Differentiating sin¯¹(x) & cos¯¹(x)

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  1. Eddie Woo
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Avoiding confusion over which sign applies to what derivative is really easy. First, draw the right triangle that satisfies y = sin^-1 (x). Next, determine dx/dy = cos (y) from your sketch. It will be sqrt{1 - x^2}. Finally, flip this result upside down according to the inverse function theorem: dy/dx = 1/sqrt{1 - x^2}. Note that there is no negative sign in this result.
If calculating the derivative dcos^-1 (x)/dx then from a similar drawing as before sin(y) = sqrt {1 - x^2}, but dx/dy = - sin(y) in this case. Therefore, dx/dy = - sqrt{1 - x^2}. Flipping this upside down gives dy/dx = - 1/sqrt{1 - x^2} ◼
Easy. No worries, no decision points just straight forward calculations.

johnnolen
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Mr Woo, for figuring out the derivative of inverse sine of x, I drew up a right-angled triangle to rewrite cos y in terms of x. It seems to avoid the plus minus drama. Is this fine, or would you say your method is better?

rafiashraf