Question

In Exercises \(1-8,\) find the derivative of \(y\) with respect to the appropriate variable. $$y=\sin ^{-1} \sqrt{2} t$$

Short answer

The derivative of \(y\) with respect to \(t\) is \(y' = \frac{\sqrt{2}}{\sqrt{1-2t^2}}\).

Step-by-step solution

Identify the Inner and Outer Functions

In the given function \(y = \sin^{-1}\sqrt{2}t\), the outer function is \(\sin^{-1}(x)\) and the inner function is \(\sqrt{2}t\).

Derivative of the Outer Function

The derivative of \(\sin^{-1}(x)\) is \(\frac{1}{\sqrt{1-x^2}}\). Therefore, the derivative of the outer function, \(\sin^{-1}(\sqrt{2}t)\) is \(\frac{1}{\sqrt{1-(\sqrt{2}t)^2}} = \frac{1}{\sqrt{1-2t^2}}\).

Derivative of the Inner Function

The derivative of \(\sqrt{2}t\) is \(\sqrt{2}\).

Application of the Chain Rule

By the chain rule, the derivative of \(y\) with respect to \(t\) is the derivative of the outer function times the derivative of the inner function. Therefore, \(y' = \frac{1}{\sqrt{1-2t^2}} \times \sqrt{2}\).