Derivative of -ln(-\Theta): Explained

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In summary, the derivative of -ln(-\Theta) with respect to \Theta is -\frac{1}{\Theta}. To find this, you need to apply the chain rule and multiply the derivative of -\frac{1}{\Theta} with -\Theta.
  • #1
roadworx
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Hi,

I'm trying to find the derivative of [tex]-ln(-\Theta)[/tex] with respect to [tex]\Theta[/tex]

The answer's [tex]-\frac{1}{\Theta}[/tex]

I'm not sure why though. Here's my working.

[tex]\frac{d}{d\Theta} -ln(-\Theta)[/tex]

[tex] = \frac{d}{d\Theta} ln(-\frac{1}{\Theta})[/tex]

[tex] = -\Theta[/tex]

Can anyone explain where I'm going wrong? Thanks.
 
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  • #2
d(-ln(-th))/dth
= - d(ln(-th))/dth
= - (-th)^(-1) d(-th)/dth
= - (-th)^(-1) (-1)
= - (1/th)
 
  • #3
You need to apply the chain rule. The derivative of [tex]-\frac{1}{\Theta}[/tex] is [tex]\frac{1}{\Theta^2}[/tex]. If you multiply this with [tex]-\Theta[/tex] you get the correct answer.
 

FAQ: Derivative of -ln(-\Theta): Explained

What is the derivative of -ln(-\Theta)?

The derivative of -ln(-\Theta) is equal to 1/\Theta. This can be derived using the chain rule and the derivative of natural logarithm.

Why is the derivative of -ln(-\Theta) equal to 1/\Theta?

The derivative of natural logarithm is equal to 1/x. Therefore, when the negative sign is added, it becomes -1/x. And since the derivative of -\Theta is -1, the two negatives cancel out, resulting in a derivative of 1/\Theta.

Can you explain the concept of the negative sign in -ln(-\Theta)?

The negative sign in -ln(-\Theta) indicates that the value of \Theta must be negative in order for the natural logarithm to be defined. This is because the natural logarithm function is only defined for positive values.

How is the derivative of -ln(-\Theta) related to the inverse function of natural logarithm?

The derivative of -ln(-\Theta) is related to the inverse function of natural logarithm by the fact that the derivative of the inverse function is equal to 1 divided by the derivative of the original function. In this case, the original function is natural logarithm and its derivative is 1/x, resulting in a derivative of 1/\Theta for the inverse function.

Are there any other ways to represent the derivative of -ln(-\Theta)?

Yes, the derivative of -ln(-\Theta) can also be represented as -1/\Theta^2. This can be derived by using the quotient rule and simplifying the expression.

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