\sin 2x}) = e^{\sin 2x} \cdot \frac{d}{dx}(\sin 2x) = e^{\sin 2x} \cdot (\cos 2x \cdot 2) = 2\cos 2x e^{\sin 2x}$.$v' = \frac{d}{dx}(\cos 2x) = -\sin 2x \cdot 2 = -2\sin 2x$.Now, apply the quotient rule:$\frac{dy}{dx} = \frac{(2\cos 2x e^{\sin 2x})(\cos 2x) - (e^{\sin 2x})(-2\sin 2x)}{(\cos 2x)^2}$$\frac{dy}{dx} = \frac{2\cos^2 2x e^{\sin 2x} + 2\sin 2x e^{\sin 2x}}{\cos^2 2x}$Factor out $2e^{\sin 2x}$ from the numerator:$\frac{dy}{dx} = \frac{2e^{\sin 2x} (\cos^2 2x + \sin 2x)}{\cos^2 2x}$Separate the fraction:$\frac{dy}{dx} = 2e^{\sin 2x} \left( \frac{\cos^2 2x}{\cos^2 2x} + \frac{\sin 2x}{\cos^2 2x} \right)$$\frac{dy}{dx} = 2e^{\sin 2x} \left( 1 + \frac{\sin 2x}{\cos 2x} \cdot \frac{1}{\cos 2x} \right)$$\frac{dy}{dx} = 2e^{\sin 2x} (1 + \tan 2x \sec 2x)$.Comparing this to the options, option C is $2e^{\sin 2x} [1 + \sec^2 2x \tan 2x]$. There appears to be a typo in option C, where $\sec 2x$ should be $\sec^2 2x$. However, given the choices, this is the closest and most likely intended answer, assuming $\sec^2 2x$ was intended to be $\sec 2x$ or the simplification was intended to produce $\sec^2 2x \tan 2x$ from a slightly different form. Based on standard differentiation, the derived form is $2e^{\sin 2x} (1 + \tan 2x \sec 2x)$. If a direct match is enforced from the options, we note the discrepancy.