Final answer:
The derivative of arcsin(1/t) is -1/(t*sqrt(t^2 - 1)), using the chain rule and the derivative of the arcsine function, and is defined for t^2 > 1.
Step-by-step explanation:
To find the derivative of the function y = arcsin(1/t), we need to use the chain rule along with the knowledge of the derivative of the arcsine function. The derivative of arcsin(u) is 1/(sqrt(1-u^2)), where u is a function of t. Applying the chain rule, we differentiate u = 1/t with respect to t, which gives -1/t^2. Combining these, the derivative of y with respect to t is:
dy/dt = -1/(t^2 * sqrt(1 - (1/t)^2))
This simplifies to:
dy/dt = -1/(t*sqrt(t^2 - 1))
It's important to note that this function is only defined when t^2 > 1, ensuring the expression under the square root is non-negative.
To find the derivative of arcsin(1/t), we can use the chain rule. Let's start by letting u = 1/t. We can rewrite the original expression as arcsin(u). The derivative of arcsin(u) with respect to u is 1/sqrt(1-u^2). Using the chain rule, the derivative of arcsin(1/t) with respect to t is (1/sqrt(1-(1/t)^2)) * (-1/t^2). Simplifying further, we get -1/(t*sqrt(t^2-1)).