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.

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, $u = \frac{x^2 + 1}{1 - x^2}$.

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

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

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

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: $\frac{dy}{dx} = \frac{4x}{2(x^4 + 1)} = \frac{2x}{x^4 + 1}.$

Final Answer:

$\frac{dy}{dx} = \frac{2x}{x^4 + 1}.$

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

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 $x = 0$.

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