Differentiate : g(x) = 5*ln x* (x^2-3x) and find g'(1).

Given the function :

f(x) =
5lnx * (x^2-3x)

We need to find f'(x) and
f'(1).

We will use the product rule to find the
derivative.

==> Let f(x) = u*v such
that:

u= 5lnx ==> u' =
5/x

v= (x^2-3x) ==> v' =
2x-3

==> f'(x) = u'v +
uv'

               = (5/x)(x^2-3x) +
(5lnx*(2x-3))

             = (5x - 15) + 10x*lnx - 15ln
x

              = 10x*lnx - 15lnx + 5x -
15

==> f'(x) = 10x*lnx - 15lnx + 5x
-15

==> f'(1) = 10*1*0 - 15*0 + 5 -15 =
-10

==> f'(1)
=-10

Then f'(x) = 10xlnx - 15lnx + 5x -15 and
f'(1) = -10