Possible derivation:
d/dr(1/(log(1+r/100)))
Using the chain rule, d/dr(1/(log(r/100+1))) = d/( du)1/u 0, where u = log(r/100+1) and ( d)/( du)(1/u) = -1/u^2:
= -(d/dr(log(1+r/100)))/(log^2(1+r/100))
Using the chain rule, d/dr(log(r/100+1)) = ( dlog(u))/( du) 0, where u = r/100+1 and ( d)/( du)(log(u)) = 1/u:
= -1/(log^2(1+r/100)) (d/dr(1+r/100))/(r/100+1)
Differentiate the sum term by term and factor out constants:
= -1/((1+r/100) log^2(1+r/100)) d/dr(1)+(d/dr(r))/100
The derivative of 1 is zero:
= -(1/100 (d/dr(r))+0)/((1+r/100) log^2(1+r/100))
Simplify the expression:
= -(d/dr(r))/(100 (1+r/100) log^2(1+r/100))
The derivative of r is 1:
Answer: |= -1/(100 (1+r/100) log^2(1+r/100)) 1
