How do you differentiate $y = {csc}^{- 1} (\frac{x}{2})$ $y=csc^-1(x/2)$ ?

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Answer 1 · 2016-11-30T02:54:11

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

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

$csc y = \frac{x}{2}$ $cscy = x/2$

$sin y = \frac{2}{x}$ $siny = 2/x$

$cos y \frac{d y}{d x} = - \frac{2}{x^{2}}$ $cosy dy/dx = -2/x^2$ (Implicit differentiation and Power rule)

$\frac{d y}{d x} = - \frac{2}{x^{2}} \cdot \frac{1}{cos y}$ $dy/dx = -2/x^2 * 1/cosy$

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

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

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

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

How do you differentiate y=csc^-1(x/2)y=csc−1(x2)y=csc^-1(x/2)?