What is the Formula for ln ex/ex?

  • Thread starter JasonX
  • Start date
  • Tags
    Formula
In summary, the conversation is about differentiating the equation y= ln e^x/e^x using the quotient rule. The formula for the quotient rule is explained and then applied to simplify the equation to 1-x/e^x. The e^x in the numerator is cancelled out with the e^x in the denominator, resulting in the final answer.
  • #1
JasonX
10
0
[SOLVED] need formula or how to...

y= ln ex/ex (x is the power)
 
Physics news on Phys.org
  • #2
uhm use the "^" to represent powers, for example, x squared is written as x^2

and please state your question better
 
  • #3
rock.freak667 said:
uhm use the "^" to represent powers, for example, x squared is written as x^2

and please state your question better

Differentiate:

y= ln e^x/e^x


Thanks!
 
  • #4
So the question is to differentiate

[tex] y=\frac{ln e^x}{e^x}[/tex] ?


if it is...what rule do you think you would use when you need to differentiate something of the form [itex]y=\frac{u}{v}[/itex] ?
 
  • #5
quotient rule.
which is: first, derivative of second, minus second, derivative of first, divided by first squared. (but, I'm still stuck. could you work it out for me?)
 
Last edited:
  • #6
Well firstly...lne^x can be simplified to xlne which is just x

so you really want to find [tex]\frac{d}{dx}(\frac{x}{e^x})[/tex]

so if u= x => [itex]\frac{du}{dx}=1[/itex]
and v=e^x => [itex]\frac{dv}{dx}=e^x[/itex]

the formula is
[tex] \frac{d}{dx}(\frac{u}{v}) =\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}[/tex]so then you'll have
[tex]\frac{d}{dx}(\frac{x}{e^x}) = \frac{(e^x)(1) -(x)(e^x)}{(e^x)^2}[/tex]

then you will simplify it
 
  • #7
thanks a lot. you made it very easy to understand. only part i didn't get is it says answer should be 1-x/e^x. what cancels out the power in the denominator?


Thanks again!
 
  • #8
...well the e^x in the numerator can be factored out and it will cancel with an e^x in the denominator...so giving you your answer
 
  • #9
oh, i see. thank you very much.. :)
 

1. What is a formula?

A formula is a mathematical expression that uses symbols and numbers to describe a relationship between different quantities. It is typically used to solve problems or make predictions in various scientific fields.

2. How do I create a formula?

To create a formula, you need to identify the variables and their relationships in the problem you are trying to solve. Then, you can use mathematical operations such as addition, subtraction, multiplication, and division to express this relationship. It is important to follow the rules of mathematical notation and use appropriate symbols.

3. Can I use a formula for any problem?

No, not all problems can be solved using a formula. Some problems may require other methods, such as experimentation or statistical analysis. It is important to understand the limitations of formulas and when they are applicable.

4. How do I know if my formula is correct?

You can check the accuracy of your formula by plugging in different values for the variables and comparing the results to known solutions or experimental data. It is also helpful to have others review your formula and calculations to ensure they are correct.

5. Can I modify an existing formula?

Yes, you can modify an existing formula to suit your specific needs or adapt it for a different problem. However, it is important to understand the original formula and its purpose before making any modifications to ensure the accuracy of your results.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
846
  • Calculus and Beyond Homework Help
Replies
14
Views
254
Replies
8
Views
823
  • Introductory Physics Homework Help
Replies
10
Views
672
  • Calculus and Beyond Homework Help
Replies
5
Views
291
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
Replies
7
Views
866
  • Calculus and Beyond Homework Help
Replies
3
Views
525
Back
Top