Question

if x=log(1+t²) and y=t- tan-¹t, then dy/DX is equal to​

Answer 1

$x = \sf \log(1 + {t}^{2} )$

$\implies \frac{dx}{dt} =( \frac{1}{1 + {t}^{2} } )(2t)$

$\therefore \frac{dx}{dt} = \frac{2t}{1 + {t}^{2} } ...(1)$

$y = t - \tan ^{ - 1} t$

$\implies\frac{dy}{dt} = 1 - \frac{1}{1 + {t}^{2} }$

$\implies\frac{dy}{dt} = \frac{1 + {t}^{2} - 1}{1 + {t}^{2} }$

$\therefore\frac{dy}{dt} = \frac{ {t}^{2} }{1 + {t}^{2} }...(2)$

$\frac{dy}{dx} = \large{ \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} } }$

$\implies \frac{dy}{dx} = \frac{ \frac{ {t}^{2} }{1 + {t}^{2} } }{ \frac{2t}{1 + {t}^{2} } } [\sf from\: (1) and (2)]$

$\implies \frac{dy}{dx} = \frac{ {t}^{2} }{2t }$

$\orange{ \therefore \frac{dy}{dx} = \frac{ t }{2 } }$

HOPE IT HELPS!!

Answer 2

Step-by-step explanation:

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