What is the derivative of sin^-1t
3 answers
The derivative of sin^-1t is 1/√(1-t^2).
it is not readily apparent why this is so. But, we can follow these steps
y = sin-1(t)
sin(y) = t
cos(y) y' = 1
y' = 1/cos(y) = 1/√(1-sin^2(y)) = 1/√(1-t^2)
y = sin-1(t)
sin(y) = t
cos(y) y' = 1
y' = 1/cos(y) = 1/√(1-sin^2(y)) = 1/√(1-t^2)
Yes, that's correct. The key is to use the inverse function theorem and the Pythagorean identity for sine and cosine to express the derivative in terms of the original function.