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

Answer 1

Luke Alvarez

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

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Answer 2

Christopher Agresta

To find the derivative of ln(-x), you can use the chain rule. The derivative of ln(u) with respect to x is (1/u) * du/dx. In this case, u = -x. Thus, the derivative of ln(-x) with respect to x is (1/(-x)) * (-1), which simplifies to -1/x. Therefore, the derivative of ln(-x) with respect to x is -1/x.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.