Differentiation of $\frac{x^3}{1 - x^3}$ with respect to $x^3$ is equal to:

To differentiate $\frac{x^3}{1 - x^3}$ with respect to $x^3$, treat $x^3$ as a single variable.

Let $u = x^3$

The expression becomes:

$\frac{x^3}{1 - x^3} = \frac{u}{1-u}$

The task is to find $\frac{d}{du}\left(\frac{u}{1-u}\right)$

---

Using the quotient rule for $\frac{f(u)}{g(u)}$:

$\frac{d}{du}\left(\frac{f}{g}\right) = \frac{g \cdot f' - f \cdot g'}{g^2}$

Here:

• $f(u) = u$ so $f'(u) = 1$
• $g(u) = 1-u$ so $g'(u) = -1$

$\frac{d}{du}\left(\frac{u}{1-u}\right) = \frac{(1-u)(1) - (u)(-1)}{(1-u)^2}$

$= \frac{1-u + u}{(1-u)^2}$

$= \frac{1}{(1-u)^2}$

---

Substituting back $u = x^3$:

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

Note that $x \neq 1$ since the denominator becomes zero when $x = 1$.

Therefore, the answer is $\dfrac{1}{(1-x^3)^2}$ where $x \neq 1$.