How do you differentiate #sin^3(x)cos(x)#?

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Answer 1 · 2016-05-20T11:17:29

#3sin^2x . cos^2x-sin^4x#

#\frac{d}{dx}(\sin ^3(x)\cos \(x))#

#(f\cdot g)^'=f^'\cdot g+f\cdot g^'#

#f=\sin ^3(x),\g=\cos(x)#

#=\frac{d}{dx}(\sin ^3(x))\cos (x)+\frac{d}{dx}\(\cos (x)\right)\sin ^3(x)#

We have,
#\frac{d}{dx}(\sin ^3\(x))=3\sin ^2(x)\cos(x)#
Also,
#\frac{d}{dx}(\cos (x))=-\sin (x)#

#=3\sin ^2(x)\cos(x)\cos (x)+(-\sin (x))\sin ^3(x)#

Simplifying it,we get,
#=3\sin ^2(x)\cos ^2(x)-\sin ^4(x)#