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

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Answer 1 · 2018-04-13T00:56:00

$d/dxln(1/x)=-1/x$

We could use the Chain Rule right away, but the properties of logarithms allow us to avoid that and make this easier.

$ln(1/x)=ln(1)-lnx=-lnx$

So,

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

Answer 2 · 2018-04-13T04:29:09

$-1/x$

Well, let's try the chain rule, which states that,

$dy/dx=dy/(du)*(du)/dx$

Let $u=1/x,:.(du)/dx=-1/x^2$ .

Then $y=lnu,:.dy/(du)=1/u$ .

Combining, we get,

$dy/dx=1/u*-1/x^2$

$=-1/(ux^2)$

Substituting back $u=1/x$ , we get,

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

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

$=-1/x$

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