Question Detail

We can rewrite ( y = e^{\sin(2x)} \cdot (\cos(2x))^{-1} ).Using the product rule ( (uv)' = u'v + uv' ):Let ( u = e^{\sin(2x)} ) and ( v = (\cos(2x))^{-1} ).( u' = e^{\sin(2x)} \cdot \frac{d}{dx}(\sin(2x)) = e^{\sin(2x)} \cdot \cos(2x) \cdot 2 )( v' = -1(\cos(2x))^{-2} \cdot \frac{d}{dx}(\cos(2x)) = -1(\cos(2x))^{-2} \cdot (-\sin(2x)) \cdot 2 = \frac{2\sin(2x)}{\cos^2(2x)} )Now, substitute into the product rule formula:( \frac{dy}{dx} = (2e^{\sin(2x)} \cos(2x)) \cdot (\cos(2x))^{-1} + e^{\sin(2x)} \cdot \frac{2\sin(2x)}{\cos^2(2x)} )( \frac{dy}{dx} = 2e^{\sin(2x)} + \frac{2e^{\sin(2x)} \sin(2x)}{\cos^2(2x)} )Factor out ( 2e^{\sin(2x)} ):( \frac{dy}{dx} = 2e^{\sin(2x)} \left[1 + \frac{\sin(2x)}{\cos^2(2x)}\right] )Recognize ( \frac{\sin(2x)}{\cos^2(2x)} = \frac{\sin(2x)}{\cos(2x)} \cdot \frac{1}{\cos(2x)} = \tan(2x) \sec(2x) ):( \frac{dy}{dx} = 2e^{\sin(2x)} [1 + \tan(2x) \sec(2x)] ).This matches option C.