Question #53041

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Answer 1 · 2017-10-17T07:49:36

$dy/dx=(-1)/(x^2sqrt(1-1/x^2)$ or $dy/dx=(-1)/(xsqrt(x^2-1)$

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

Therefore,

$y=sin^1(1/x)$

We will differentiate both sides using the chain rule.

$d/dxy=d/dx(sin^-1(1/x))$

$dy/dx=1/sqrt(1-(1/x)^2)d/dx(1/x)$

$dy/dx=1/sqrt(1-1/x^2)d/dx(x^-1)$

$dy/dx1/sqrt(1-1/x^2)((-1)/x^2)$

$dy/dx=(-1)/(x^2sqrt(1-1/x^2)$