f-x-1-x-2-x-2-1-f-2-

Question Number 51446 by gunawan last updated on 27/Dec/18

f (x) = \sqrt{\frac{1 - x^{2}}{x^{2} + 1}}

f^{^{'}} (- 2) = \dots

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Dec/18

y^{2} = \frac{1 - x^{2}}{1 + x^{2}}

2 y . \frac{d y}{d x} = \frac{(1 + x^{2}) (- 2 x) - (1 - x^{2}) (2 x)}{{(1 + x^{2})}^{2}}

\frac{d y}{d x} = 2 \sqrt{\frac{1 - x^{2}}{1 + x^{2}}} [\frac{- 2 x - 2 x^{3} - 2 x + 2 x^{3}}{{(1 + x^{2})}^{2}}]

\frac{d y}{d x} = 2 \sqrt{\frac{1 - x^{2}}{1 + x^{2}}} [\frac{- 4 x}{{(1 + x^{2})}^{2}}]

{(\frac{d y}{d x})}_{a t x = - 2} = 2 \sqrt{\frac{1 - 4}{5}} (\frac{8}{25}) = \frac{2 \times \sqrt{3} i \times 8}{25 \sqrt{5}} = \frac{16 \sqrt{3} i}{25 \sqrt{5}}