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To differentiate the function \( \sin(3x + y^3) \), we use the chain rule. If \( u = 3x + y^3 \), then the derivative of \( \sin(u) \) with respect to \( x \) is \( \cos(u) \cdot \frac{du}{dx} \). We find: \[ \frac{du}{dx} = 3 + 3y^2 \frac{dy}{dx} \] Thus, applying the chain rule gives us: \[ \frac{d}{dx} \sin(3x + y^3) = \cos(3x + y^3)\left(3 + 3y^2 \frac{dy}{dx}\right) \] This result reflects how both \( x \) and \( y \) contribute to changes in the sine function! To further clarify, let’s assume \( y \) is implicitly defined as a function of \( x \). When you differentiate implicitly, you must include \( \frac{dy}{dx} \) — which indicates how \( y \) changes with \( x \). This is why you see \( 3y^2 \frac{dy}{dx} \) in the derivative! Happy differentiating!
