To find the derivative of the function $f(x) = \arcsin(2x - 1)$, we will use the chain rule combined with the derivative of the arcsine function.

Step-by-step solution:

1. Recall the derivative formula for arcsin: $\frac{d}{dx}[\arcsin(u)] = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}$ where $u$ is a function of $x$.

2. Apply the chain rule: In our case, $u = 2x - 1$, so we need to find $\frac{du}{dx}$ first: $\frac{du}{dx} = 2$

3. Differentiate $f(x) = \arcsin(2x - 1)$: Now, using the chain rule: $f'(x) = \frac{1}{\sqrt{1 - (2x - 1)^2}} \cdot 2$

4. Simplify the expression inside the square root: $(2x - 1)^2 = 4x^2 - 4x + 1$ So the derivative becomes: $f'(x) = \frac{2}{\sqrt{1 - (4x^2 - 4x + 1)}}$

5. Simplify the denominator: $1 - (4x^2 - 4x + 1) = -4x^2 + 4x$ Hence, the derivative is: $f'(x) = \frac{2}{\sqrt{4x - 4x^2}}$

6. Final simplified form: Factoring the denominator: $f'(x) = \frac{2}{2\sqrt{x(1 - x)}} = \frac{1}{\sqrt{x(1 - x)}}$

Thus, the derivative of $f(x) = \arcsin(2x - 1)$ is: $f'(x) = \frac{1}{\sqrt{x(1 - x)}}$

Would you like further details or have any questions?

Here are some related questions:

1. What is the derivative of $\arccos(2x - 1)$?
2. How would you find the second derivative of $f(x) = \arcsin(2x - 1)$?
3. What is the domain of $f(x) = \arcsin(2x - 1)$?
4. How does the derivative change if the function were $\arcsin(3x - 2)$?
5. What is the derivative of $\arcsin(x^2)$?

Tip: When dealing with inverse trigonometric functions, always pay attention to the domain of the input to ensure the function and its derivative are well-defined.