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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Find \(\large \frac{dy}{dx}\) if $ x=\large\frac{1+\log\:t}{t^2}$$,y= \large\frac{3+2log\: t}{t} $

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Step 1:
$x=\large\frac{1+\log t}{t^2}$
$y=\large\frac{3+2\log t}{t}$
Differentiating with respect to $x$ we get,
$\large\frac{dx}{dt}=\frac{t^2(1/t)-(1+\log t)(2t)}{(t^2)^2}$
$\qquad=\large\frac{t-2t-2t\log t}{t^4}$
$\qquad=\large\frac{-1-2\log t}{t^3}$
$\large\frac{dy}{dt}=\large\frac{t.\large\frac{2}{t}-(3+2\log t).1}{t^2}$
$\qquad=\large\frac{2-3-2\log t}{t^2}$
$\qquad=\large\frac{-1-2\log t}{t^2}$
Step 2:
$\therefore\large\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}$
$\large\frac{dy}{dx}=-\frac{(1+2\log t)}{t^2}$$\times \large\frac{t^3}{-(1+2\log t)}$
$\qquad=t$
answered Nov 7, 2013 by sreemathi.v
 

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