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

Question

1783 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-31T03:59:00

The derivative is $\frac{d}{d x} [ln (x) + i π] = \frac{1}{x}$ $d/(dx)[ln(x)+ipi]=1/x$

We can use the following relationship discovered by
Euler.

$e^{i π} + 1 = 0$ $e^(ipi)+1=0$

Subtracting $1$ $1$ from both sides

$e^{i π} = - 1$ $e^(ipi)=-1$

Now take the natural logarithm
of both sides

${ln e}^{i π} = ln (- 1)$ $lne^(ipi)=ln(-1)$

Using rule of logarithms we can rewrite the left hand side

$(i π) ln e = ln (- 1)$ $(ipi)lne=ln(-1)$

Recall that $ln e = 1$ $lne=1$

So $ln (- 1) = i π$ $ln(-1)=ipi$

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

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

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

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

From above $ln (- 1) = i π$ $ln(-1)=ipi$

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

The derivative is $\frac{d}{d x} [ln (x) + i π] = \frac{1}{x}$ $d/(dx)[ln(x)+ipi]=1/x$

$i π$ $ipi$ is a constant

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