In Exercises 21–48, find the derivative of y with respect to the appropriate variable.
23. y=arcsin(√2t)

Verified step by step guidance

1. Identify the function to differentiate: \(y = \arcsin(\sqrt{2t})\) where the variable is \(t\).
2. Recall the derivative formula for the inverse sine function: \(\frac{d}{dx} \arcsin(u) = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}\), where \(u\) is a function of \(x\).
3. Set \(u = \sqrt{2t} = (2t)^{1/2}\). Find the derivative of \(u\) with respect to \(t\): \(\frac{du}{dt} = \frac{1}{2} (2t)^{-1/2} \cdot 2\) using the chain rule.
4. Substitute \(u\) and \(\frac{du}{dt}\) into the derivative formula: \(\frac{dy}{dt} = \frac{1}{\sqrt{1 - (\sqrt{2t})^2}} \cdot \frac{du}{dt}\).
5. Simplify the expression inside the square root and the derivative to write the final derivative in terms of \(t\) (without calculating the numeric value).

Key Concepts

Derivative of Inverse Trigonometric Functions

The derivative of arcsin(x) with respect to x is 1 divided by the square root of (1 minus x squared), i.e., d/dx[arcsin(x)] = 1/√(1 - x²). This formula is essential when differentiating functions involving arcsin.

Chain Rule

The chain rule is used to differentiate composite functions. If y = f(g(t)), then dy/dt = f'(g(t)) * g'(t). Here, arcsin(√2t) is a composite function requiring the chain rule to differentiate the outer and inner functions.

Derivative of Square Root Functions

The derivative of √u, where u is a function of t, is (1/(2√u)) * du/dt. Recognizing that √2t can be written as (2t)^(1/2) helps in applying the power rule combined with the chain rule.