We can differentiate the two terms separately, since we have
[tex] [f(x)+g(x)]' = f'(x)+g'(x)[/tex]
The first term comes in the form [tex]k\cdot f(x)[/tex], where [tex]k=3[/tex] and [tex]f(x)=\sin(x)[/tex]
So, we have
[tex][k\cdot f(x)]' = kf'(x)[/tex]
The derivative of sine is cosine, so we have
[tex][3\sin(x)]' = 3\cos(x)[/tex]
And the first term is done. For the second term, we have to use the power rule
[tex][f^n(x)]' = nf^{n-1}(x)\cdot f'(x)[/tex]
Since the derivative of cosine is negative sine, we have
[tex][4\cos^3(x)]'= 4\cdot 3\cos^2(x)(-\sin(x)) = -12\cos^2(x)\sin(x)[/tex]
So, the whole derivative is the sum of the derivatives of the two terms, and we have
[tex][3\sin(x) + 4\cos^3(x)]'=3\cos(x)-12\cos^2(x)\sin(x)[/tex]
