Question

If x= √1+t and y= √1-t ,Find dy/dx​

Answer 1

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