To derive the expression $\frac{-2x}{(1+x^2)^2}$, we will use the quotient rule for derivatives, as it is a ratio of two functions. The quotient rule states that if you have a function of the form $\frac{u(x)}{v(x)}$, then its derivative is:

$\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$

Here, $u(x) = -2x$ and $v(x) = (1 + x^2)^2$. Now, we compute the derivatives of $u(x)$ and $v(x)$.

1. Derivative of $u(x) = -2x$:

$u'(x) = \frac{d}{dx}(-2x) = -2$

2. Derivative of $v(x) = (1 + x^2)^2$:

To find the derivative of $v(x)$, we need to use the chain rule. Let $w(x) = 1 + x^2$, so $v(x) = w(x)^2$. Applying the chain rule:

$v'(x) = 2w(x) \cdot w'(x)$

Now, compute $w'(x)$:

$w'(x) = \frac{d}{dx}(1 + x^2) = 2x$

So,

$v'(x) = 2(1 + x^2) \cdot 2x = 4x(1 + x^2)$

Now, apply the quotient rule:

$\frac{d}{dx} \left(\frac{-2x}{(1 + x^2)^2}\right) = \frac{-2 \cdot (1 + x^2)^2 - (-2x) \cdot 4x(1 + x^2)}{(1 + x^2)^4}$

Simplify the numerator:

$= \frac{-2(1 + x^2)^2 + 8x^2(1 + x^2)}{(1 + x^2)^4}$

Factor $(1 + x^2)$ from both terms in the numerator:

$= \frac{(1 + x^2) \left[-2(1 + x^2) + 8x^2\right]}{(1 + x^2)^4}$

Simplify inside the brackets:

$-2(1 + x^2) + 8x^2 = -2 - 2x^2 + 8x^2 = -2 + 6x^2$

So the expression becomes:

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

Cancel one factor of $(1 + x^2)$ from the numerator and denominator:

$= \frac{-2 + 6x^2}{(1 + x^2)^3}$

Thus, the derivative of $\frac{-2x}{(1 + x^2)^2}$ is:

$\frac{d}{dx} \left(\frac{-2x}{(1 + x^2)^2}\right) = \frac{-2 + 6x^2}{(1 + x^2)^3}$

Would you like further details on any of the steps?

---

Here are 5 related questions:

1. How do you apply the chain rule to more complex functions?
2. Can you derive $\frac{x}{1+x^2}$ using the quotient rule?
3. What are some common applications of the quotient rule in calculus?
4. How do you differentiate composite functions using the chain rule?
5. How do you simplify derivatives that involve both the product and quotient rules?

Tip: Always simplify your expressions before differentiating, as it may make the derivative easier to compute.