Purpose of differentiating both sides

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Homework Help Overview

The discussion revolves around the differentiation of the equation y=e^x and its rearrangement to ln y=x. The original poster is exploring the implications of differentiating both sides of the equation after it has been altered.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand the necessity of differentiating both sides of an equation after rearranging it. They express confusion about the consequences of this process and whether it is a general requirement. Other participants provide insights into the differentiation process and the use of the chain rule.

Discussion Status

Participants have engaged in clarifying the differentiation process and its requirements. Some have attempted to address the original poster's questions, but there remains a focus on understanding the reasoning behind differentiating both sides of an altered equation.

Contextual Notes

The original poster indicates a lack of familiarity with calculus concepts, which may affect their understanding of the differentiation process and the implications of rearranging equations.

nobahar
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Hey guys,
I've just recently started calculus, and just trying to get to grips with it.
I'll use an example for my question; if [tex]y=e^x[/tex], then the source suggests rearranging to [tex]\ln y=x[/tex], the differentiating with respect to x yields [tex]\frac{\delta}{\delta{x}} (ln\ y)=1[/tex].

I understand that for my original function I wouldn't be able to find [tex]\delta{y}[/tex], so it's rearranged, but what are the consequences of this? As it appears that the requirement is to DIFFERENTIATE BOTH SIDES. Do I always do this when the equation has been altered? I can't piece togeather why this is happening. Hopefully someone can help.
Thanks.
Nobahar.
 
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[tex]y=e^x \Rightarrow lny=x[/tex]


[tex]\frac{d}{dx}(lny)=\frac{d}{dx}(x)[/tex]

[tex]= \frac{dy}{dx}\frac{d}{dy}(lny)=1[/tex]

I didn't fully understand what you were asking so I tried my best to alleviate your confusion.
 
rock.freak667 said:
[tex]y=e^x \Rightarrow lny=x[/tex]


[tex]\frac{d}{dx}(lny)=\frac{d}{dx}(x)[/tex]

[tex]= \frac{dy}{dx}\frac{d}{dy}(lny)=1[/tex]

I didn't fully understand what you were asking so I tried my best to alleviate your confusion.

As to the purpose, if you continue with this, since y=e^x then dy/dx=y. So you've shown y*d(ln(y))/dy=1. So d(ln(y))/dy=1/y. You've used the fact you know how to differentiate the exp function to figure out how to differentiate it's inverse function, ln.
 
Cheers guys! :smile:
 
nobahar said:
Hey guys,
I've just recently started calculus, and just trying to get to grips with it.
I'll use an example for my question; if [tex]y=e^x[/tex], then the source suggests rearranging to [tex]\ln y=x[/tex], the differentiating with respect to x yields [tex]\frac{\delta}{\delta{x}} (ln\ y)=1[/tex].
Other folks showed you what to do, but I don't think they answered your questions, so I'll take a shot at it.
Your goal here was to differentiate e^x, but didn't have a formula to do so.
What you called "rearranging" the equation is to rewrite the original equation, where y is a function of x, as a new equation where x is a different function of y. In fact, the different function is the inverse of the exponential function.

Apparently you already know the derivative of ln(x), so using the chain rule, you can differentiate ln(y) with respect to x (i.e., find d/dx(ln(y)), which as Dick pointed out is 1/y * dy/dx.
nobahar said:
I understand that for my original function I wouldn't be able to find [tex]\delta{y}[/tex], so it's rearranged, but what are the consequences of this? As it appears that the requirement is to DIFFERENTIATE BOTH SIDES. Do I always do this when the equation has been altered?
It doesn't have anything to do with whether an equation has been altered. Whenever you differentiate one side of an equation, you have to differentiate the other side as well. And the differentiation has to be with respect to the same variable. IOW, if you differentiate one side with respect to x, you have to differentiate the other side with respect to x as well.
nobahar said:
I can't piece togeather why this is happening. Hopefully someone can help.
Thanks.
Nobahar.
 
That's pretty much answered my questions fully!
Thanks!
 

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