How do you differentiate # g(x) = x^2-xsin3x #?

1 Answer
Dec 11, 2015

#2x-sin(3x) - 3xcos(3x)#

Explanation:

You must use three rules for differentiation:

  1. 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^2-xsin(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:

  2. 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:

  3. 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

#2x-sin(3x) - 3xcos(3x)#