0000:00:00:00:00Start28Number 28ExplanationThe first derivative of y=cos3xy = \cos^3 xy=cos3x is ....3cosx⋅sinx3\cos x \cdot \sin x3cosx⋅sinx−3cosx⋅sinx-3\cos x \cdot \sin x−3cosx⋅sinx2cosx⋅sin2x2\cos x \cdot \sin 2x2cosx⋅sin2x−6cosx⋅sin2x-6\cos x \cdot \sin 2x−6cosx⋅sin2x−32cosx⋅sin2x\frac{-3}{2}\cos x \cdot \sin 2x2−3cosx⋅sin2xExplanationGiven y=cos3xy = \cos^3 xy=cos3x Derivative Using Chain Rule Use the chain rule to differentiate the composite function y′=3(cos2x)⋅(−sinx)y' = 3(\cos^2 x) \cdot (-\sin x)y′=3(cos2x)⋅(−sinx)=−3cos2xsinx= -3\cos^2 x \sin x=−3cos2xsinx=−3cosxcosxsinx= -3\cos x \cos x \sin x=−3cosxcosxsinx Using Trigonometric Identity Use the identity 2sinxcosx=sin2x2\sin x \cos x = \sin 2x2sinxcosx=sin2x =−32cosx(2sinxcosx)= \frac{-3}{2}\cos x (2\sin x \cos x)=2−3cosx(2sinxcosx)=−32cosx⋅sin2x= \frac{-3}{2}\cos x \cdot \sin 2x=2−3cosx⋅sin2x Therefore, the first derivative of y=cos3xy = \cos^3 xy=cos3x is y′=−32cosx⋅sin2xy' = \frac{-3}{2}\cos x \cdot \sin 2xy′=2−3cosx⋅sin2x
28Number 28ExplanationThe first derivative of y=cos3xy = \cos^3 xy=cos3x is ....3cosx⋅sinx3\cos x \cdot \sin x3cosx⋅sinx−3cosx⋅sinx-3\cos x \cdot \sin x−3cosx⋅sinx2cosx⋅sin2x2\cos x \cdot \sin 2x2cosx⋅sin2x−6cosx⋅sin2x-6\cos x \cdot \sin 2x−6cosx⋅sin2x−32cosx⋅sin2x\frac{-3}{2}\cos x \cdot \sin 2x2−3cosx⋅sin2xExplanationGiven y=cos3xy = \cos^3 xy=cos3x Derivative Using Chain Rule Use the chain rule to differentiate the composite function y′=3(cos2x)⋅(−sinx)y' = 3(\cos^2 x) \cdot (-\sin x)y′=3(cos2x)⋅(−sinx)=−3cos2xsinx= -3\cos^2 x \sin x=−3cos2xsinx=−3cosxcosxsinx= -3\cos x \cos x \sin x=−3cosxcosxsinx Using Trigonometric Identity Use the identity 2sinxcosx=sin2x2\sin x \cos x = \sin 2x2sinxcosx=sin2x =−32cosx(2sinxcosx)= \frac{-3}{2}\cos x (2\sin x \cos x)=2−3cosx(2sinxcosx)=−32cosx⋅sin2x= \frac{-3}{2}\cos x \cdot \sin 2x=2−3cosx⋅sin2x Therefore, the first derivative of y=cos3xy = \cos^3 xy=cos3x is y′=−32cosx⋅sin2xy' = \frac{-3}{2}\cos x \cdot \sin 2xy′=2−3cosx⋅sin2x
