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We are tasked with solving for \( 2f(\cot(x)) \) where the function \( f(x) = \frac{x^2}{\sqrt{1 + x^2}} \). Let's proceed step by step:

### Step 1: Substitute \( x = \cot(x) \) into \( f(x) \)

We are given \( f(x) = \frac{x^2}{\sqrt{1 + x^2}} \), and we need to find \( f(\cot(x)) \).

So, we substitute \( \cot(x) \) into the equation:
\[
f(\cot(x)) = \frac{\cot^2(x)}{\sqrt{1 + \cot^2(x)}}
\]

### Step 2: Simplify the expression

Using the identity \( 1 + \cot^2(x) = \csc^2(x) \), we can simplify the square root term:
\[
f(\cot(x)) = \frac{\cot^2(x)}{\sqrt{\csc^2(x)}} = \frac{\cot^2(x)}{\csc(x)}
\]

Now, use the identity \( \cot(x) = \frac{\cos(x)}{\sin(x)} \) and \( \csc(x) = \frac{1}{\sin(x)} \):
\[
f(\cot(x)) = \frac{\left( \frac{\cos^2(x)}{\sin^2(x)} \right)}{\frac{1}{\sin(x)}} = \frac{\cos^2(x)}{\sin(x)}
\]

### Step 3: Multiply by 2

We are asked to find \( 2f(\cot(x)) \):
\[
2f(\cot(x)) = 2 \cdot \frac{\cos^2(x)}{\sin(x)} = \frac{2\cos^2(x)}{\sin(x)}
\]

### Step 4: Compare with the options

The final expression \( \frac{2\cos^2(x)}{\sin(x)} \) matches option **D**, which is \( 2\cos(x) \cdot \cot(x) \), because:
\[
\frac{2\cos^2(x)}{\sin(x)} = 2\cos(x) \cdot \frac{\cos(x)}{\sin(x)} = 2\cos(x) \cdot \cot(x)
\]

Thus, the correct answer is **D**.

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Would you like more details on any part of the solution?

Here are 5 related questions to practice:

1. What is the derivative of \( f(x) = \frac{x^2}{\sqrt{1 + x^2}} \)?
2. How do you prove the identity \( 1 + \cot^2(x) = \csc^2(x) \)?
3. Can you express \( f(\tan(x)) \) for the same function \( f(x) \)?
4. What is the integral of \( \cot(x) \)?
5. What is the range of the function \( f(x) = \frac{x^2}{\sqrt{1 + x^2}} \)?

**Tip:** Always remember common trigonometric identities like \( \sin^2(x) + \cos^2(x) = 1 \) and \( 1 + \cot^2(x) = \csc^2(x) \) as they are frequently used in problems involving trigonometric functions.

Agar f(x) = x^2 / sqrt(1 + x^2) bo'lsa, 2f(ctgx) ni toping.

Learn how to solve for 2f(ctgx) given the function f(x) = x^2 / sqrt(1 + x^2) by substituting trigonometric identities and simplifying step-by-step. Ideal for high school students in grades 10-12 studying advanced trigonometry and functions.

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