How do you differentiate #g(x) = sin(3x)(x^2-4)# using the product rule?

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Answer 1 · 2016-01-22T16:49:07

#g'(x)=3(x^2-4)cos(3x)+2xsin(3x)#

The product rule states that

#g'(x)=(x^2-4)d/dx[sin(3x)]+sin(3x)d/dx[x^2+4]#

Find each derivative.

The first will require the chain rule: #d/dx[sin(u)]=u'*cos(u)#, and we have #u=3x#.

#d/dx[sin(3x)]=d/dx[3x]*cos(3x)=3cos(3x)#

The next simply requires power rule.

#d/dx[x^2-4]=2x#

Plug these back in to determine #g'(x)#.

#g'(x)=3(x^2-4)cos(3x)+2xsin(3x)#