Answer:
[tex]\frac{dy}{dx}\text{ = }3x^2\cdot e^{-x^3}(1-x^3)\text{ }[/tex]
Explanation:
Here, we want to find the derivative of the given function
To do this, we are going to use the product rule
Mathematically, we have the product rule as follows:
[tex]\frac{dy}{dx}\text{ = u}\frac{dv}{dx}\text{ + v}\frac{du}{dx}[/tex]
where:
[tex]\begin{gathered} u=x^3 \\ v=e^{-x^3} \end{gathered}[/tex]
We proceed as follows to find the unit derivatives:
[tex]\begin{gathered} \frac{du}{dx}=3x^2 \\ \\ \text{for }\frac{dv}{dx} \\ \\ \text{let w = -x}^3 \\ v=e^w \\ \frac{dw}{dx}=-3x^2 \\ \frac{dv}{dw}=e^w \\ \\ \frac{dv}{dx}\text{ = }\frac{dw}{dx}\times\frac{dv}{dw} \\ \\ =-3x^2\times e^w \\ =-3x^2\text{ }\times e^{-x^3} \end{gathered}[/tex]
We put together the final answer as follows:
[tex]\begin{gathered} \frac{dy}{dx}=x^3\times(-3x^2\times e^{-x^3})+e^{-x^3}(3x^2) \\ \\ \\ \frac{dy}{dx}=3x^2\cdot e^{-x^3}(-x^3+1) \\ \frac{dy}{dx}\text{ = }3x^2\cdot e^{-x^3}(1-x^3)\text{ } \end{gathered}[/tex]
