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

$$f\left(x\right)=\sqrt{\frac{1-x^2}{x^2+1}}$$$$f^{{}^{\prime }}(-2)=...$$

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

$$y^2=\frac{1-x^2}{1+x^2}$$$$2y.\frac{dy}{dx}=\frac{(1+x^2)(-2x)-(1-x^2)\left(2x\right)}{{(1+x^2)}^2}$$$$\frac{dy}{dx}=2\sqrt{\frac{1-x^2}{1+x^2}}\left[\frac{-2x-2x^3-2x+2x^3}{{(1+x^2)}^2}\right]$$$$\frac{dy}{dx}=2\sqrt{\frac{1-x^2}{1+x^2}}\left[\frac{-4x}{{(1+x^2)}^2}\right]$$$${\left(\frac{dy}{dx}\right)}_{atx=-2}=2\sqrt{\frac{1-4}{5}}\left(\frac{8}{25}\right)=\frac{2\times \sqrt{3}i\times 8}{25\sqrt{5}}=\frac{16\sqrt{3}i}{25\sqrt{5}}$$

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