To differentiate the function \( \frac{x}{(1-x)^2} \), we will use the quotient rule, which states that if we have a function \( \frac{u(x)}{v(x)} \), then its derivative is given by:

\[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \]

Here, \( u(x) = x \) and \( v(x) = (1-x)^2 \).

First, we find the derivatives of \( u(x) \) and \( v(x) \):

\[ u'(x) = 1 \]

To find \( v'(x) \), we use the chain rule. Let \( w = 1 - x \), so \( v(x) = w^2 \). Then:

\[ \frac{d}{dx} (w^2) = 2w \cdot \frac{dw}{dx} = 2(1-x) \cdot (-1) = -2(1-x) \]

Now we apply the quotient rule:

\[ \left( \frac{x}{(1-x)^2} \right)' = \frac{(1) \cdot (1-x)^2 - x \cdot (-2(1-x))}{((1-x)^2)^2} \]

Simplify the numerator:

\[ (1-x)^2 + 2x(1-x) \]

Expand and combine like terms:

\[ (1 - 2x + x^2) + (2x - 2x^2) = 1 - 2x + x^2 + 2x - 2x^2 = 1 - x^2 \]

Thus, the derivative is:

\[ \frac{1 - x^2}{(1-x)^4} \]

This simplifies to:

\[ \frac{1 - x^2}{(1-x)^4} = \frac{1 - x^2}{(1-x)^2 (1-x)^2} = \frac{(1-x)(1+x)}{(1-x)^4} = \frac{1+x}{(1-x)^3} \]

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

\[ \frac{1 + x}{(1 - x)^3} \]

Would you like details or have any questions? Here are some questions you might consider next:

1. How do you integrate the function \( \frac{x}{(1-x)^2} \)?
2. Can you explain the quotient rule in more detail?
3. What is the derivative of \( \frac{1}{(1-x)^2} \)?
4. How do you apply the chain rule in differentiation?
5. What is the second derivative of \( \frac{x}{(1-x)^2} \)?

**Tip:** When dealing with quotient functions, always consider the quotient rule, but also explore simplifications that might make the differentiation process easier.

Learn how to differentiate the function x/(1-x)^2 using the quotient rule and chain rule. This problem explores advanced differentiation techniques and is suitable for high school students studying calculus.

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