We are tasked with finding the derivative of the function:

$h(t) = \sin^{-1}(2t)$

Step-by-Step Solution:

1. Recall the derivative of the inverse sine function: The derivative of $\sin^{-1}(x)$ with respect to $x$ is:

   $\frac{d}{dx} \sin^{-1}(x) = \frac{1}{\sqrt{1 - x^2}}$

2. Chain Rule Application: Since $h(t) = \sin^{-1}(2t)$, we have an inner function $g(t) = 2t$. To differentiate, we'll apply the chain rule.

   The chain rule states: $\frac{d}{dt} \sin^{-1}(g(t)) = \frac{1}{\sqrt{1 - [g(t)]^2}} \cdot g'(t)$

3. Differentiate $g(t) = 2t$: $g'(t) = 2$

4. Substitute $g(t) = 2t$ and $g'(t) = 2$ into the chain rule: $h'(t) = \frac{1}{\sqrt{1 - (2t)^2}} \cdot 2$

5. Simplify: $h'(t) = \frac{2}{\sqrt{1 - 4t^2}}$

Final Answer:

$h'(t) = \frac{2}{\sqrt{1 - 4t^2}}$

Let me know if you'd like any further clarification!

Related Questions:

1. What is the derivative of $h(t) = \cos^{-1}(3t)$?
2. How does the chain rule apply to composite trigonometric functions?
3. How do you find the derivative of $h(t) = \tan^{-1}(5t)$?
4. What is the domain of the derivative of $h(t) = \sin^{-1}(2t)$?
5. How can you differentiate $h(t) = \sin^{-1}(t^2)$ using the chain rule?

Tip:

When differentiating inverse trigonometric functions, always remember the domain restrictions of the original function to avoid errors.