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;

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

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

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