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.

Learn how to find the derivative of the function f(x) = arcsin(2x - 1) using the chain rule and the derivative formula for arcsine. This solution breaks down each step, including applying the chain rule, simplifying, and factoring. Suitable for Grades 11-12 and early college students studying calculus.

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