What is the derivative of sin^-1t
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)
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.
To find the derivative of the inverse sine function, denoted as sin^(-1)(t) or arcsin(t), we can use the chain rule.
The derivative of the inverse sine function can be calculated as follows:
1. Let y = sin^(-1)(t).
2. Rewrite the equation as t = sin(y).
3. Differentiate both sides of the equation with respect to y: dt/dy = d(sin(y))/dy.
4. Recall that the derivative of sin(y) with respect to y is equal to cos(y): dt/dy = cos(y).
5. To solve for dy/dt (the derivative of y with respect to t), we take the reciprocal of both sides: dy/dt = 1/(dt/dy).
6. Substituting cos(y) for dt/dy, we have dy/dt = 1/cos(y).
To express the derivative in terms of t, we use the trigonometric identity: cos^2(y) + sin^2(y) = 1.
Since t = sin(y), we can substitute sin^2(y) with (t)^2 to get cos^2(y) = 1 - (t)^2.
Plugging this into the expression for dy/dt, we have:
dy/dt = 1/cos(y) = 1/√(1 - (t)^2).
Therefore, the derivative of sin^(-1)(t) with respect to t is equal to 1/√(1 - (t)^2).