How do you find the derivative of $arcsin (\frac{2 x}{1 + x^{2}})$ $arcsin((2x)/(1+x^2))$ ?

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Answer 1 · 2017-01-24T19:44:13

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

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

$sin y = \frac{2 x}{1 + x^{2}}$ $siny=(2x)/(1+x^2)$

$\frac{d y}{d x} = \frac{1}{\frac{d x}{d y}}$ $dy/dx=1/(dx/dy)$

$\frac{d x}{d y} = cos y$ $dx/dy=cosy$

$\frac{d y}{d x} = \frac{1}{cos y}$ $dy/dx=1/cosy$

${sin}^{2} y + {cos}^{2} y = 1$ $sin^2y+cos^2y=1$

${cos}^{2} y = 1 - {sin}^{2} y$ $cos^2y=1-sin^2y$

$cos y = \sqrt{1 - {sin}^{2} y} = \sqrt{1 - \frac{4 x^{2}}{{(x^{2} + 1)}^{2}}}$ $cosy=sqrt(1-sin^2y)=sqrt(1-(4x^2)/(x^2+1)^2$

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

How do you find the derivative of arcsin(2x1+x2)arcsin((2x)/(1+x^2))?