Given that y=((3x+1)^2)*cos(3x), find dy/dx.
As why is in the for y=uv where u and v are funtions of x, dy/dx=u'v+v'u (where ' implies the derivative) u=(3x+1)2, v=cos(3x) therefore using the chain rule u'=23(3x+1)=18x+6 and v'=-3sin(3x). Using this, dy/dy=(18x+6)*cos(3x)-3(3x+1)2*sin(3x)
WR
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