Let's go through each step of the problem using the product rule and calculating the derivatives as provided. I'll break down the calculations clearly to ensure full understanding.

Step 1: Product Rule

Given a function $y = u(x) \cdot v(x)$, the product rule states that the derivative of $y$ with respect to $x$ is:

$\frac{d}{dx} (u(x) \cdot v(x)) = u'(x) \cdot v(x) + u(x) \cdot v'(x).$

Here, $u(x)$ and $v(x)$ are two functions of $x$. To apply the product rule, we first need to compute the derivatives of $u(x)$ and $v(x)$.

Step 2: Derivative of $u(x)$

Let's find the derivative of $u(x) = 2 \cdot \tan^{-1}(x)$.

1. Function $u(x) = 2 \cdot \tan^{-1}(x)$:

   • $\tan^{-1}(x)$ is the inverse tangent function, also known as the arctangent.

2. Derivative of $\tan^{-1}(x)$:

   • The derivative of $\tan^{-1}(x)$ with respect to $x$ is $\frac{1}{x^2 + 1}$.

3. Using the Chain Rule:

   • Since $u(x) = 2 \cdot \tan^{-1}(x)$, we apply the constant multiple rule and chain rule:

   $u'(x) = 2 \cdot \frac{d}{dx} (\tan^{-1}(x)) = 2 \cdot \frac{1}{x^2 + 1}.$

Thus, the derivative of $u(x)$ is:

$u'(x) = \frac{2}{x^2 + 1}.$

Step 3: Derivative of $v(x)$

Given $v(x) = \frac{1}{x^2 + 1}$, let's find its derivative.

1. Function $v(x) = \frac{1}{x^2 + 1}$:

   • This is a rational function of the form $\frac{1}{g(x)}$, where $g(x) = x^2 + 1$.

2. Derivative using the Quotient Rule:

   • To differentiate $\frac{1}{x^2 + 1}$, we rewrite it as $(x^2 + 1)^{-1}$.

   • Using the chain rule, the derivative of $(x^2 + 1)^{-1}$ is:

     $v'(x) = -\frac{1 \cdot 2x}{(x^2 + 1)^2}.$

   This simplifies to:

   $v'(x) = -\frac{2x}{(x^2 + 1)^2}.$

Step 4: Applying the Product Rule

Now, substitute the derivatives $u'(x)$ and $v'(x)$ into the product rule:

$\frac{d}{dx} (u(x) \cdot v(x)) = u'(x) \cdot v(x) + u(x) \cdot v'(x).$

Substituting the values:

$\frac{d}{dx} (2 \cdot \tan^{-1}(x) \cdot \frac{1}{x^2 + 1}) = \left(\frac{2}{x^2 + 1}\right) \cdot \left(\frac{1}{x^2 + 1}\right) + \left(2 \cdot \tan^{-1}(x)\right) \cdot \left(-\frac{2x}{(x^2 + 1)^2}\right).$

Simplify each term:

1. First term: $\frac{2}{x^2 + 1} \cdot \frac{1}{x^2 + 1} = \frac{2}{(x^2 + 1)^2}.$

2. Second term: $2 \cdot \tan^{-1}(x) \cdot \left(-\frac{2x}{(x^2 + 1)^2}\right) = -\frac{4x \cdot \tan^{-1}(x)}{(x^2 + 1)^2}.$

Thus, the complete derivative is:

$\frac{d}{dx} (u(x) \cdot v(x)) = \frac{2}{(x^2 + 1)^2} - \frac{4x \cdot \tan^{-1}(x)}{(x^2 + 1)^2}.$

Summary of Steps:

1. Apply the product rule: $u'(x) \cdot v(x) + u(x) \cdot v'(x)$.
2. Find $u'(x)$ using the chain rule: $u'(x) = \frac{2}{x^2 + 1}$.
3. Find $v'(x)$ using the quotient rule: $v'(x) = -\frac{2x}{(x^2 + 1)^2}$.
4. Substitute and simplify the expression.

Would you like to go over any part of this process in more detail or need help with related questions? Here are some follow-up questions:

1. How would the result change if $u(x)$ or $v(x)$ were different?
2. What are the implications of the negative term in the product rule application?
3. How can we simplify this expression further for specific values of $x$?
4. What is the interpretation of the product rule in real-world applications?
5. How would we integrate the resulting derivative expression?

Tip: When dealing with products of functions, always ensure each function's derivative is accurately computed before applying the product rule to avoid mistakes.