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To determine the derivative \( \frac{dy}{dx} \) for \( y = \tan^{-1} \left( \frac{x^2 + 1}{1 - x^2} \right) \), let us solve this step-by-step.

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### Step 1: Recall the derivative formula for \(\tan^{-1}(u)\)
The derivative of \(\tan^{-1}(u)\) with respect to \(x\) is:
\[
\frac{d}{dx} \tan^{-1}(u) = \frac{1}{1 + u^2} \cdot \frac{du}{dx}.
\]

Here, \( u = \frac{x^2 + 1}{1 - x^2} \).

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### Step 2: Differentiate \( u \)
We need the derivative of:
\[
u = \frac{x^2 + 1}{1 - x^2}.
\]

Using the quotient rule:
\[
\frac{du}{dx} = \frac{(1 - x^2)(2x) - (x^2 + 1)(-2x)}{(1 - x^2)^2}.
\]

Simplify the numerator:
\[
(1 - x^2)(2x) - (x^2 + 1)(-2x) = 2x - 2x^3 + 2x^3 + 2x = 4x.
\]

Thus:
\[
\frac{du}{dx} = \frac{4x}{(1 - x^2)^2}.
\]

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### Step 3: Plug \( u \) and \( \frac{du}{dx} \) into the formula
Now substitute \( u = \frac{x^2 + 1}{1 - x^2} \) and \( \frac{du}{dx} = \frac{4x}{(1 - x^2)^2} \) into the derivative formula for \(\tan^{-1}(u)\):
\[
\frac{dy}{dx} = \frac{1}{1 + \left( \frac{x^2 + 1}{1 - x^2} \right)^2} \cdot \frac{4x}{(1 - x^2)^2}.
\]

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### Step 4: Simplify the expression
Simplify \( 1 + \left( \frac{x^2 + 1}{1 - x^2} \right)^2 \):
\[
\left( \frac{x^2 + 1}{1 - x^2} \right)^2 = \frac{(x^2 + 1)^2}{(1 - x^2)^2}.
\]

So:
\[
1 + \left( \frac{x^2 + 1}{1 - x^2} \right)^2 = \frac{(1 - x^2)^2 + (x^2 + 1)^2}{(1 - x^2)^2}.
\]

Expand both terms in the numerator:
\[
(1 - x^2)^2 = 1 - 2x^2 + x^4,
\]
\[
(x^2 + 1)^2 = x^4 + 2x^2 + 1.
\]

Adding them:
\[
(1 - x^2)^2 + (x^2 + 1)^2 = (1 - 2x^2 + x^4) + (x^4 + 2x^2 + 1) = 2x^4 + 2.
\]

Thus:
\[
1 + \left( \frac{x^2 + 1}{1 - x^2} \right)^2 = \frac{2x^4 + 2}{(1 - x^2)^2}.
\]

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### Step 5: Final expression
Substitute back into \( \frac{dy}{dx} \):
\[
\frac{dy}{dx} = \frac{1}{\frac{2x^4 + 2}{(1 - x^2)^2}} \cdot \frac{4x}{(1 - x^2)^2}.
\]

Simplify:
\[
\frac{dy}{dx} = \frac{(1 - x^2)^2}{2x^4 + 2} \cdot \frac{4x}{(1 - x^2)^2}.
\]

Cancel \((1 - x^2)^2\):
\[
\frac{dy}{dx} = \frac{4x}{2x^4 + 2}.
\]

Factor \(2\) from the denominator:
\[
\frac{dy}{dx} = \frac{4x}{2(x^4 + 1)} = \frac{2x}{x^4 + 1}.
\]

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### Final Answer:
\[
\frac{dy}{dx} = \frac{2x}{x^4 + 1}.
\]

Let me know if you want additional explanations or have any questions!  

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Here are 5 related questions for practice:
1. What is the derivative of \( y = \tan^{-1}(x^2) \)?
2. Find the derivative of \( y = \tan^{-1}\left(\frac{1}{x}\right) \).
3. How would you differentiate \( y = \cot^{-1}(x) \)?
4. What is the second derivative of \( y = \tan^{-1}(x) \)?
5. Evaluate the derivative of \( y = \tan^{-1}(e^x) \) at \( x = 0 \).

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**Tip:** When differentiating inverse trigonometric functions, always identify the argument (u) and apply the chain rule carefully!

Determine the derivative dy/dx if y = tan^(-1)((x^2 + 1) / (1 - x^2)). Simplify the expression of the derivative.

Step-by-step solution to find the derivative of y = tan^(-1)((x^2 + 1) / (1 - x^2)) using the chain rule, quotient rule, and properties of inverse trigonometric functions. Simplify the derivative into its final form.

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