Given that y=t^2sin^-1t find the derivative of the function y
Taking the derivative of y with respect to t, we apply the product rule:
y = t^2sin^-1t
y' = 2t sin^-1t + t^2(d/dt)(sin^-1t)
We can find the derivative of sin^-1t using the chain rule:
(d/dt)(sin^-1t) = 1 / √(1-t^2)
So substituting back into the equation for y':
y' = 2t sin^-1t + t^2(1 / √(1-t^2))
Simplifying this expression, we get:
y' = 2t sin^-1t + t / √(1-t^2)
Thus, the derivative of the function y is y' = 2t sin^-1t + t / √(1-t^2).