Question

if x = log (1+t2), y= t - tan -1 t show that day/dx = √ ex-1/2.​

Answer 1

Step-by-step explanation:

Given :

$x = log(1 + {t}^{2} ) ; \: y \: = t \: - \: { tan}^{ - 1}t$

To prove :

$\frac{dy}{dx} = \frac{1}{2} \sqrt{ {e}^{x} - 1}$

Solution :

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

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

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

Now,

$Since, \: x = log(1 + {t}^{2} ) \\ \therefore \: (1 + {t}^{2} ) = {e}^{x} \\ \therefore \: {t}^{2} = {e}^{x} - 1 \\ \therefore \: t \: = \sqrt{ {e}^{x} - 1} \\ \therefore \: \frac{t}{2} = \frac{1}{2} \sqrt{ {e}^{x} - 1} \\ \implies \: \frac{dy}{dx} \: = \frac{1}{2} \sqrt{ {e}^{x} - 1}$

Hence, proved.