Question

If x = e${}^{sin3t}$ , y = e${}^{cos3t}$ , show that $\frac{dy}{dx}=-\frac{y\ln x}{x\ln y}$

Answer 1

$x=e^{\sin 3t}\Rightarrow \ln x=\sin 3t$

$\Rightarrow \frac{1}{x}\frac{dx}{dt}=3\cos 3t$

$\Rightarrow \frac{dx}{dt}=3x\cos 3t$

$y=e^{\cos 3t}\Rightarrow \ln y=\cos 3t$

$\Rightarrow \frac{1}{y}\frac{dy}{dt}=-3\sin 3t$

$\Rightarrow \frac{dy}{dt}=-3y\sin 3t$

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

$=\frac{-3y\sin 3t}{3x\cos 3t}$

$=\frac{-y\sin 3t}{x\cos 3t}$

$=\frac{-y\ln x}{x\ln y}$ where $\ln x=\sin 3t$ and $\ln y=\cos 3t$