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

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  • 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}$$=\large\frac{\Large\frac{dy}{dt}}{\Large\frac{dx}{dt}}$
Step 1:
$y=t-\large\frac{1}{t}$
$\Rightarrow 1-t^{-1}$
$x=t+\large\frac{1}{t}$
$\Rightarrow 1+t^{-1}$
$\large\frac{dy}{dt}$$=1-(-1t^{-2})$
$\quad=1+\large\frac{1}{t^2}$
$\large\frac{dx}{dt}$$=1+(-1t^{-2})$
$\quad=1-\large\frac{1}{t^2}$
Step 2:
Therefore $\large\frac{dy}{dx}=\large\frac{dy}{dt}\times\large\frac{dt}{dx}$
$\qquad\qquad\;\;\;=\large\frac{1+\large\frac{1}{t^2}}{1-\large\frac{1}{t^2}}$
On simplifying we get,
$\large\frac{dy}{dx}=\frac{t^2+1}{t^2-1}$
answered Jul 1, 2013 by sreemathi.v
 

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