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Q)

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

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

• 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$
$f'(t)=4$
$y=\large\frac{4}{t}$
$\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}$
$\quad\;=\large\frac{-1}{t^2}$