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)
the correct answer is 2t sin^-1(t)+t^2/√(1-t^2)
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.
To find the derivative of the function y = t^2 sin^(-1)t, we need to apply the chain rule.
The chain rule states that if we have a composite function y = f(g(t)), then the derivative dy/dt is given by:
dy/dt = f'(g(t)) * g'(t)
Let's break down the given function y = t^2 sin^(-1)t and apply the chain rule step by step.
Step 1: Identify the outer function f(u) and the inner function g(t).
In this case, the outer function is f(u) = u^2, where u = sin^(-1)t.
The inner function is g(t) = sin^(-1)t.
Step 2: Find the derivatives of the outer and inner functions.
To differentiate f(u) = u^2, we have f'(u) = 2u.
To differentiate g(t) = sin^(-1)t, we have g'(t) = 1/sqrt(1 - t^2).
Step 3: Apply the chain rule formula and substitute the derivatives.
According to the chain rule, dy/dt = f'(g(t)) * g'(t).
Plugging in the derivatives, we get dy/dt = 2u * (1/sqrt(1 - t^2)).
Step 4: Substitute u back into the equation.
Our original function was y = t^2 sin^(-1)t, where u = sin^(-1)t.
So, u = sin^(-1)t implies t = sin(u).
Substituting t back into the equation, we get dy/dt = 2sin^(-1)t * (1/sqrt(1 - (sin(u))^2)).
Next, we need to express sin(u) in terms of t using the trigonometric identity sin^2(u) + cos^2(u) = 1. Rearranging, we have sin(u) = sqrt(1 - cos^2(u)).
Substituting this expression into the equation, we get dy/dt = 2sin^(-1)t * (1/sqrt(1 - (sqrt(1 - cos^2(u)))^2)).
Simplifying the equation gives dy/dt = (2sin^(-1)t) / sqrt(1 - (1 - cos^2(u))).
Further simplification leads to dy/dt = (2sin^(-1)t) / sqrt(cos^2(u)).
Since cos(u) = t, we can rewrite the equation as dy/dt = (2sin^(-1)t) / |cos(u)|.
Now, cos(u) = cos(sin^(-1)t), and cos(sin^(-1)t) = sqrt(1 - t^2).
So, we have dy/dt = (2sin^(-1)t) / |sqrt(1 - t^2)|.
Finally, we can rewrite |sqrt(1 - t^2)| as sqrt(1 - t^2).
Therefore, the derivative of y = t^2 sin^(-1)t is dy/dt = (2sin^(-1)t) / sqrt(1 - t^2).
Hence, the correct answer is (2sin^(-1)t) / sqrt(1 - t^2).
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)).