How would you differentiate 25000 - 20000 ln(1 - (x/100))?

Question

How would you differentiate 25000 - 20000 ln(1 - (x/100))?

Answer 1

if y=25000-20000ln(1-x/100)

  y=25000-20000ln(1-x100^-1)  let u=1-x100^-1

y=25000-20000ln(u)

y=25000-ln(u)^20000

then dy/du=(u^1999)/(u^20000)=1/u

then du/dx=.1x100^-2

                =-(x/100^2)

du/dx*dy/du=dy/dx

dy/dx=(-x100+x6^2/)(1000000)

Answer 2

first derivative of a constant is zero so you have a sum derivative of 25000=0

it remains derivative of -20000 ln (1-(x/100))

derivative ln u = u'/u

u= 1-x/100=(100-x)/100    u'= -1/100

u'/u=-1/(100-x)

final answer 20000/(100-x)    

Answer 3

-20000 / (x-100)