How do you find the first and second derivative of #ln(x/2)#?
1 Answer
Nov 12, 2016
# dy/dx= 1/x#
# (d^2y)/(dx^2) = -1/x^2 #
Explanation:
Let
We can use the law of logarithms to write:
# y= lnx - ln2 #
Differentiating wrt
# dy/dx= 1/x#
We can rewrite this as:
# dy/dx= x^-1 #
Differentiating again wrt
# (d^2y)/(dx^2) = (-1)x^-2 #
# (d^2y)/(dx^2) = -1/x^2 #