How do you find the first and second derivative of #ln(x/2)#?

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Answer 1 · 2016-11-12T00:06:21

# dy/dx= 1/x#

# (d^2y)/(dx^2) = -1/x^2 #

Let # y = ln(x/2) #

We can use the law of logarithms to write:

# y= lnx - ln2 #

Differentiating wrt #x# we have;

# dy/dx= 1/x#

We can rewrite this as:

# dy/dx= x^-1 #

Differentiating again wrt #x# we have;

# (d^2y)/(dx^2) = (-1)x^-2 #

# (d^2y)/(dx^2) = -1/x^2 #