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To find \( f'(x) \) for the function \( f(x) = \tan^{-1} \left( \sqrt{5x^2 - 1} \right) \), we will use the chain rule.

1. **Differentiate the outer function**: The derivative of \( \tan^{-1}(u) \) with respect to \( u \) is \( \frac{1}{1 + u^2} \).
2. **Differentiate the inner function**: The inner function here is \( u = \sqrt{5x^2 - 1} \). 

Let's proceed step-by-step.

### Step 1: Differentiate the outer function
Let \( u = \sqrt{5x^2 - 1} \), so:
\[
f(x) = \tan^{-1}(u)
\]
Then:
\[
f'(x) = \frac{1}{1 + u^2} \cdot u'
\]

### Step 2: Substitute \( u = \sqrt{5x^2 - 1} \) and find \( u' \)
Since \( u = \sqrt{5x^2 - 1} \), we have:
\[
u^2 = 5x^2 - 1
\]
Thus, \( 1 + u^2 = 1 + (5x^2 - 1) = 5x^2 \).

Now, we need to differentiate \( u = \sqrt{5x^2 - 1} \) with respect to \( x \):
\[
u' = \frac{d}{dx} \left( \sqrt{5x^2 - 1} \right) = \frac{1}{2\sqrt{5x^2 - 1}} \cdot (10x) = \frac{5x}{\sqrt{5x^2 - 1}}
\]

### Step 3: Substitute everything back
Now we can substitute \( u = \sqrt{5x^2 - 1} \) and \( u' = \frac{5x}{\sqrt{5x^2 - 1}} \) into the expression for \( f'(x) \):
\[
f'(x) = \frac{1}{5x^2} \cdot \frac{5x}{\sqrt{5x^2 - 1}}
\]

Simplifying this:
\[
f'(x) = \frac{5x}{5x^2 \sqrt{5x^2 - 1}} = \frac{x}{x^2 \sqrt{5x^2 - 1}}
\]

Thus:


Let f(x) = tan^(-1)(√(5x^2 - 1)). Find f'(x).

Learn how to find the derivative of f(x) = tan^(-1)(√(5x^2 - 1)) using the chain rule. This problem covers key calculus concepts, including inverse trigonometric functions and differentiation. Ideal for undergraduate calculus students.

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