Mixing
Problem in evaluating logarithm derivatives
$begingroup$
Given the property of the logarithm that $log{xy} = log{x} + log{y}$, how would one take the 'derivative' of this?
To be more clear,
$log{xy} = log{x} + log{y}$ (property of $log$)
$D(log{xy}) = D(log{x} + log{y})$ (i) (Take derivative on both sides)
Now, $D(log{x} + log{y}) = D(log{x}) + D(log{y})$ (ii) (Derivative of sum is sum of derivatives)
Combining (i) and (ii): $D(log{xy}) = D(log{x}) + D(log{y})$ (iii)
Implies: $frac{1}{xy} = frac{1}{x} + frac{1}{y}$ (Evaluate derivative of logarithm using $D(log{x}) = frac{1}{x}$
$frac{1}{xy} = frac{1}{x} + frac{1}{y}$ looks false to me; e.g. while $log{6}$ does equal $log{2} + log{3}$, $frac{1}{6}$ does not equal $frac{1}{2} + frac{1}{3}$.
My first guess was that the issue was related to what variable I take the derivative with respect to, but I'd like to understand this a little more formally if someone could guide me.
What's going wrong in this example?
real-analysis calculus logarithms
edited Jan 5 at 4:33
David G. Stork
10.7k31332
asked Jan 5 at 4:14
user2192320user2192320
595
$endgroup$
add a comment |
$begingroup$
Given the property of the logarithm that $log{xy} = log{x} + log{y}$, how would one take the 'derivative' of this?
To be more clear,
$log{xy} = log{x} + log{y}$ (property of $log$)
$D(log{xy}) = D(log{x} + log{y})$ (i) (Take derivative on both sides)
Now, $D(log{x} + log{y}) = D(log{x}) + D(log{y})$ (ii) (Derivative of sum is sum of derivatives)
Combining (i) and (ii): $D(log{xy}) = D(log{x}) + D(log{y})$ (iii)
Implies: $frac{1}{xy} = frac{1}{x} + frac{1}{y}$ (Evaluate derivative of logarithm using $D(log{x}) = frac{1}{x}$
$frac{1}{xy} = frac{1}{x} + frac{1}{y}$ looks false to me; e.g. while $log{6}$ does equal $log{2} + log{3}$, $frac{1}{6}$ does not equal $frac{1}{2} + frac{1}{3}$.
My first guess was that the issue was related to what variable I take the derivative with respect to, but I'd like to understand this a little more formally if someone could guide me.
What's going wrong in this example?
real-analysis calculus logarithms
edited Jan 5 at 4:33
David G. Stork
10.7k31332
asked Jan 5 at 4:14
user2192320user2192320
595
$endgroup$
$begingroup$
Like the comment above said- you can’t take derivatives with respect to all different variables and expect them to be equal.
$endgroup$
– MathIsLife12
Jan 5 at 4:19
$begingroup$
(This comment was from me -- deleted when reading the fact that ayt the end of your question you acknowledged something was sketchy in the variables you differentiated with respect to. Essentially, I was saying you were "differentiating simultaneously wrt 3 variables," in different places: $xy$, $x$, $y$ -- and that's not allowed.)
$endgroup$
– Clement C.
Jan 5 at 4:21
$begingroup$
Thanks guys; this clears things up! I appreciate the fast responses.
$endgroup$
– user2192320
Jan 5 at 4:24
$begingroup$
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
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– John Doe
Jan 5 at 5:01
add a comment |
$begingroup$
Given the property of the logarithm that $log{xy} = log{x} + log{y}$, how would one take the 'derivative' of this?
To be more clear,
$log{xy} = log{x} + log{y}$ (property of $log$)
$D(log{xy}) = D(log{x} + log{y})$ (i) (Take derivative on both sides)
Now, $D(log{x} + log{y}) = D(log{x}) + D(log{y})$ (ii) (Derivative of sum is sum of derivatives)
Combining (i) and (ii): $D(log{xy}) = D(log{x}) + D(log{y})$ (iii)
Implies: $frac{1}{xy} = frac{1}{x} + frac{1}{y}$ (Evaluate derivative of logarithm using $D(log{x}) = frac{1}{x}$
$frac{1}{xy} = frac{1}{x} + frac{1}{y}$ looks false to me; e.g. while $log{6}$ does equal $log{2} + log{3}$, $frac{1}{6}$ does not equal $frac{1}{2} + frac{1}{3}$.
My first guess was that the issue was related to what variable I take the derivative with respect to, but I'd like to understand this a little more formally if someone could guide me.
What's going wrong in this example?
real-analysis calculus logarithms
edited Jan 5 at 4:33
David G. Stork
10.7k31332
asked Jan 5 at 4:14
user2192320user2192320
595
$endgroup$
Given the property of the logarithm that $log{xy} = log{x} + log{y}$, how would one take the 'derivative' of this?
To be more clear,
$log{xy} = log{x} + log{y}$ (property of $log$)
$D(log{xy}) = D(log{x} + log{y})$ (i) (Take derivative on both sides)
Now, $D(log{x} + log{y}) = D(log{x}) + D(log{y})$ (ii) (Derivative of sum is sum of derivatives)
Combining (i) and (ii): $D(log{xy}) = D(log{x}) + D(log{y})$ (iii)
Implies: $frac{1}{xy} = frac{1}{x} + frac{1}{y}$ (Evaluate derivative of logarithm using $D(log{x}) = frac{1}{x}$
$frac{1}{xy} = frac{1}{x} + frac{1}{y}$ looks false to me; e.g. while $log{6}$ does equal $log{2} + log{3}$, $frac{1}{6}$ does not equal $frac{1}{2} + frac{1}{3}$.
My first guess was that the issue was related to what variable I take the derivative with respect to, but I'd like to understand this a little more formally if someone could guide me.
What's going wrong in this example?
real-analysis calculus logarithms
real-analysis calculus logarithms
edited Jan 5 at 4:33
David G. Stork
10.7k31332
asked Jan 5 at 4:14
user2192320user2192320
595
edited Jan 5 at 4:33
David G. Stork
10.7k31332
asked Jan 5 at 4:14
user2192320user2192320
595
edited Jan 5 at 4:33
David G. Stork
10.7k31332
edited Jan 5 at 4:33
David G. Stork
10.7k31332
edited Jan 5 at 4:33
David G. Stork
10.7k31332
10.7k31332
asked Jan 5 at 4:14
user2192320user2192320
595
asked Jan 5 at 4:14
user2192320user2192320
595
asked Jan 5 at 4:14
user2192320user2192320
595
595
$begingroup$
Like the comment above said- you can’t take derivatives with respect to all different variables and expect them to be equal.
$endgroup$
– MathIsLife12
Jan 5 at 4:19
$begingroup$
(This comment was from me -- deleted when reading the fact that ayt the end of your question you acknowledged something was sketchy in the variables you differentiated with respect to. Essentially, I was saying you were "differentiating simultaneously wrt 3 variables," in different places: $xy$, $x$, $y$ -- and that's not allowed.)
$endgroup$
– Clement C.
Jan 5 at 4:21
$begingroup$
Thanks guys; this clears things up! I appreciate the fast responses.
$endgroup$
– user2192320
Jan 5 at 4:24
$begingroup$
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
$endgroup$
– John Doe
Jan 5 at 5:01
add a comment |
$begingroup$
Like the comment above said- you can’t take derivatives with respect to all different variables and expect them to be equal.
$endgroup$
– MathIsLife12
Jan 5 at 4:19
$begingroup$
(This comment was from me -- deleted when reading the fact that ayt the end of your question you acknowledged something was sketchy in the variables you differentiated with respect to. Essentially, I was saying you were "differentiating simultaneously wrt 3 variables," in different places: $xy$, $x$, $y$ -- and that's not allowed.)
$endgroup$
– Clement C.
Jan 5 at 4:21
$begingroup$
Thanks guys; this clears things up! I appreciate the fast responses.
$endgroup$
– user2192320
Jan 5 at 4:24
$begingroup$
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
$endgroup$
– John Doe
Jan 5 at 5:01
$begingroup$
Like the comment above said- you can’t take derivatives with respect to all different variables and expect them to be equal.
$endgroup$
– MathIsLife12
Jan 5 at 4:19
$begingroup$
Like the comment above said- you can’t take derivatives with respect to all different variables and expect them to be equal.
$endgroup$
– MathIsLife12
Jan 5 at 4:19
$begingroup$
(This comment was from me -- deleted when reading the fact that ayt the end of your question you acknowledged something was sketchy in the variables you differentiated with respect to. Essentially, I was saying you were "differentiating simultaneously wrt 3 variables," in different places: $xy$, $x$, $y$ -- and that's not allowed.)
$endgroup$
– Clement C.
Jan 5 at 4:21
$begingroup$
(This comment was from me -- deleted when reading the fact that ayt the end of your question you acknowledged something was sketchy in the variables you differentiated with respect to. Essentially, I was saying you were "differentiating simultaneously wrt 3 variables," in different places: $xy$, $x$, $y$ -- and that's not allowed.)
$endgroup$
– Clement C.
Jan 5 at 4:21
$begingroup$
Thanks guys; this clears things up! I appreciate the fast responses.
$endgroup$
– user2192320
Jan 5 at 4:24
$begingroup$
Thanks guys; this clears things up! I appreciate the fast responses.
$endgroup$
– user2192320
Jan 5 at 4:24
$begingroup$
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
$endgroup$
– John Doe
Jan 5 at 5:01
$begingroup$
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
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– John Doe
Jan 5 at 5:01
add a comment |
2 Answers
2
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oldest
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Yes, as you noted, it matters which variable you take a derivative with respect to. Since we have two variables, if we assume they are independent, then we need to use partial derivatives. If we chose $x$, then $$frac{partial log(xy)}{partial x}=frac{y}{xy}=frac1x$$and $$frac{partial}{partial x}(log x+log y)=frac1x+0=frac1x$$So the answers agree. (the same thing happens if we chose $y$ instead).
Alternatively, you could take a total derivative. $$mathrm d(log xy)=frac{partial log xy}{partial x}mathrm dx+frac{partial log xy}{partial y}mathrm dy=frac1xmathrm dx+frac1y mathrm dy$$This agrees with $$d(log x+log y)=frac1xmathrm dx+frac1y mathrm dy$$
So there are no inconsistencies.
answered Jan 5 at 4:21
John DoeJohn Doe
11.1k11238
$endgroup$
$begingroup$
Thank you for the quick response; this is my favorite answer posted so far.
$endgroup$
– user2192320
Jan 5 at 4:25
$begingroup$
No problem, happy to help!
$endgroup$
– John Doe
Jan 5 at 5:02
2
$begingroup$
@user2192320 you may want to accept it then by clicking the tick mark on the left of the answer.
$endgroup$
– Ruslan
Jan 5 at 16:24
$begingroup$
Hey Ruslan, sorry for the late follow up; I accepted the response and checked the tick mark. Thanks for the guidance.
$endgroup$
– user2192320
Jan 16 at 20:10
add a comment |
$begingroup$
The issue is that your differential operator $D$ does not behave in the way you think it does!
$$
D(log(xy))=frac{1}{xy}D(xy)=frac{1}{xy}(ydx+xdy)=frac{1}{x}dx+frac{1}{y}dy=D(log(x))+D(log(y))
$$
At no point in this computation do we actually have $frac{1}{xy}=frac{1}{x}+frac{1}{y}$ -- we are working with differentials, and not derivatives.
answered Jan 5 at 4:19
ItsJustTranscendenceBroItsJustTranscendenceBro
1712
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
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active
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$begingroup$
Yes, as you noted, it matters which variable you take a derivative with respect to. Since we have two variables, if we assume they are independent, then we need to use partial derivatives. If we chose $x$, then $$frac{partial log(xy)}{partial x}=frac{y}{xy}=frac1x$$and $$frac{partial}{partial x}(log x+log y)=frac1x+0=frac1x$$So the answers agree. (the same thing happens if we chose $y$ instead).
Alternatively, you could take a total derivative. $$mathrm d(log xy)=frac{partial log xy}{partial x}mathrm dx+frac{partial log xy}{partial y}mathrm dy=frac1xmathrm dx+frac1y mathrm dy$$This agrees with $$d(log x+log y)=frac1xmathrm dx+frac1y mathrm dy$$
So there are no inconsistencies.
answered Jan 5 at 4:21
John DoeJohn Doe
11.1k11238
$endgroup$
$begingroup$
Thank you for the quick response; this is my favorite answer posted so far.
$endgroup$
– user2192320
Jan 5 at 4:25
$begingroup$
No problem, happy to help!
$endgroup$
– John Doe
Jan 5 at 5:02
2
$begingroup$
@user2192320 you may want to accept it then by clicking the tick mark on the left of the answer.
$endgroup$
– Ruslan
Jan 5 at 16:24
$begingroup$
Hey Ruslan, sorry for the late follow up; I accepted the response and checked the tick mark. Thanks for the guidance.
$endgroup$
– user2192320
Jan 16 at 20:10
add a comment |
$begingroup$
Yes, as you noted, it matters which variable you take a derivative with respect to. Since we have two variables, if we assume they are independent, then we need to use partial derivatives. If we chose $x$, then $$frac{partial log(xy)}{partial x}=frac{y}{xy}=frac1x$$and $$frac{partial}{partial x}(log x+log y)=frac1x+0=frac1x$$So the answers agree. (the same thing happens if we chose $y$ instead).
Alternatively, you could take a total derivative. $$mathrm d(log xy)=frac{partial log xy}{partial x}mathrm dx+frac{partial log xy}{partial y}mathrm dy=frac1xmathrm dx+frac1y mathrm dy$$This agrees with $$d(log x+log y)=frac1xmathrm dx+frac1y mathrm dy$$
So there are no inconsistencies.
answered Jan 5 at 4:21
John DoeJohn Doe
11.1k11238
$endgroup$
$begingroup$
Thank you for the quick response; this is my favorite answer posted so far.
$endgroup$
– user2192320
Jan 5 at 4:25
$begingroup$
No problem, happy to help!
$endgroup$
– John Doe
Jan 5 at 5:02
2
$begingroup$
@user2192320 you may want to accept it then by clicking the tick mark on the left of the answer.
$endgroup$
– Ruslan
Jan 5 at 16:24
$begingroup$
Hey Ruslan, sorry for the late follow up; I accepted the response and checked the tick mark. Thanks for the guidance.
$endgroup$
– user2192320
Jan 16 at 20:10
add a comment |
$begingroup$
Yes, as you noted, it matters which variable you take a derivative with respect to. Since we have two variables, if we assume they are independent, then we need to use partial derivatives. If we chose $x$, then $$frac{partial log(xy)}{partial x}=frac{y}{xy}=frac1x$$and $$frac{partial}{partial x}(log x+log y)=frac1x+0=frac1x$$So the answers agree. (the same thing happens if we chose $y$ instead).
Alternatively, you could take a total derivative. $$mathrm d(log xy)=frac{partial log xy}{partial x}mathrm dx+frac{partial log xy}{partial y}mathrm dy=frac1xmathrm dx+frac1y mathrm dy$$This agrees with $$d(log x+log y)=frac1xmathrm dx+frac1y mathrm dy$$
So there are no inconsistencies.
answered Jan 5 at 4:21
John DoeJohn Doe
11.1k11238
$endgroup$
Yes, as you noted, it matters which variable you take a derivative with respect to. Since we have two variables, if we assume they are independent, then we need to use partial derivatives. If we chose $x$, then $$frac{partial log(xy)}{partial x}=frac{y}{xy}=frac1x$$and $$frac{partial}{partial x}(log x+log y)=frac1x+0=frac1x$$So the answers agree. (the same thing happens if we chose $y$ instead).
Alternatively, you could take a total derivative. $$mathrm d(log xy)=frac{partial log xy}{partial x}mathrm dx+frac{partial log xy}{partial y}mathrm dy=frac1xmathrm dx+frac1y mathrm dy$$This agrees with $$d(log x+log y)=frac1xmathrm dx+frac1y mathrm dy$$
So there are no inconsistencies.
answered Jan 5 at 4:21
John DoeJohn Doe
11.1k11238
answered Jan 5 at 4:21
John DoeJohn Doe
11.1k11238
answered Jan 5 at 4:21
John DoeJohn Doe
11.1k11238
answered Jan 5 at 4:21
John DoeJohn Doe
11.1k11238
11.1k11238
$begingroup$
Thank you for the quick response; this is my favorite answer posted so far.
$endgroup$
– user2192320
Jan 5 at 4:25
$begingroup$
No problem, happy to help!
$endgroup$
– John Doe
Jan 5 at 5:02
2
$begingroup$
@user2192320 you may want to accept it then by clicking the tick mark on the left of the answer.
$endgroup$
– Ruslan
Jan 5 at 16:24
$begingroup$
Hey Ruslan, sorry for the late follow up; I accepted the response and checked the tick mark. Thanks for the guidance.
$endgroup$
– user2192320
Jan 16 at 20:10
add a comment |
$begingroup$
Thank you for the quick response; this is my favorite answer posted so far.
$endgroup$
– user2192320
Jan 5 at 4:25
$begingroup$
No problem, happy to help!
$endgroup$
– John Doe
Jan 5 at 5:02
2
$begingroup$
@user2192320 you may want to accept it then by clicking the tick mark on the left of the answer.
$endgroup$
– Ruslan
Jan 5 at 16:24
$begingroup$
Hey Ruslan, sorry for the late follow up; I accepted the response and checked the tick mark. Thanks for the guidance.
$endgroup$
– user2192320
Jan 16 at 20:10
$begingroup$
Thank you for the quick response; this is my favorite answer posted so far.
$endgroup$
– user2192320
Jan 5 at 4:25
$begingroup$
Thank you for the quick response; this is my favorite answer posted so far.
$endgroup$
– user2192320
Jan 5 at 4:25
$begingroup$
No problem, happy to help!
$endgroup$
– John Doe
Jan 5 at 5:02
$begingroup$
No problem, happy to help!
$endgroup$
– John Doe
Jan 5 at 5:02
2
2
$begingroup$
@user2192320 you may want to accept it then by clicking the tick mark on the left of the answer.
$endgroup$
– Ruslan
Jan 5 at 16:24
$begingroup$
@user2192320 you may want to accept it then by clicking the tick mark on the left of the answer.
$endgroup$
– Ruslan
Jan 5 at 16:24
$begingroup$
Hey Ruslan, sorry for the late follow up; I accepted the response and checked the tick mark. Thanks for the guidance.
$endgroup$
– user2192320
Jan 16 at 20:10
$begingroup$
Hey Ruslan, sorry for the late follow up; I accepted the response and checked the tick mark. Thanks for the guidance.
$endgroup$
– user2192320
Jan 16 at 20:10
add a comment |
$begingroup$
The issue is that your differential operator $D$ does not behave in the way you think it does!
$$
D(log(xy))=frac{1}{xy}D(xy)=frac{1}{xy}(ydx+xdy)=frac{1}{x}dx+frac{1}{y}dy=D(log(x))+D(log(y))
$$
At no point in this computation do we actually have $frac{1}{xy}=frac{1}{x}+frac{1}{y}$ -- we are working with differentials, and not derivatives.
answered Jan 5 at 4:19
ItsJustTranscendenceBroItsJustTranscendenceBro
1712
$endgroup$
add a comment |
$begingroup$
The issue is that your differential operator $D$ does not behave in the way you think it does!
$$
D(log(xy))=frac{1}{xy}D(xy)=frac{1}{xy}(ydx+xdy)=frac{1}{x}dx+frac{1}{y}dy=D(log(x))+D(log(y))
$$
At no point in this computation do we actually have $frac{1}{xy}=frac{1}{x}+frac{1}{y}$ -- we are working with differentials, and not derivatives.
answered Jan 5 at 4:19
ItsJustTranscendenceBroItsJustTranscendenceBro
1712
$endgroup$
add a comment |
$begingroup$
The issue is that your differential operator $D$ does not behave in the way you think it does!
$$
D(log(xy))=frac{1}{xy}D(xy)=frac{1}{xy}(ydx+xdy)=frac{1}{x}dx+frac{1}{y}dy=D(log(x))+D(log(y))
$$
At no point in this computation do we actually have $frac{1}{xy}=frac{1}{x}+frac{1}{y}$ -- we are working with differentials, and not derivatives.
answered Jan 5 at 4:19
ItsJustTranscendenceBroItsJustTranscendenceBro
1712
$endgroup$
The issue is that your differential operator $D$ does not behave in the way you think it does!
$$
D(log(xy))=frac{1}{xy}D(xy)=frac{1}{xy}(ydx+xdy)=frac{1}{x}dx+frac{1}{y}dy=D(log(x))+D(log(y))
$$
At no point in this computation do we actually have $frac{1}{xy}=frac{1}{x}+frac{1}{y}$ -- we are working with differentials, and not derivatives.
answered Jan 5 at 4:19
ItsJustTranscendenceBroItsJustTranscendenceBro
1712
answered Jan 5 at 4:19
ItsJustTranscendenceBroItsJustTranscendenceBro
1712
answered Jan 5 at 4:19
ItsJustTranscendenceBroItsJustTranscendenceBro
1712
answered Jan 5 at 4:19
ItsJustTranscendenceBroItsJustTranscendenceBro
1712
1712
add a comment |
add a comment |
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Like the comment above said- you can’t take derivatives with respect to all different variables and expect them to be equal.
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– MathIsLife12
Jan 5 at 4:19
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(This comment was from me -- deleted when reading the fact that ayt the end of your question you acknowledged something was sketchy in the variables you differentiated with respect to. Essentially, I was saying you were "differentiating simultaneously wrt 3 variables," in different places: $xy$, $x$, $y$ -- and that's not allowed.)
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– Clement C.
Jan 5 at 4:21
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Thanks guys; this clears things up! I appreciate the fast responses.
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– user2192320
Jan 5 at 4:24
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After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
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– John Doe
Jan 5 at 5:01