Differentiate Sin4X Over X3 048029

1. **State the problem:** Differentiate the function $$f(x) = \frac{\sin 4x}{x^3}$$ with respect to $x$. 2. **Formula used:** We use the quotient rule for differentiation, which states: $$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$$ where $u = \sin 4x$ and $v = x^3$. 3. **Find derivatives of numerator and denominator:** - $\frac{du}{dx} = \frac{d}{dx} (\sin 4x) = 4 \cos 4x$ - $\frac{dv}{dx} = \frac{d}{dx} (x^3) = 3x^2$ 4. **Apply quotient rule:** $$\frac{d}{dx} \left( \frac{\sin 4x}{x^3} \right) = \frac{x^3 (4 \cos 4x) - \sin 4x (3x^2)}{(x^3)^2}$$ 5. **Simplify numerator:** $$= \frac{4x^3 \cos 4x - 3x^2 \sin 4x}{x^6}$$ 6. **Factor out common term $x^2$ in numerator:** $$= \frac{x^2 (4x \cos 4x - 3 \sin 4x)}{x^6}$$ 7. **Cancel common factors:** $$= \frac{\cancel{x^2} (4x \cos 4x - 3 \sin 4x)}{x^{\cancel{2}} x^4} = \frac{4x \cos 4x - 3 \sin 4x}{x^4}$$ **Final answer:** $$\boxed{\frac{d}{dx} \left( \frac{\sin 4x}{x^3} \right) = \frac{4x \cos 4x - 3 \sin 4x}{x^4}}$$