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derivative of log(y) function. [on hold]

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2

1

$begingroup$

I am trying to find derivative of log function.

How would I apply the chain rule on a basic example?

$$ln(y)=m+bx$$

I think it is

$$fracdydx=e^m+bx *b.$$
Is this correct?

derivatives logarithms

share|cite|improve this question

edited Mar 11 at 20:54

thesmallprint

2,6711618

asked Mar 11 at 20:42

O B.O B.

183

$endgroup$

put on hold as off-topic by uniquesolution, Lee David Chung Lin, Leucippus, Vinyl_cape_jawa, Parcly Taxel Mar 12 at 3:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – uniquesolution, Lee David Chung Lin, Leucippus, Vinyl_cape_jawa, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For clarification: by $log(y)$ do you mean the base-$e$ logarithm, i.e. natural logarithm? And also, the derivative of $y$ with respect to $x$, or the other way around?
  $endgroup$
  – Eevee Trainer
  Mar 11 at 20:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You might also find your question more well-received if you include your own attempts on the problem in the body of the question, what you understand, and what you're stuck on
  $endgroup$
  – Eevee Trainer
  Mar 11 at 20:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I updated the question with base-e logarithm and what I thought the derivative would be, sorry. @EeveeTrainer
  $endgroup$
  – O B.
  Mar 11 at 20:50
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  yes, you are correct.
  $endgroup$
  – thesmallprint
  Mar 11 at 20:52
  

  
  
  
  

add a comment | 

2

1

$begingroup$

I am trying to find derivative of log function.

How would I apply the chain rule on a basic example?

$$ln(y)=m+bx$$

I think it is

$$fracdydx=e^m+bx *b.$$
Is this correct?

derivatives logarithms

share|cite|improve this question

edited Mar 11 at 20:54

thesmallprint

2,6711618

asked Mar 11 at 20:42

O B.O B.

183

$endgroup$

put on hold as off-topic by uniquesolution, Lee David Chung Lin, Leucippus, Vinyl_cape_jawa, Parcly Taxel Mar 12 at 3:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – uniquesolution, Lee David Chung Lin, Leucippus, Vinyl_cape_jawa, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For clarification: by $log(y)$ do you mean the base-$e$ logarithm, i.e. natural logarithm? And also, the derivative of $y$ with respect to $x$, or the other way around?
  $endgroup$
  – Eevee Trainer
  Mar 11 at 20:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You might also find your question more well-received if you include your own attempts on the problem in the body of the question, what you understand, and what you're stuck on
  $endgroup$
  – Eevee Trainer
  Mar 11 at 20:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I updated the question with base-e logarithm and what I thought the derivative would be, sorry. @EeveeTrainer
  $endgroup$
  – O B.
  Mar 11 at 20:50
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  yes, you are correct.
  $endgroup$
  – thesmallprint
  Mar 11 at 20:52
  

  
  
  
  

add a comment | 

$begingroup$

I am trying to find derivative of log function.

How would I apply the chain rule on a basic example?

$$ln(y)=m+bx$$

I think it is

$$fracdydx=e^m+bx *b.$$
Is this correct?

derivatives logarithms

share|cite|improve this question

edited Mar 11 at 20:54

thesmallprint

2,6711618

asked Mar 11 at 20:42

O B.O B.

183

$endgroup$

share|cite|improve this question

edited Mar 11 at 20:54

thesmallprint

2,6711618

asked Mar 11 at 20:42

O B.O B.

183

share|cite|improve this question

edited Mar 11 at 20:54

thesmallprint

2,6711618

asked Mar 11 at 20:42

O B.O B.

183

edited Mar 11 at 20:54

thesmallprint

2,6711618

edited Mar 11 at 20:54

thesmallprint

2,6711618

asked Mar 11 at 20:42

O B.O B.

183

asked Mar 11 at 20:42

O B.O B.

183

2 Answers
2

active

oldest

votes

1

$begingroup$

$ln y = m + bx; tag 1$

differentiate with respect to $x$:

$dfracy'y = b; tag 2$

$y' = by = be^m + bx, tag 3$

since from (1)

$y = e^m + bx. tag 4$

Our OP thesmallprint's result is thus correct.

share|cite|improve this answer

answered Mar 11 at 21:00

Robert LewisRobert Lewis

48.1k23067

$endgroup$

add a comment | 

0

$begingroup$

Using the definition of a derivative, $$fracddylog y=lim_epsilonto 0fracln(1+fracepsilony)epsilon.$$ Since logarithms are monotonic, the derivative doesn't vanish at $0$, so a Taylor-series argument implies a constant $k$ exists for which $fracddylog y=fracky$. The value of $k$ depends on the logarithm's base. One definition of $e$ is as the basis obtaining the $k=1$ natural logarithm $ln y$.

share|cite|improve this answer

answered Mar 11 at 20:55

J.G.J.G.

30.1k23148

$endgroup$

add a comment | 

1

$begingroup$

$ln y = m + bx; tag 1$

differentiate with respect to $x$:

$dfracy'y = b; tag 2$

$y' = by = be^m + bx, tag 3$

since from (1)

$y = e^m + bx. tag 4$

Our OP thesmallprint's result is thus correct.

share|cite|improve this answer

answered Mar 11 at 21:00

Robert LewisRobert Lewis

48.1k23067

$endgroup$

add a comment | 

1

$begingroup$

$ln y = m + bx; tag 1$

differentiate with respect to $x$:

$dfracy'y = b; tag 2$

$y' = by = be^m + bx, tag 3$

since from (1)

$y = e^m + bx. tag 4$

Our OP thesmallprint's result is thus correct.

share|cite|improve this answer

answered Mar 11 at 21:00

Robert LewisRobert Lewis

48.1k23067

$endgroup$

add a comment | 

$begingroup$

$ln y = m + bx; tag 1$

differentiate with respect to $x$:

$dfracy'y = b; tag 2$

$y' = by = be^m + bx, tag 3$

since from (1)

$y = e^m + bx. tag 4$

Our OP thesmallprint's result is thus correct.

share|cite|improve this answer

answered Mar 11 at 21:00

Robert LewisRobert Lewis

48.1k23067

$endgroup$

share|cite|improve this answer

answered Mar 11 at 21:00

Robert LewisRobert Lewis

48.1k23067

answered Mar 11 at 21:00

Robert LewisRobert Lewis

48.1k23067

answered Mar 11 at 21:00

Robert LewisRobert Lewis

48.1k23067

0

$begingroup$

Using the definition of a derivative, $$fracddylog y=lim_epsilonto 0fracln(1+fracepsilony)epsilon.$$ Since logarithms are monotonic, the derivative doesn't vanish at $0$, so a Taylor-series argument implies a constant $k$ exists for which $fracddylog y=fracky$. The value of $k$ depends on the logarithm's base. One definition of $e$ is as the basis obtaining the $k=1$ natural logarithm $ln y$.

share|cite|improve this answer

answered Mar 11 at 20:55

J.G.J.G.

30.1k23148

$endgroup$

add a comment | 

0

$begingroup$

Using the definition of a derivative, $$fracddylog y=lim_epsilonto 0fracln(1+fracepsilony)epsilon.$$ Since logarithms are monotonic, the derivative doesn't vanish at $0$, so a Taylor-series argument implies a constant $k$ exists for which $fracddylog y=fracky$. The value of $k$ depends on the logarithm's base. One definition of $e$ is as the basis obtaining the $k=1$ natural logarithm $ln y$.

share|cite|improve this answer

answered Mar 11 at 20:55

J.G.J.G.

30.1k23148

$endgroup$

add a comment | 

$begingroup$

Using the definition of a derivative, $$fracddylog y=lim_epsilonto 0fracln(1+fracepsilony)epsilon.$$ Since logarithms are monotonic, the derivative doesn't vanish at $0$, so a Taylor-series argument implies a constant $k$ exists for which $fracddylog y=fracky$. The value of $k$ depends on the logarithm's base. One definition of $e$ is as the basis obtaining the $k=1$ natural logarithm $ln y$.

share|cite|improve this answer

answered Mar 11 at 20:55

J.G.J.G.

30.1k23148

$endgroup$

share|cite|improve this answer

answered Mar 11 at 20:55

J.G.J.G.

30.1k23148

answered Mar 11 at 20:55

J.G.J.G.

30.1k23148

answered Mar 11 at 20:55

J.G.J.G.

30.1k23148