To differentiate the function [tex]\( y = \frac{x^4}{1 - x^3} \)[/tex], we will use the quotient rule for differentiation. The quotient rule states that for two functions [tex]\( u(x) \)[/tex] and [tex]\( v(x) \)[/tex], where [tex]\( y = \frac{u(x)}{v(x)} \)[/tex], the derivative [tex]\( \frac{dy}{dx} \)[/tex] is given by:
[tex]\[ \frac{dy}{dx} = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \][/tex]
In our case, let's identify [tex]\( u(x) \)[/tex] and [tex]\( v(x) \)[/tex]:
- [tex]\( u(x) = x^4 \)[/tex]
- [tex]\( v(x) = 1 - x^3 \)[/tex]
First, differentiate [tex]\( u(x) \)[/tex] and [tex]\( v(x) \)[/tex] with respect to [tex]\( x \)[/tex]:
- [tex]\( \frac{du}{dx} = \frac{d}{dx}(x^4) = 4x^3 \)[/tex]
- [tex]\( \frac{dv}{dx} = \frac{d}{dx}(1 - x^3) = -3x^2 \)[/tex]
Now, apply the quotient rule:
[tex]\[ \frac{dy}{dx} = \frac{(4x^3)(1 - x^3) - (x^4)(-3x^2)}{(1 - x^3)^2} \][/tex]
Simplify the numerator:
[tex]\[ \begin{aligned}
\text{Numerator: } & 4x^3(1 - x^3) + 3x^6 \\
& = 4x^3 - 4x^6 + 3x^6 \\
& = 4x^3 - x^6
\end{aligned} \][/tex]
So the expression for [tex]\( \frac{dy}{dx} \)[/tex] becomes:
[tex]\[ \frac{dy}{dx} = \frac{4x^3 - x^6}{(1 - x^3)^2} \][/tex]
Factor [tex]\( x^3 \)[/tex] out of the numerator:
[tex]\[ \frac{dy}{dx} = \frac{x^3(4 - x^3)}{(1 - x^3)^2} \][/tex]
Therefore, the derivative of [tex]\( y = \frac{x^4}{1 - x^3} \)[/tex] is:
[tex]\[ \frac{dy}{dx} = \frac{x^3(4 - x^3)}{(x^3 - 1)^2} \][/tex]
Based on this result, the correct answer is:
[tex]\[ \frac{x^3(4 - x^3)}{(x^3 - 1)^2} \][/tex]
This corresponds to the third option given:
[tex]\[ \boxed{\frac{x^3 \left(4 - x^3 \right)}{\left(x^3 - 1\right)^2}} \][/tex]
