If \(x = e^{\cos 2t}, y = e^{\sin 2t}\), then \(\displaystyle\frac{dy}{dx}\) equals to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle -\frac{y\log_e x}{x\log_e y}\)
By parametric differentiation:
\(\displaystyle \frac{dx}{dt} = -2\sin 2t\,e^{\cos 2t} = -2x\sin 2t\)
\(\displaystyle \frac{dy}{dt} = 2\cos 2t\,e^{\sin 2t} = 2y\cos 2t\)
\(\displaystyle \frac{dy}{dx} = \frac{2y\cos 2t}{-2x\sin 2t} = -\frac{y\cos 2t}{x\sin 2t}\)
Taking logarithms of the given equations, \(\cos 2t = \log_e x\) and \(\sin 2t = \log_e y\). So
\(\displaystyle \frac{dy}{dx} = -\frac{y\log_e x}{x\log_e y}\)
NCERT Exemplar:
Continuity and Differentiability
