1. **Problem:** Find the derivative $\frac{d}{dx}(\sec^2(x^3))$. 2. **Formula and rules:** Use the chain rule: if $y = [u(x)]^2$, then $\frac{dy}{dx} = 2u(x) \cdot u'(x)$. Also, $\frac{d}{dx}(\sec x) = \sec x \tan x$. 3. **Step-by-step solution:** - Let $u = \sec(x^3)$, so the function is $y = u^2$. - Then $\frac{dy}{dx} = 2u \cdot \frac{du}{dx}$. - Find $\frac{du}{dx}$: since $u = \sec(v)$ with $v = x^3$, then $\frac{du}{dx} = \sec v \tan v \cdot \frac{dv}{dx}$. - Compute $\frac{dv}{dx} = 3x^2$. - Substitute back: $\frac{du}{dx} = \sec(x^3) \tan(x^3) \cdot 3x^2$. 4. **Combine all:** $$\frac{dy}{dx} = 2 \sec(x^3) \cdot \sec(x^3) \tan(x^3) \cdot 3x^2 = 6x^2 \sec^2(x^3) \tan(x^3)$$ **Final answer:** $$\boxed{\frac{d}{dx}(\sec^2(x^3)) = 6x^2 \sec^2(x^3) \tan(x^3)}$$