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