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

*print*Print*list*Cite

### 2 Answers

We have g(x) = 5*ln x*(x^2 - 3x). I'm sure you want the value of g'(1) not f'(1).

g'(x) = 5*(1/x)(x^2 - 3x) + 5*ln x*(2x - 3)

g'(1) = 5*(1/1)(1^2 - 3*1) + 5*ln 1*(2*1 - 3)

=> 5*(1 - 3) + 5*0

=> 5*-2

=> -10

**g'(x) = 5*(1/x)(x^2 - 3x) + 5*ln x*(2x - 3) and t****he value of g'(1) = -10**

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**