Calculate the derivative of the given function. Show that sin-1x=tan-1(x/sqrt1+x^2) if x is smaller than 1.   Where is the function continous and differentiable?

Expert Answers

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I am not sure what function you want to take the derivative of but I can help with the formula you want verified.  So I will do the second part.

We know that if sin^(-1)(x) = y then x = sin(y)

`tan(y) = sin(y)/cos(y) = sin(y)/(sqrt(1-sin^2(y)))`

Substituting x for sin(y) we get

`tan(y) = x/(sqrt(1-x^2))`

Taking the inverse tangent of both sides we get and noting that `tan^(-1)(tan(y)) = y`

`y = tan^(-1)(x/sqrt(1-x^2))` and since `y = sin^(-1)(x)` we have

`sin^(-1)(x) = tan^(-1)(x/sqrt(1-x^2))`

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