\[\frac{d}{dx} \sin^{2}x\cos^{3}x\]

1
Use Product Rule to find the derivative of sin2xcos3x\sin^{2}x\cos^{3}x. The product rule states that (fg)=fg+fg(fg)'=f'g+fg'.
(ddxsin2x)cos3x+sin2x(ddxcos3x)(\frac{d}{dx} \sin^{2}x)\cos^{3}x+\sin^{2}x(\frac{d}{dx} \cos^{3}x)

2
Use Chain Rule on ddxsin2x\frac{d}{dx} \sin^{2}x. Let u=sinxu=\sin{x}. Use Power Rule: dduun=nun1\frac{d}{du} {u}^{n}=n{u}^{n-1}.
2sinx(ddxsinx)cos3x+sin2x(ddxcos3x)2\sin{x}(\frac{d}{dx} \sin{x})\cos^{3}x+\sin^{2}x(\frac{d}{dx} \cos^{3}x)

3
Use Trigonometric Differentiation: the derivative of sinx\sin{x} is cosx\cos{x}.
2sinxcos4x+sin2x(ddxcos3x)2\sin{x}\cos^{4}x+\sin^{2}x(\frac{d}{dx} \cos^{3}x)

4
Use Chain Rule on ddxcos3x\frac{d}{dx} \cos^{3}x. Let u=cosxu=\cos{x}. Use Power Rule: dduun=nun1\frac{d}{du} {u}^{n}=n{u}^{n-1}.
2sinxcos4x+sin2x×3cos2x(ddxcosx)2\sin{x}\cos^{4}x+\sin^{2}x\times 3\cos^{2}x(\frac{d}{dx} \cos{x})

5
Use Trigonometric Differentiation: the derivative of cosx\cos{x} is sinx-\sin{x}.
2sinxcos4x3sin3xcos2x2\sin{x}\cos^{4}x-3\sin^{3}x\cos^{2}x

Done

How can we make this solution more helpful?