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Find $\large \frac{dy}{dx}$ of the function expressed in parametric form $x=\large\frac{1+log t}{t^2},$$y=\large\frac{3+2log t}{t}$

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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}$$=\large\frac{\Large\frac{dy}{dt}}{\Large\frac{dx}{dt}}$
Step 1:
Given $x=\large\frac{1+\log t}{t^2}$
$y=\large\frac{3+2\log t}{t}$
Consider $x=\large\frac{1+\log t}{t^2}$
This can be written as $x=t^{-2}(1+\log t)$
Differentiating w.r.t $x$ we get,(By applying product rule)
Let $u=t^{-2}$
$\large\frac{du}{dt}=$$-2t^{-3}$
Let $v=(1+\log t)$
$\large\frac{dv}{dt}=\frac{1}{t}$
Therefore $\large\frac{dx}{dt}=$$u.\large\frac{dv}{dt}+v.\large\frac{du}{dt}$
$\qquad\qquad\;\;\;\;=t^{-2}\big(\large\frac{1}{t}\big)$$+(1+\log t).(-2t^{-3})$
$\qquad\qquad\;\;\;\;=\large\frac{1}{t^3}-\frac{2}{t^3}$$(1+\log t)$
$\qquad\qquad\;\;\;\;=\large\frac{-1}{t^3}$$\big(1+2\log t)$
Step 2:
Consider $y=\large\frac{3+2\log t}{t}$
This can be written as
$y=(t^{-1})(3+2\log t)$
Now differentiating w.r.t $t$ by applying product rule,
Let $u=t^{-1}$
$\large\frac{du}{dt}=-t^{-2}$
Let $v=3+2\log t$
$\large\frac{dv}{dt}$$=2.\large\frac{1}{t}$
$\large\frac{dy}{dt}$$=t^{-1}.3\big(\large\frac{1}{t}\big)$$+(3+2\log t).1$
$\quad\;\;\;=\large\frac{-(3+2\log t)}{t^2}+\frac{2}{t^2}$
$\quad\;\;\;=\large\frac{-1}{t^2}$$(1+2\log t)$
Step 3:
Therefore $\large\frac{dy}{dx}$$=\large\frac{\Large\frac{dy}{dt}}{\Large\frac{dx}{dt}}$
$\qquad\qquad=\large\frac{-1/t^2(1+2\log t)}{-1/t^3(1+2\log t)}$
$\qquad\qquad=t$
$\large\frac{dy}{dx}$$=t$
answered Jul 1, 2013 by sreemathi.v
 

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