differentiate y=tan^-1((1+x)^1+2-(1-x)^1+2+(1+x)^1+2+(1-x)^1+2) with respects to x

To differentiate the function \( y = \tan^{-1} \left( \frac{(1+x)^2 - (1-x)^2}{(1+x)^2 + (1-x)^2} \right) \) with respect to \( x \), we will first simplify the expression inside the arctangent function and then apply the chain rule for differentiation.

Step 1: Simplifying the Expression

Let's start by simplifying the fraction inside the arctangent. We have:

• Numerator: \( (1+x)^2 - (1-x)^2 \)
• Denominator: \( (1+x)^2 + (1-x)^2 \)

Calculating the numerator:

• Expand \( (1+x)^2 = 1 + 2x + x^2 \)
• Expand \( (1-x)^2 = 1 - 2x + x^2 \)
• Thus, \( (1+x)^2 - (1-x)^2 = (1 + 2x + x^2) - (1 - 2x + x^2) = 4x \)

Now for the denominator:

• Using the expansions again, we find:
• Denominator: \( (1+x)^2 + (1-x)^2 = (1 + 2x + x^2) + (1 - 2x + x^2) = 2 + 2x^2 \)

Putting it all together, we can rewrite \( y \) as:

\( y = \tan^{-1} \left( \frac{4x}{2 + 2x^2} \right) = \tan^{-1} \left( \frac{2x}{1 + x^2} \right) \)

Step 2: Differentiating Using the Chain Rule

Now, we differentiate \( y \) with respect to \( x \). The derivative of \( \tan^{-1}(u) \) is given by:

\( \frac{dy}{dx} = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \)

Here, \( u = \frac{2x}{1 + x^2} \). We need to find \( \frac{du}{dx} \) first.

Finding \( \frac{du}{dx} \)

Using the quotient rule, where \( u = \frac{f(x)}{g(x)} \) with \( f(x) = 2x \) and \( g(x) = 1 + x^2 \), we have:

• \( f'(x) = 2 \)
• \( g'(x) = 2x \)

Applying the quotient rule:

\( \frac{du}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} = \frac{2(1 + x^2) - 2x(2x)}{(1 + x^2)^2} \)

This simplifies to:

\( \frac{du}{dx} = \frac{2 + 2x^2 - 4x^2}{(1 + x^2)^2} = \frac{2 - 2x^2}{(1 + x^2)^2} = \frac{2(1 - x^2)}{(1 + x^2)^2} \)

Step 3: Putting It All Together

Now we can substitute \( u \) and \( \frac{du}{dx} \) back into the derivative of \( y \):

\( \frac{dy}{dx} = \frac{1}{1 + \left( \frac{2x}{1 + x^2} \right)^2} \cdot \frac{2(1 - x^2)}{(1 + x^2)^2} \)

Calculating \( 1 + \left( \frac{2x}{1 + x^2} \right)^2 \):

\( = 1 + \frac{4x^2}{(1 + x^2)^2} = \frac{(1 + x^2)^2 + 4x^2}{(1 + x^2)^2} = \frac{1 + 2x^2 + x^4 + 4x^2}{(1 + x^2)^2} = \frac{1 + 6x^2 + x^4}{(1 + x^2)^2} \)

Thus, the derivative becomes:

\( \frac{dy}{dx} = \frac{2(1 - x^2)}{(1 + x^2)^2} \cdot \frac{(1 + x^2)^2}{1 + 6x^2 + x^4} = \frac{2(1 - x^2)}{1 + 6x^2 + x^4} \)

Final Result

In conclusion, the derivative of the given function is:

\( \frac{dy}{dx} = \frac{2(1 - x^2)}{1 + 6x^2 + x^4} \)

This result captures how the function \( y \) changes with respect to \( x \) based on the simplifications and differentiation steps we've performed. If you have any further questions or need clarification on any part, feel free to ask!