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If $x$ and $y$ are connected parametrically by the equations given in $x = 4t, \: y = \large\frac{4}{t} $ without eliminating the parameter, Find $\large\frac{dy}{dx}$

$\begin{array}{1 1} \frac{-1}{t^3} \\\frac{1}{t^2} \\\frac{-1}{t^2} \\\frac{1}{t^3}\end{array} $

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Toolbox:
  • By chain rule we have $\large\frac{dy}{dx}=\frac{dy}{dt}$$\times\large\frac{dt}{dx}$
Step 1:
Given :
$x=4t$
$\large\frac{dx}{dt}=$$4$
$f'(t)=4$
Step 2:
$y=\large\frac{4}{t}$
$\large\frac{dy}{dt}$$=4t^{-1}$
$\quad\;=-1(4)(t^{-2})$
$\quad\;=\large\frac{-4}{t^2}$
$g'(t)=\large\frac{-4}{t^2}$
Step 3:
$\large\frac{dy}{dx}=\frac{dy}{dt}$$\times\large\frac{dt}{dx}$
$\quad\;=\large\frac{-4}{t^2}$$\times \large\frac{1}{4}$
$\quad\;=\large\frac{-1}{t^2}$
answered May 10, 2013 by sreemathi.v
 

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