Using the product rule:

y2=ddx(u(x)⋅v(x))=u′(x)⋅v(x)+u(x)⋅v′(x)*y*2​=*dxd*​(*u*(*x*)⋅*v*(*x*))=*u*′(*x*)⋅*v*(*x*)+*u*(*x*)⋅*v*′(*x*)

Now, calculate the derivatives:

u′(x)=ddx(2⋅tan⁡−1(x))=2x2+1*u*′(*x*)=*dxd*​(2⋅tan−1(*x*))=*x*2+12​

v(x)=1x2+1*v*(*x*)=*x*2+11​, so v′(x)=ddx(1x2+1)=−2x(x2+1)2*v*′(*x*)=*dxd*​(*x*2+11​)=−(*x*2+1)22*x*​
explain each step

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.

Explore how to use the product rule for differentiation step-by-step with examples. Learn the chain rule and quotient rule applications in calculus. Suitable for Grades 11-12.

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