\(sin^{2}x \ cos^{2} x= \frac{1}{2} (1-cos2x) \times \frac{1}{2} (1+cos2x) \\ = \frac{1}{4} (1- cos^{2} 2x) \\ = \frac{1}{4} ( sin^{2} 2x +cos^{2} 2x- cos^{2} 2x) \\ = \frac{1}{4} ( sin^{2} 2x) \\ = \frac{1}{4} ( \frac{1}{2} (1-cos2(2x))) \\ = \frac{1}{8} (1-cos4x) \\ \frac{d}{dx} (sin^{2}x \ cos^{2} x)= \frac{1}{8} \frac{d}{dx} (1-cos4x) \\ = \frac{1}{8} [-4(-sin4x) \\ \frac{1}{2} sin4x\)
