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




Question Number 211944 by liuxinnan last updated on 25/Sep/24
  f(x)=(((√(1+x))−(√(1−x)))/( (√(1+x))+(√(1−x))))    f^′ (x)=?
$$ \\ $$$${f}\left({x}\right)=\frac{\sqrt{\mathrm{1}+{x}}−\sqrt{\mathrm{1}−{x}}}{\:\sqrt{\mathrm{1}+{x}}+\sqrt{\mathrm{1}−{x}}}\:\:\:\:{f}^{'} \left({x}\right)=? \\ $$$$ \\ $$$$ \\ $$
Answered by efronzo1 last updated on 25/Sep/24
 f(x)= ((((√(1+x))−(√(1−x)))^2 )/(2x))    f(x)= ((2−2(√(1−x^2 )))/(2x))= (1/x)−(√((1/x^2 )−1))    f ′(x)=−(1/x^2 ) −((−(2/x^3 ))/(2(√((1/x^2 )−1))))
$$\:\mathrm{f}\left(\mathrm{x}\right)=\:\frac{\left(\sqrt{\mathrm{1}+\mathrm{x}}−\sqrt{\mathrm{1}−\mathrm{x}}\right)^{\mathrm{2}} }{\mathrm{2x}} \\ $$$$\:\:\mathrm{f}\left(\mathrm{x}\right)=\:\frac{\mathrm{2}−\mathrm{2}\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}{\mathrm{2x}}=\:\frac{\mathrm{1}}{\mathrm{x}}−\sqrt{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }−\mathrm{1}} \\ $$$$\:\:\mathrm{f}\:'\left(\mathrm{x}\right)=−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:−\frac{−\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{3}} }}{\mathrm{2}\sqrt{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }−\mathrm{1}}} \\ $$

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