Question #53041

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Question #53041

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Answer 1 · 2017-10-17T07:49:36

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

Explanation:

We know the derivative of $({sin}^{- 1} x)$ $(sin^-1x)$ is $(\frac{1}{\sqrt{1 - x^{2}}})$ $(1/sqrt(1-x^2))$

Therefore,

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

We will differentiate both sides using the chain rule.

$\frac{d}{d x} y = \frac{d}{d x} ({sin}^{- 1} (\frac{1}{x}))$ $d/dxy=d/dx(sin^-1(1/x))$

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

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

$\frac{d y}{d x} \frac{1}{\sqrt{1 - \frac{1}{x^{2}}}} (\frac{- 1}{x^{2}})$ $dy/dx1/sqrt(1-1/x^2)((-1)/x^2)$

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

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Question #53041

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