How do you differentiate y=sin^-1(1/x)?

1 Answer
Steve M
Oct 27, 2016

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

Explanation:

The easiest way is to rewrite y=sin^-1(1/x) as siny=1/x

:. siny=x^-1

Then, differentiating simplicity gives:

cosydy/dx=-x^-2
:. dy/dx=(-1)/(x^2cosy)

And, using the trig odentity sin^2A+cos^2A-=1 we have
cosy=sqrt(1-sin^2y)
:. cosy=sqrt(1-(1/x)^2)
:. cosy=sqrt(x^2/x^2-1/x^2)
:. cosy=sqrt((x^2-1)/x^2)
:. cosy=1/xsqrt(x^2-1)

And, so substituting into the derivative above we get;
:. dy/dx=(-1)/(x^2(1/xsqrt(x^2-1)))
Hence, dy/dx=(-1)/(xsqrt(x^2-1))

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