How do you differentiate # g(x) = x^2xsin3x #?
1 Answer
Explanation:
You must use three rules for differentiation:

The derivative of sum (or difference) of two or more functions is the sum (or difference) of each single derivative. So, in this case, we have
#d/dx (x^2xsin(3x)) = d/dx x^2  d/dx xsin(3x)#
At this point, we can easily derive the first term, since the derivative of#x^2# is#2x# . Now let's work on the rest: 
The derivative of a product
#f*g# follows this rule:
#d/dx(f(x) * g(x)) = (d/dx f(x)) * g(x) + f(x) * (d/dx g(x))#
In your case, we have
#(d/dx x) * sin(3x) + x * (d/dx sin(3x))#
Again, the derivative of#x# is#1# , so the first term is simply#sin(3x)# for the second term, we need the last rule: 
The derivative of a composite function
#f(g(x))# follows this rule:
#d/dx f(g(x)) = (d/dx f)(g(x)) * d/dx g(x)#
In your case, we have that
#d/dx f = d/dx sin = cos# , and so#(d/dx f)(g(x)) = cos(3x)#
while#d/dx g(x)=3#
Put everything back together, and the result is