Question #d4611

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Question #d4611

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Answer 1 · 2017-10-08T01:09:32

$\frac{d y}{d x} = - \frac{2}{x^{2} + 1}$ $dy/dx=-2/(x^2+1)$ or $\frac{- 2 x^{- 1}}{x + x^{- 1}}$ $(-2x^(-1))/[x+x^(-1)]$

Explanation:

$y = {cos}^{- 1} [\frac{x - x^{- 1}}{x + x^{- 1}}]$ $y=cos^(-1) [(x-x^(-1))/(x+x^(-1))]$

= ${cos}^{- 1} [\frac{x^{2} - 1}{x^{2} + 1}]$ $cos^(-1) [(x^2-1)/(x^2+1)]$

Take cosine both sides,

$cos y = \frac{x^{2} - 1}{x^{2} + 1}$ $cosy=(x^2-1)/(x^2+1)$

Take differentiation both sides,

$- sin y \cdot d y = [2 x \cdot (x^{2} + 1) - 2 x \cdot (x^{2} - 1)] \cdot \frac{d x}{{(x^{2} + 1)}^{2}}$ $-siny*dy=[2x*(x^2+1)-2x*(x^2-1)]*dx/(x^2+1)^2$

$- \sqrt{1 - {(cos y)}^{2}} \cdot d y = \frac{4 x \cdot d x}{{(x^{2} + 1)}^{2}}$ $-sqrt[1-(cosy)^2]*dy=(4x*dx)/(x^2+1)^2$

$- \sqrt{1 - {[\frac{x^{2} - 1}{x^{2} + 1}]}^{2}} \cdot d y = \frac{4 x \cdot d x}{{(x^{2} + 1)}^{2}}$ $-sqrt(1-[(x^2-1)/(x^2+1)]^2)*dy=(4x*dx)/(x^2+1)^2$

$- \sqrt{\frac{4 x^{2}}{{(x^{2} + 1)}^{2}}} \cdot d y = \frac{4 x \cdot d x}{{(x^{2} + 1)}^{2}}$ $-sqrt[(4x^2)/(x^2+1)^2]*dy=(4x*dx)/(x^2+1)^2$

$- \frac{2 x \cdot d y}{x^{2} + 1} = \frac{4 x \cdot d x}{{(x^{2} + 1)}^{2}}$ $-(2x*dy)/(x^2+1)=(4x*dx)/(x^2+1)^2$

$\frac{d y}{d x} = \frac{4 x \cdot d x}{{(x^{2} + 1)}^{2}} \cdot - \frac{x^{2} + 1}{2 x}$ $dy/dx=(4x*dx)/(x^2+1)^2*-(x^2+1)/(2x)$

$\frac{d y}{d x} = - \frac{2}{x^{2} + 1}$ $dy/dx=-2/(x^2+1)$ or $\frac{- 2 x^{- 1}}{x + x^{- 1}}$ $(-2x^(-1))/[x+x^(-1)]$

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Question #d4611

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