Question

1. (\frac{d}{dx}cos^2(x^3) =)

Explanation:

Step1: Apply Chain Rule (Outer Function)

Let \( u = \cos(x^3) \), then the function becomes \( y = u^2 \). The derivative of \( y \) with respect to \( u \) is \( \frac{dy}{du} = 2u \). Substituting back \( u = \cos(x^3) \), we get \( \frac{dy}{du} = 2\cos(x^3) \).

Step2: Apply Chain Rule (Middle Function)

Now, we need the derivative of \( u = \cos(x^3) \) with respect to \( v = x^3 \). The derivative of \( \cos(v) \) with respect to \( v \) is \( -\sin(v) \), so \( \frac{du}{dv} = -\sin(x^3) \).

Step3: Apply Chain Rule (Inner Function)

Next, the derivative of \( v = x^3 \) with respect to \( x \) is \( \frac{dv}{dx} = 3x^2 \).

Step4: Multiply the Derivatives

By the chain rule, \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx} \). Substituting the values we found: \( \frac{dy}{dx} = 2\cos(x^3) \cdot (-\sin(x^3)) \cdot 3x^2 \).

Step5: Simplify the Expression

Using the double - angle identity \( \sin(2\theta)=2\sin\theta\cos\theta \), we can rewrite \( 2\cos(x^3)\sin(x^3)=\sin(2x^3) \). So the expression becomes \( - 6x^2\cos(x^3)\sin(x^3)=-3x^2\sin(2x^3) \) (or we can also leave it as \( - 6x^2\cos(x^3)\sin(x^3) \)).

Answer:

\( - 6x^2\cos(x^3)\sin(x^3) \) (or \( - 3x^2\sin(2x^3) \))