Find y' for y = sin²(x³)

Find $y'$ for $y = \sin^{2}\bigl(x^{3}\bigr)$.

Step 1: Identify the composition

Reading from the outside in, the function is a chain of three: \[y = u^{2}, \qquad u = \sin v, \qquad v = x^{3}.\]

Step 2: Apply the chain rule

\[ y' = \frac{dy}{du}\cdot\frac{du}{dv}\cdot\frac{dv}{dx} = 2u\cdot\cos v\cdot 3x^{2} = 2\sin\bigl(x^{3}\bigr)\cos\bigl(x^{3}\bigr)\cdot 3x^{2}. \]

Step 3: Apply the double-angle identity

With $2\sin\theta\cos\theta = \sin 2\theta$ and $\theta = x^{3}$: $y' = 3x^{2}\sin\bigl(2x^{3}\bigr)$.

Step 4: Verification

\[x = 0:\quad y = \sin^{2}0 = 0,\quad y'(0) = 0, \text{ a minimum} \;\;\checkmark\]

\[x = 1:\quad y' = 3\sin 2 \approx 2.72789; \qquad \frac{y(1+10^{-5})-y(1-10^{-5})}{2\times 10^{-5}} \approx 2.72789 \;\;\checkmark\]