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If $x=e^{\large\cos 2t}\; and\;y=e^{\large\sin 2t},prove\;that\;\large \frac{dy}{dx}=\frac{-ylog x}{xlog y}$

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Toolbox:
  • To find $\large\frac{dy}{dx}$ in the case of parametric functions,if $x=\phi(t)$ and $y=\psi(t)$,then $\large\frac{dy}{dx}$$=\large\frac{\Large\frac{dy}{dt}}{\Large\frac{dx}{dt}}$
Step 1:
$x=e^{\large\cos 2t}$
$\log x=\cos 2t$
$y=e^{\large\sin 2t}$
$\log y=\sin 2t$
Consider $x=e^{\large\cos 2t}$
Let us differentiate this w.r.t $t$
$\large\frac{dx}{dt}$$=e^{\large \cos 2t}\times (-\sin 2t)\times 2$
$\quad\;\;=-2\sin 2t\times e^{\large\cos 2t}$
Step 2:
Consider $y=e^{\sin 2t}$
Let us differentiate this w.r.t $t$,
$\large\frac{dy}{dt}=$$e^{\large\sin 2t}\times \cos 2t\times 2$
$\quad\;\;=2\cos 2t.e^{\large\sin 2t}$
Therefore $\large\frac{dy}{dx}$$=\large\frac{\Large\frac{dy}{dt}}{\Large\frac{dx}{dt}}=\large\frac{2\cos 2t e^{\Large\sin 2t}}{-2\sin 2te^{\Large\cos 2t}}$
$\qquad\qquad\;\;=\large\frac{\cos 2te^{\Large\sin 2t}}{\sin 2t.e^{\Large\cos 2t}}$
Step 3:
But $e^{\large\sin 2t}$$=y$
$e^{\large\cos 2t}=x$
$\log x=\cos 2t$
$\log y=\sin t$
Therefore $\large\frac{dy}{dx}=\frac{-y}{x}\frac{\log x}{\log y}$
Hence proved.
answered Jul 1, 2013 by sreemathi.v
 

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