If x= 1+ t^2 ,y= 3+2 t , find dy dx - Vidyadip

Question

If $\text{x}=\frac{1+\log\text{t}}{\text{t}^2},\text{y}=\frac{3+2\log\text{t}}{\text{t}},$ find $\frac{\text{dy}}{\text{dx}}$

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Answer

$\text{x}=\frac{1+\log\text{t}}{\text{t}^2},\text{y}=\frac{3+2\log\text{t}}{\text{t}}$
$\frac{\text{dy}}{\text{dt}}=\frac{\text{t}^2\big(\frac{1}{\text{t}}\big)-(1+\log\text{t})(2\text{t})}{\text{t}^4} \\ =\frac{\text{t}-2\text{t}-2\text{t}\log\text{t}}{\text{t}^4}=\frac{-2\log\text{t}-1}{\text{t}^3}$
$\frac{\text{dy}}{\text{dt}}=\frac{\text{t}\big(\frac{2}{\text{t}}\big)-(3+2\log\text{t})(1)}{\text{t}^2} \\ =\frac{2-3-2\log\text{t}}{\text{t}^2}=\frac{-2\log\text{t}-1}{\text{t}^2}$
$\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\frac{-2\log\text{t}-1}{\text{t}^2}}{\frac{-2\log\text{t}-1}{\text{t}^3}}=\text{t}$

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