What is the derivative of $arcsin (\frac{1}{x})$ $arcsin(1/x)$ ?

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What is the derivative of $arcsin (\frac{1}{x})$ $arcsin(1/x)$ ?

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Answer 1 · 2018-02-16T13:27:30

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

Explanation:

To find $\frac{d}{d x} (arcsin (\frac{1}{x}))$ $d/dx(arcsin(1/x))$
$arcsin (\frac{1}{x}) = {sin}^{- 1} (\frac{1}{x})$ $arcsin(1/x)=sin^-1(1/x)$
Let $y = {sin}^{=} 1 (\frac{1}{x})$ $y=sin^=1(1/x)$
Let $u = \frac{1}{x}$ $u=1/x$
$\frac{d u}{d x} = \frac{- 1}{x^{2}}$ $(du)/dx=(-1)/x^2$
$y = {sin}^{- 1} u$ $y=sin^-1u$
$\frac{d y}{d x} = \frac{1}{\sqrt{1 - u^{2}}} \frac{d u}{d x}$ $(dy)/dx=1/sqrt(1-u^2)(du)/dx$
Substituting the values of $u and \frac{d u}{d x}$ $u and (du)/dx$
we have

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

Simplifying

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

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

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What is the derivative of $arcsin (\frac{1}{x})$ $arcsin(1/x)$ ?

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