Chain rule

1,868
0
1. The problem statement, all variables and given/known data
Hi.

I want to solve:

[tex]
\frac{d\ln(-x)}{dx}.
[/tex]

When using the chain rule I get:

[tex]
\frac{d \ln(-x)}{dx} = \frac{d\ln(-x)}{d(-x)}\frac{d(-x)}{dx} = -\frac{d\ln(-x)}{d(-x)}.
[/tex]

But how do I find the last derivative? I know by experience that it is -x-1, but how is the derivation done?

Thanks in advance.
 
The last derivate is of the same form as:

[tex]\frac{d}{dx}\ln x[/tex]

Which is a simple solution, no?
 
31,927
3,893
The last derivative works out to - 1/(-x) = 1/x, not -x^(-1) as you have.

For the same reason that [itex]\int dx/x = ln |x| + C[/itex], d/dx(ln |x|) = 1/x.

This means that d/dx(ln(x)) = d/dx(ln(-x)) = 1/x.
 
1,868
0
The last derivative works out to - 1/(-x) = 1/x, not -x^(-1) as you have.
If this is true, then we obtain a total of -1/x. And jazznaz is telling me the opposite of you?

Thanks for replying.
 
156
0
If this is true, then we obtain a total of -1/x. And jazznaz is telling me the opposite of [Mark]?
Well, no, but it probably helps to replace x with u in what Jazznaz wrote.

When using the chain rule I get:

[tex]
\frac{d \ln(-x)}{dx} = \frac{d\ln(-x)}{d(-x)}\frac{d(-x)}{dx} = -\frac{d\ln(-x)}{d(-x)}.
[/tex]

But how do I find the last derivative?
To derive

[tex]\frac{d}{du} \ln{u} = \frac{1}{u}[/tex]

first let [tex]y = \ln{u}[/tex]. Then [tex]e^y = u[/tex].


Now use implicit differentiation to find [tex]\frac{dy}{du}[/tex] (which is [tex]\frac{d}{du} \ln{u}[/tex] ) in terms of u.
 

HallsofIvy

Science Advisor
41,626
821
1. The problem statement, all variables and given/known data
Hi.

I want to solve:

[tex]
\frac{d\ln(-x)}{dx}.
[/tex]

When using the chain rule I get:

[tex]
\frac{d \ln(-x)}{dx} = \frac{d\ln(-x)}{d(-x)}\frac{d(-x)}{dx} = -\frac{d\ln(-x)}{d(-x)}.
[/tex]

But how do I find the last derivative? I know by experience that it is -x-1, but how is the derivation done?

Thanks in advance.
If you let u= -x, your last expresion is [tex]-\frac{d ln(u)}{du}[/tex]. Does that make more sense?
 
1,868
0
If you let u= -x, your last expresion is [tex]-\frac{d ln(u)}{du}[/tex]. Does that make more sense?
Yes, that does make sense.

I understand it now. Thanks to everybody for helping.
 

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