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Find $\large \frac{dy}{dx}$ of the function expressed in parametric form $\sin x=\large\frac{2t}{1+t^2},$$\tan y=\large\frac{2t}{1-t^2}$

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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}$$=\large\frac{\Large\frac{dy}{dt}}{\Large\frac{dx}{dt}}$
Step 1:
Given : $\sin x=\large\frac{2t}{1+t^2}$
$\tan y=\large\frac{2t}{1-t^2}$
Consider $\sin x=\large\frac{2t}{1+t^2}$
$\Rightarrow x=\sin^{-1}\big(\large\frac{2t}{1+t^2}\big)$
Put $t=\tan\theta$
$\theta=\tan^{-1}(t)$
$\Rightarrow x=\sin^{-1}\big(\large\frac{2\tan\theta}{1+\tan^2\theta}\big)$
But $\large\frac{2\tan \theta}{1+\tan^2\theta}=$$\sin 2\theta$
$x=\sin^{-1}(\sin 2\theta)$
Therefore $x=2\theta$
Substituting for $\theta$ we get,
$x=2\tan^{-1}(t)$
Step 2:
Differentiating w.r.t to $t$ we get,
$\large\frac{dx}{dt}$$=2.\large\frac{1}{1+t^2}=\frac{2}{1+t^2}$
Consider $\tan y=\large\frac{2t}{1+t^2}$
$\Rightarrow y=\tan^{-1}\big(\large\frac{2t}{1-t^2}\big)$
Put $t=\tan \theta$
$\Rightarrow \theta=\tan^{-1}(t)$
$y=\tan^{-1}\big(\large\frac{2\tan\theta}{1-\tan^2\theta}\big)$
But $\large\frac{2\tan\theta}{1-\tan^2\theta}=$$\tan 2\theta$
Therefore $y=\tan^{-1}(\tan 2\theta)$
$\qquad\qquad\;\;=2\theta$
Step 3:
Substituting for $\theta$ we get,
$y=2\tan^{-1}t$
Differentiating w.r.t $t$ we get,
$\large\frac{dy}{dt}=$$2.\large\frac{1}{1+t^2}$
$\Rightarrow \large\frac{2}{1+t^2}$
$\large\frac{dy}{dx}$$=\large\frac{\Large\frac{dy}{dt}}{\Large\frac{dx}{dt}}=\large\frac{2/1+t^2}{2/1+t^2}$
Therefore $\large\frac{dy}{dx}$$=1$
answered Jul 1, 2013 by sreemathi.v
 

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