If $x = e^{\cos 2t}, y = e^{\sin 2t}$, then $\frac{dy}{dx}$ equals to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $-\frac{y\log_ex}{x\log_ey}$
Given:
\(y = e^{\sin 2t}\)
\(\Rightarrow \ln y = \sin 2t\)
Also,
\[ \frac{dy}{dx} = - e^{\sin 2t - \cos 2t} \cdot \cot 2t = - \frac{e^{\sin 2t}}{e^{\cos 2t}} \cdot \cot 2t = - \frac{y}{e^{\cos 2t}} \cdot \cot 2t \]
From \(x = e^{\cos 2t}\),
\(\ln x = \cos 2t\)
So,
\[ \frac{dy}{dx} = - \frac{y}{x} \cdot \cot 2t \]
Express \(\cot 2t\) in terms of \(\sin 2t = \ln y\) and \(\cos 2t = \ln x\):
\[ \cot 2t = \frac{\cos 2t}{\sin 2t} = \frac{\ln x}{\ln y} \]
Therefore,
\({\frac{dy}{dx} = - \frac{y}{x} \cdot \frac{\ln x}{\ln y}}\)