Question

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

Answer 1

`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)`