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′
 ( 
5
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 ) in simplest form?

The given function is \( f(x) = \sin^{-1}(x) \), which is the inverse sine function (also written as arcsin). We are tasked with finding \( f'(x) \), the derivative of \( f(x) \), and then evaluating it at \( x = \frac{2}{5} \).

### Step 1: Find the derivative of \( f(x) = \sin^{-1}(x) \)
The derivative of the inverse sine function is well-known:

\[
f'(x) = \frac{1}{\sqrt{1 - x^2}}
\]

### Step 2: Evaluate \( f'(x) \) at \( x = \frac{2}{5} \)

Substitute \( x = \frac{2}{5} \) into the derivative formula:

\[
f'\left( \frac{2}{5} \right) = \frac{1}{\sqrt{1 - \left(\frac{2}{5}\right)^2}}
\]

### Step 3: Simplify the expression

First, calculate \( \left(\frac{2}{5}\right)^2 \):

\[
\left(\frac{2}{5}\right)^2 = \frac{4}{25}
\]

Now, substitute that into the square root expression:

\[
f'\left( \frac{2}{5} \right) = \frac{1}{\sqrt{1 - \frac{4}{25}}}
\]

Simplify the expression inside the square root:

\[
1 - \frac{4}{25} = \frac{25}{25} - \frac{4}{25} = \frac{21}{25}
\]

Now substitute this back into the expression:

\[
f'\left( \frac{2}{5} \right) = \frac{1}{\sqrt{\frac{21}{25}}} = \frac{1}{\frac{\sqrt{21}}{5}} = \frac{5}{\sqrt{21}}
\]

### Step 4: Rationalize the denominator (optional)

To rationalize the denominator:

\[
f'\left( \frac{2}{5} \right) = \frac{5}{\sqrt{21}} \cdot \frac{\sqrt{21}}{\sqrt{21}} = \frac{5\sqrt{21}}{21}
\]

Thus, the value of \( f'\left( \frac{2}{5} \right) \) in its simplest form is:

\[
f'\left( \frac{2}{5} \right) = \frac{5\sqrt{21}}{21}
\]

---

Would you like more details on any steps? Here are five related questions:

1. What is the derivative of \( f(x) = \cos^{-1}(x) \)?
2. How do you differentiate inverse trigonometric functions in general?
3. What is the geometric interpretation of the derivative of an inverse sine function?
4. How would the calculation change if the function were \( \sin^{-1}(2x) \) instead of \( \sin^{-1}(x) \)?
5. Can you explain why we rationalize the denominator, and when it's necessary?

**Tip:** Always double-check your algebra when simplifying expressions, especially when dealing with square roots and fractions!

If f(x) = sin^(-1)(x), then what is the value of f'(2/5) in simplest form?

Learn how to find the derivative of the inverse sine function, f(x) = sin^(-1)(x), and evaluate it at x = 2/5. This detailed solution involves using the derivative formula for inverse trigonometric functions, simplifying square roots and fractions. Suitable for advanced high school and college-level calculus students.

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