How do you differentiate (3x+cos(x))(2+4sin(3x))?
Here we have a product of two things, so we will be using the product rule of differentiation. This is: for y=u(x)v(x), where u(x) and v(x) are funtions of x, dy/dx = u'(x)v(x) + u(x)v'(x). So in this case let u(x) = 3x+cos(x) and let v(x) = 2+4sin(3x). We need to find u'(x). u'(x) = 3-sin(x) as we differentiate u(x). v'(x) = 12cos(3x) as we diferentiate v(x). Then using the product rule sated, dy/dx = (3-sin(x))(2+4sin(3x)) + (3x+cos(x))(12cos(3x)).
JP
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