Question

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

Answer 1

`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))`