Our first term will have to have the product rule applied. The product rule states that given #f(x)=g(x)h(x), f'(x) = g'(x)h(x) + g(x)h'(x)#
Thus our first term...
#d/dx (3xsin^2(4x)) = 3sin^2(4x) + 3x d/dx (sin^2 (4x))#
The derivative for this second term will require use of the chain rule, and of substitution. If we declare #u(x) = sin(4x)#, then we have #(du)/dx = 4 cos(4x) sin^2(4x) = u^2#, meaning #d/dx u^2 = d/du (u^2) * (du)/dx = 2u*(du)/dx. = 2 sin(4x) * 4cos(4x) = 8sin(4x)cos(4x)#
This means:
#d/dx (3xsin^2(4x)) = 3sin^2(4x) + 3x(8 sin(4x)cos(4x)) = 3sin^2(4x) + 24xsin(4x)cos(4x)#
Meanwhile, the derivative of #sec(x)# can be found using the quotient rule and the definition of the secant.
#sec(x) = 1/cos(x) -> d/dx (1/cos x) = ((0*cosx) - (-sin x))/(cos^2x) = sin x / (cos^2x) = tan(x)* 1/cos(x) = tanx secx#
Thus we have...
#(dg)/dx = 3sin^2(4x) + 24xsin(4x)cos(4x)+tan(x)sec(x)#
