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.

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.