Given that y=t^2sin^-1t,find the derivative of the function y
We can use the product rule and the chain rule to differentiate y:
y = t^2sin^-1t
First, let's find the derivative of sin^-1t:
f(t) = sin^-1t
f'(t) = 1/√(1-t^2) (using the formula for the derivative of sin^-1x)
Now, using the product rule, we have:
y' = (t^2)'sin^-1t + t^2(sin^-1t)'
y' = 2tsin^-1t + t^2(1/√(1-t^2))
Simplifying this expression, we get:
y' = t(2sin^-1t + t/√(1-t^2))
Therefore, the derivative of the function y is given by y' = t(2sin^-1t + t/√(1-t^2)).