How to differentiate ln#x/2# and ln(#√(1-#1/x#)#) respect to x?
1 Answer
May 13, 2018
a) Simplify using logarithm rules:
#ln(x/2) = ln(x) - ln(2)#
#d/dx(ln (x/2)) = 1/x - 0 = 1/x#
b) Once again apply logarithm laws to your advantage
#lnsqrt((1 - 1/x)) = 1/2ln(1 - 1/x) #
Now by the chain rule:
#d/dx(lnsqrt(1 - 1/x)) = (1/2(1/x^2))/(1 - 1/x) = (1/(2x^2))/((x -1)/x) = 1/(2x(x -1))#
Hopefully this helps!
