Question

solve find dy/dx isko solve karo

Answer 1

Hi...☺

Here is your answer...✌

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

Now,

$\frac{dy}{dx} = \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} } = \frac{ \frac{ - 1 - 2 \log t}{ {t}^{2} } }{ \frac{- 1 - 2 \log t}{ {t}^{3} } } \\ \\ \implies \frac{dy}{dx} = t$

Answer 2

Answer:  dy/dx=t

Explanation:

x= 1+logt/t²

dx/dt= 1/t ×t²-2t(1+logt)/(t²)²

        = t-2t(1+logt)/(t²)

        = 1-2-2㏒t/t³

        = -1-2log t / t³

y= 3+2logt /t

dy/dx= 2/t×t-1(3+2logt)/t

         = 2-3-2log t/ t²

         = -1-2logt/t²

dy/dx = -1-2log t /t² × t³/-1-2log t

         = t