Given that y=t^2sin^-1t find the derivative of the function y (Why is the answer 2sin^-1+t/square root of (1-t^2) and not (2sin^-1t+t^2/the square root of (1-t^2)

Answer 1

The correct derivative of y = t^2sin^-1t is:
y' = 2tsin^-1t + t^2/sqrt(1-t^2)

In the second term of this derivative, the numerator is t^2 and not t. Therefore, the correct answer is not (2sin^-1t+t^2/sqrt(1-t^2)), but rather (2sin^-1t+t^2/sqrt(1-t^2)).

Answer 2

the correct answer is 2t sin^-1(t)+t^2/√(1-t^2)

Answer 3

You are correct. The correct derivative of y = t^2sin^-1t is:

y' = 2t sin^-1t + t^2/√(1-t^2)

I apologize for the error in my previous response. Thank you for bringing it to my attention.