# Chain rule

## Homework Statement

Hi.

I want to solve:

$$\frac{d\ln(-x)}{dx}.$$

When using the chain rule I get:

$$\frac{d \ln(-x)}{dx} = \frac{d\ln(-x)}{d(-x)}\frac{d(-x)}{dx} = -\frac{d\ln(-x)}{d(-x)}.$$

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

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The last derivate is of the same form as:

$$\frac{d}{dx}\ln x$$

Which is a simple solution, no?

Mark44
Mentor
The last derivative works out to - 1/(-x) = 1/x, not -x^(-1) as you have.

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

This means that d/dx(ln(x)) = d/dx(ln(-x)) = 1/x.

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?

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:

$$\frac{d \ln(-x)}{dx} = \frac{d\ln(-x)}{d(-x)}\frac{d(-x)}{dx} = -\frac{d\ln(-x)}{d(-x)}.$$

But how do I find the last derivative?
To derive

$$\frac{d}{du} \ln{u} = \frac{1}{u}$$

first let $$y = \ln{u}$$. Then $$e^y = u$$.

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

HallsofIvy
Homework Helper

## Homework Statement

Hi.

I want to solve:

$$\frac{d\ln(-x)}{dx}.$$

When using the chain rule I get:

$$\frac{d \ln(-x)}{dx} = \frac{d\ln(-x)}{d(-x)}\frac{d(-x)}{dx} = -\frac{d\ln(-x)}{d(-x)}.$$

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

If you let u= -x, your last expresion is $$-\frac{d ln(u)}{du}$$. Does that make more sense?
If you let u= -x, your last expresion is $$-\frac{d ln(u)}{du}$$. Does that make more sense?