Certainly! Let's differentiate the function [tex]\( y = -4 \sin(1 - x^3) \)[/tex] with respect to [tex]\( x \)[/tex]. We will use the chain rule to accomplish this.
1. Identify the outer function and the inner function:
- The outer function is [tex]\( -4 \sin(u) \)[/tex] where [tex]\( u = 1 - x^3 \)[/tex].
2. Differentiate the outer function with respect to the inner function:
- The derivative of [tex]\( -4 \sin(u) \)[/tex] with respect to [tex]\( u \)[/tex] is [tex]\( -4 \cos(u) \)[/tex].
3. Differentiate the inner function with respect to [tex]\( x \)[/tex]:
- The inner function is [tex]\( u = 1 - x^3 \)[/tex].
- The derivative of [tex]\( 1 - x^3 \)[/tex] with respect to [tex]\( x \)[/tex] is [tex]\( -3x^2 \)[/tex].
4. Apply the chain rule:
- According to the chain rule, the derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex] is the derivative of the outer function with respect to the inner function multiplied by the derivative of the inner function with respect to [tex]\( x \)[/tex].
Let's put it all together:
[tex]\[ \frac{dy}{dx} = \frac{d}{du} \left( -4 \sin(u) \right) \cdot \frac{du}{dx} \][/tex]
Plugging in the respective derivatives we have:
[tex]\[ \frac{dy}{dx} = \left( -4 \cos(u) \right) \cdot \left( -3x^2 \right) \][/tex]
5. Substitute back in the inner function [tex]\( u = 1 - x^3 \)[/tex]:
[tex]\[ \frac{dy}{dx} = (-4 \cos(1 - x^3)) \cdot (-3x^2) \][/tex]
Simplifying this expression:
[tex]\[ \frac{dy}{dx} = 12x^2 \cos(x^3 - 1) \][/tex]
Therefore, the derivative of [tex]\( y = -4 \sin(1 - x^3) \)[/tex] with respect to [tex]\( x \)[/tex] is:
[tex]\[ \boxed{12x^2 \cos(x^3 - 1)} \][/tex]
