What is the derivative of $ln(1/x)$ ?

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Answer 1 · 2016-09-01T14:25:05

The derivative is $-1/x$

Here are two ways to find the derivative:

Method 1 relies on knowing about exponents and logarithms.

We'll use $1/x = x^-1$ and $ln(a^b) = blna$ and $d/dx(cf(x)) = c d/dx(f(x))$

We get:

$d/dx(ln(1/x)) = d/dx(ln(x^-1))$

$= d/dx(-1lnx) = -1 d/dx(lnx)$

$= -1/x$ .

Method 2 uses the chain rule.

$d/dx(lnu) = 1/u * (du)/dx$ and

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

We get

$d/dx(ln(1/x)) = 1/(1/x) * d/dx(1/x)$

$= x * (-1/x^2)$

$= -1/x$

What is the derivative of ln(1/x)ln(1/x)?