How do you find the derivative of $ln( -x)$ ?

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Answer 1 · 2016-12-31T03:59:00

The derivative is $d/(dx)[ln(x)+ipi]=1/x$

We can use the following relationship discovered by
Euler.

$e^(ipi)+1=0$

Subtracting $1$ from both sides

$e^(ipi)=-1$

Now take the natural logarithm
of both sides

$lne^(ipi)=ln(-1)$

Using rule of logarithms we can rewrite the left hand side

$(ipi)lne=ln(-1)$

Recall that $lne=1$

So $ln(-1)=ipi$

Now we can rewrite $ln(-x)$ as follows

$ln(x(-1))$

Now we have the logarithm of product which
we can rewrite as follows

$ln(x)+ln(-1)$

From above $ln(-1)=ipi$

$ln(-x)=ln(x)+ipi$

The derivative is $d/(dx)[ln(x)+ipi]=1/x$

$ipi$ is a constant

How do you find the derivative of ln( -x)ln( -x)?