# Solve this

Question:

Find $\frac{d y}{d x}$, when

If $x=e^{\cos 2 t}$ and $y=e^{\sin 2 t}$, prove that $\frac{d y}{d x}=-\frac{y \log x}{x \log y}$

Solution:

Here, $x=e^{\cos 2 t}$

Differentiating it with respect to $\theta$ using chain rule,

$\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{e}^{\cos 2 \mathrm{t}}\right)$

$=e^{\cos 2 t} \frac{d}{d t}(\cos 2 t)$

$=e^{\cos 2 t}(-\sin 2 t) \frac{d}{d t}(2 t)$

$=e^{\cos 2 t}(-\sin 2 t)(2)$

$\frac{d x}{d t}=-2 \sin 2 t e^{\cos 2 t}$     .....(1)

And, $y=e^{\sin 2 t}$

Differentiating it with respect to $\theta$ using chain rule,

$\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{e}^{\sin 2 \mathrm{t}}\right)$

$=e^{\sin 2 t} \frac{d}{d t}(\sin 2 t)$

$=e^{\sin 2 t} \cos 2 t \frac{d}{d t}(2 t)$

$=e^{\sin 2 t} \cos 2 t(2)$

$\frac{d y}{d t}=2 \cos 2 t e^{\sin 2 t}$        ...(2)

dividing equation (2)by (1),

$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{2 \cos 2 t e^{\sin 2 t}}{-2 \sin 2 t e^{\cos 2 t}}$

$\frac{d y}{d x}=-\frac{y \log x}{x \log y}\left[\right.$ since $\left.x=e^{\cos 2 t} \Rightarrow \log x=\cos 2 t\right]$

$\left[y=e^{\sin 2 t} \Rightarrow \log y=\sin 2 t\right]$