Derivative of $a^x$ for $a<0$ The Next CEO of Stack OverflowUsing L'hopital's rule to evaluate the limit $lim_x to 0 left(e^x+xright)^frac1x$How to choose a numeric approach for derivatesWhy is this limit $frace^xx^x-1$coming out wrong?Partial Derivative of the one variable functionHow to prove the limit of “the exponential of a sequence”Proof of the derivative of ln(x)I need help with this limit: $lim_nto inftysum_k=2^n frac1klog k$ No idea how to approach it.Step-by-step derivative of $left ( fracc_1 xc_2 x + c_3 + c_4 sqrtc_5 x right)^c_6x + c_7 + c_8 sqrtc_9 x$Value of limit $(1+e^-x)^2^x logx$Limit of $lim_x to inftyfracC_3a^xD_1+(ka)^xD_2$
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Derivative of $a^x$ for $a
The Next CEO of Stack OverflowUsing L'hopital's rule to evaluate the limit $lim_x to 0 left(e^x+xright)^frac1x$How to choose a numeric approach for derivatesWhy is this limit $frace^xx^x-1$coming out wrong?Partial Derivative of the one variable functionHow to prove the limit of “the exponential of a sequence”Proof of the derivative of ln(x)I need help with this limit: $lim_nto inftysum_k=2^n frac1klog k$ No idea how to approach it.Step-by-step derivative of $left ( fracc_1 xc_2 x + c_3 + c_4 sqrtc_5 x right)^c_6x + c_7 + c_8 sqrtc_9 x$Value of limit $(1+e^-x)^2^x logx$Limit of $lim_x to inftyfrac^A+a^xD_1+(ka)^xD_2$
$begingroup$
Is it correct to write that
$$ fracmathrmdmathrmdx left( a^x right)
= a^x log(a) $$
also for $a < 0$, where $log$ is the natural logarithm?
I was trying to compute the derivative of $(-10)^x$. WolframAlpha gave the following result:
$$ fracmathrmdmathrmdx bigl( (-10)^x bigr)
= (-10)^x (log(10) + mathrmipi),$$
but I am not sure if it is correct, because in this case the argument of logarithm is negative.
The reason why I am asking this question is becasue I want to calculate the following limit
$$lim_x to inftyfrac^A+a^xD_1+(ka)^xD_2$$
knowing that $-1<a<1$, $-1<k<1$, $1<A<2$ and $x in mathbbN$.
Since for $x to infty$ we have $left[frac00right]$, I use the de L'Hopital's rule.
$$lim_x to inftyfrac^A+a^xD_1+(ka)^xD_2=left[frac00right]=lim_x to inftyfracC_1a^x+C_2(ka)^xa^xlog(a)D_1+(ka)^xlog(ka)D_2\fracC_1a^x+C_2(ka)^x-C_3a^xlog(a)D_1+(ka)^xlog(ka)D_2$$
Now, I take the expression $a^x$ in front of both nominator and denominator and reduce it obtaining
$$lim_x to inftyfracAlog(a)D_1+(k)^xlog(ka)D_2\frac-Alog(a)D_1+(k)^xlog(ka)D_2$$
and finally I obtain $frac^A-1sgn(C_3)C_1log(a)log(a)D_1$ as the result.
At some point, I calculate the derivative of $a^x$ which is equal to $a^xlog(a)$ and I am sure that it is correct in the case of $0<a<1$, but what if $-1<a<0$? Can I write that the derivative of $a^x$ is $a^xlog(a)$ when $-1<a<0$? If not, how should I calculate this limit in the case of $-1<a<0$?
derivatives logarithms exponential-function
asked yesterday
GateyGatey
163
New contributor
Gatey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
|
show 5 more comments
$begingroup$
Is it correct to write that
$$ fracmathrmdmathrmdx left( a^x right)
= a^x log(a) $$
also for $a < 0$, where $log$ is the natural logarithm?
I was trying to compute the derivative of $(-10)^x$. WolframAlpha gave the following result:
$$ fracmathrmdmathrmdx bigl( (-10)^x bigr)
= (-10)^x (log(10) + mathrmipi),$$
but I am not sure if it is correct, because in this case the argument of logarithm is negative.
The reason why I am asking this question is becasue I want to calculate the following limit
$$lim_x to inftyfrac^A+a^xD_1+(ka)^xD_2$$
knowing that $-1<a<1$, $-1<k<1$, $1<A<2$ and $x in mathbbN$.
Since for $x to infty$ we have $left[frac00right]$, I use the de L'Hopital's rule.
$$lim_x to inftyfrac^A+a^xD_1+(ka)^xD_2=left[frac00right]=lim_x to inftyfracC_1a^x+C_2(ka)^xa^xlog(a)D_1+(ka)^xlog(ka)D_2\fracC_1a^x+C_2(ka)^x-C_3a^xlog(a)D_1+(ka)^xlog(ka)D_2$$
Now, I take the expression $a^x$ in front of both nominator and denominator and reduce it obtaining
$$lim_x to inftyfracAlog(a)D_1+(k)^xlog(ka)D_2\frac-Alog(a)D_1+(k)^xlog(ka)D_2$$
and finally I obtain $frac^A-1sgn(C_3)C_1log(a)log(a)D_1$ as the result.
At some point, I calculate the derivative of $a^x$ which is equal to $a^xlog(a)$ and I am sure that it is correct in the case of $0<a<1$, but what if $-1<a<0$? Can I write that the derivative of $a^x$ is $a^xlog(a)$ when $-1<a<0$? If not, how should I calculate this limit in the case of $-1<a<0$?
derivatives logarithms exponential-function
asked yesterday
GateyGatey
163
New contributor
Gatey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
6
$begingroup$
For $a<0$, there isn't even an obvious way to define that function, let alone compute its derivative.
$endgroup$
– Jack M
yesterday
1
$begingroup$
It would help a potential answer-writer give you a useful answer if you explained your motivation for asking this. The short answer is that (1) there is a way to make sense of the identity that makes it true (by regarding it as a complex-valued function), but (2) given what little context there is in the question it doesn't seem like that's what you want, and (3) even then there are issues about what one means by $a^x$ for $a$ not a positive real number.
$endgroup$
– Travis
yesterday
1
$begingroup$
"In this case the argument of logarithm is negative." Note that Wolfram Alpha already told you that $log(-10) = log(10) + mathrmi pi$.
$endgroup$
– L. F.
yesterday
1
$begingroup$
Thank you. Now, I explained my motivation - the question is edited.
$endgroup$
– Gatey
yesterday
1
$begingroup$
The edit fails to address all the concerns. For example, you seem to think that $C_1a^x+C_2(ka)^x$ has a sign, when either $a$ or $ka$ may be negative?
$endgroup$
– Jyrki Lahtonen
yesterday
|
show 5 more comments
$begingroup$
Is it correct to write that
$$ fracmathrmdmathrmdx left( a^x right)
= a^x log(a) $$
also for $a < 0$, where $log$ is the natural logarithm?
I was trying to compute the derivative of $(-10)^x$. WolframAlpha gave the following result:
$$ fracmathrmdmathrmdx bigl( (-10)^x bigr)
= (-10)^x (log(10) + mathrmipi),$$
but I am not sure if it is correct, because in this case the argument of logarithm is negative.
The reason why I am asking this question is becasue I want to calculate the following limit
$$lim_x to inftyfrac^A+a^xD_1+(ka)^xD_2$$
knowing that $-1<a<1$, $-1<k<1$, $1<A<2$ and $x in mathbbN$.
Since for $x to infty$ we have $left[frac00right]$, I use the de L'Hopital's rule.
$$lim_x to inftyfrac^A+a^xD_1+(ka)^xD_2=left[frac00right]=lim_x to inftyfracC_1a^x+C_2(ka)^xa^xlog(a)D_1+(ka)^xlog(ka)D_2\fracC_1a^x+C_2(ka)^x-C_3a^xlog(a)D_1+(ka)^xlog(ka)D_2$$
Now, I take the expression $a^x$ in front of both nominator and denominator and reduce it obtaining
$$lim_x to inftyfracAlog(a)D_1+(k)^xlog(ka)D_2\frac-Alog(a)D_1+(k)^xlog(ka)D_2$$
and finally I obtain $frac^A-1sgn(C_3)C_1log(a)log(a)D_1$ as the result.
At some point, I calculate the derivative of $a^x$ which is equal to $a^xlog(a)$ and I am sure that it is correct in the case of $0<a<1$, but what if $-1<a<0$? Can I write that the derivative of $a^x$ is $a^xlog(a)$ when $-1<a<0$? If not, how should I calculate this limit in the case of $-1<a<0$?
derivatives logarithms exponential-function
asked yesterday
GateyGatey
163
New contributor
Gatey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Is it correct to write that
$$ fracmathrmdmathrmdx left( a^x right)
= a^x log(a) $$
also for $a < 0$, where $log$ is the natural logarithm?
I was trying to compute the derivative of $(-10)^x$. WolframAlpha gave the following result:
$$ fracmathrmdmathrmdx bigl( (-10)^x bigr)
= (-10)^x (log(10) + mathrmipi),$$
but I am not sure if it is correct, because in this case the argument of logarithm is negative.
The reason why I am asking this question is becasue I want to calculate the following limit
$$lim_x to inftyfrac^A+a^xD_1+(ka)^xD_2$$
knowing that $-1<a<1$, $-1<k<1$, $1<A<2$ and $x in mathbbN$.
Since for $x to infty$ we have $left[frac00right]$, I use the de L'Hopital's rule.
$$lim_x to inftyfrac^A+a^xD_1+(ka)^xD_2=left[frac00right]=lim_x to inftyfracC_1a^x+C_2(ka)^xa^xlog(a)D_1+(ka)^xlog(ka)D_2\fracC_1a^x+C_2(ka)^x-C_3a^xlog(a)D_1+(ka)^xlog(ka)D_2$$
Now, I take the expression $a^x$ in front of both nominator and denominator and reduce it obtaining
$$lim_x to inftyfracAlog(a)D_1+(k)^xlog(ka)D_2\frac-Alog(a)D_1+(k)^xlog(ka)D_2$$
and finally I obtain $frac^A-1sgn(C_3)C_1log(a)log(a)D_1$ as the result.
At some point, I calculate the derivative of $a^x$ which is equal to $a^xlog(a)$ and I am sure that it is correct in the case of $0<a<1$, but what if $-1<a<0$? Can I write that the derivative of $a^x$ is $a^xlog(a)$ when $-1<a<0$? If not, how should I calculate this limit in the case of $-1<a<0$?
derivatives logarithms exponential-function
derivatives logarithms exponential-function
asked yesterday
GateyGatey
163
New contributor
Gatey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
GateyGatey
163
New contributor
Gatey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Gatey
asked yesterday
GateyGatey
163
New contributor
Gatey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
GateyGatey
163
asked yesterday
GateyGatey
163
163
New contributor
Gatey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Gatey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Gatey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
6
$begingroup$
For $a<0$, there isn't even an obvious way to define that function, let alone compute its derivative.
$endgroup$
– Jack M
yesterday
1
$begingroup$
It would help a potential answer-writer give you a useful answer if you explained your motivation for asking this. The short answer is that (1) there is a way to make sense of the identity that makes it true (by regarding it as a complex-valued function), but (2) given what little context there is in the question it doesn't seem like that's what you want, and (3) even then there are issues about what one means by $a^x$ for $a$ not a positive real number.
$endgroup$
– Travis
yesterday
1
$begingroup$
"In this case the argument of logarithm is negative." Note that Wolfram Alpha already told you that $log(-10) = log(10) + mathrmi pi$.
$endgroup$
– L. F.
yesterday
1
$begingroup$
Thank you. Now, I explained my motivation - the question is edited.
$endgroup$
– Gatey
yesterday
1
$begingroup$
The edit fails to address all the concerns. For example, you seem to think that $C_1a^x+C_2(ka)^x$ has a sign, when either $a$ or $ka$ may be negative?
$endgroup$
– Jyrki Lahtonen
yesterday
|
show 5 more comments
6
$begingroup$
For $a<0$, there isn't even an obvious way to define that function, let alone compute its derivative.
$endgroup$
– Jack M
yesterday
1
$begingroup$
It would help a potential answer-writer give you a useful answer if you explained your motivation for asking this. The short answer is that (1) there is a way to make sense of the identity that makes it true (by regarding it as a complex-valued function), but (2) given what little context there is in the question it doesn't seem like that's what you want, and (3) even then there are issues about what one means by $a^x$ for $a$ not a positive real number.
$endgroup$
– Travis
yesterday
1
$begingroup$
"In this case the argument of logarithm is negative." Note that Wolfram Alpha already told you that $log(-10) = log(10) + mathrmi pi$.
$endgroup$
– L. F.
yesterday
1
$begingroup$
Thank you. Now, I explained my motivation - the question is edited.
$endgroup$
– Gatey
yesterday
1
$begingroup$
The edit fails to address all the concerns. For example, you seem to think that $C_1a^x+C_2(ka)^x$ has a sign, when either $a$ or $ka$ may be negative?
$endgroup$
– Jyrki Lahtonen
yesterday
6
6
$begingroup$
For $a<0$, there isn't even an obvious way to define that function, let alone compute its derivative.
$endgroup$
– Jack M
yesterday
$begingroup$
For $a<0$, there isn't even an obvious way to define that function, let alone compute its derivative.
$endgroup$
– Jack M
yesterday
1
1
$begingroup$
It would help a potential answer-writer give you a useful answer if you explained your motivation for asking this. The short answer is that (1) there is a way to make sense of the identity that makes it true (by regarding it as a complex-valued function), but (2) given what little context there is in the question it doesn't seem like that's what you want, and (3) even then there are issues about what one means by $a^x$ for $a$ not a positive real number.
$endgroup$
– Travis
yesterday
$begingroup$
It would help a potential answer-writer give you a useful answer if you explained your motivation for asking this. The short answer is that (1) there is a way to make sense of the identity that makes it true (by regarding it as a complex-valued function), but (2) given what little context there is in the question it doesn't seem like that's what you want, and (3) even then there are issues about what one means by $a^x$ for $a$ not a positive real number.
$endgroup$
– Travis
yesterday
1
1
$begingroup$
"In this case the argument of logarithm is negative." Note that Wolfram Alpha already told you that $log(-10) = log(10) + mathrmi pi$.
$endgroup$
– L. F.
yesterday
$begingroup$
"In this case the argument of logarithm is negative." Note that Wolfram Alpha already told you that $log(-10) = log(10) + mathrmi pi$.
$endgroup$
– L. F.
yesterday
1
1
$begingroup$
Thank you. Now, I explained my motivation - the question is edited.
$endgroup$
– Gatey
yesterday
$begingroup$
Thank you. Now, I explained my motivation - the question is edited.
$endgroup$
– Gatey
yesterday
1
1
$begingroup$
The edit fails to address all the concerns. For example, you seem to think that $C_1a^x+C_2(ka)^x$ has a sign, when either $a$ or $ka$ may be negative?
$endgroup$
– Jyrki Lahtonen
yesterday
$begingroup$
The edit fails to address all the concerns. For example, you seem to think that $C_1a^x+C_2(ka)^x$ has a sign, when either $a$ or $ka$ may be negative?
$endgroup$
– Jyrki Lahtonen
yesterday
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Of course $a^x$ where $a<0$ and $x$ is irrational is a complex number, not a real number. So let us define
$$
mathrmLog; a
$$
to be the principal value of the multivalued function $log a$; that is the solution $z$ of $e^z=a$ with argument in $(-pi,pi]$. Then define
$$
a^x = exp(x;mathrmLog;a)
$$
So, when $a$ is real and negative, we have $mathrmLog;a = log|a|+ipi$ and
$$
a^x = exp(x;mathrmLog;a) = exp(x; (log|a|+ipi))
=e^xlog; e^ipi x = |a|^x;big(cos(pi x)+isin(pi x)big)
$$
Now we are in a position to consider the derivative formula.
Use the product rule,
$$
fracddxbig(a^xbig) = |a|^x;big(-pisin(pi x)+i pi cos(pi x)big)
+|a|^x;log|a|;big(cos(pi x)+isin(pi x)big)
$$
and on the other hand
beginalign
a^x;mathrmLog;a &=
|a|^x;big(cos(pi x)+isin(pi x)big);big(log|a|+ipibig)
\
&=|a|^x;big(cos(pi x)+isin(pi x)big);big(log|a|big)
+|a|^x;big(cos(pi x)+isin(pi x)big);big(ipibig)
endalign
They agree.
answered yesterday
GEdgarGEdgar
63.3k268172
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Of course $a^x$ where $a<0$ and $x$ is irrational is a complex number, not a real number. So let us define
$$
mathrmLog; a
$$
to be the principal value of the multivalued function $log a$; that is the solution $z$ of $e^z=a$ with argument in $(-pi,pi]$. Then define
$$
a^x = exp(x;mathrmLog;a)
$$
So, when $a$ is real and negative, we have $mathrmLog;a = log|a|+ipi$ and
$$
a^x = exp(x;mathrmLog;a) = exp(x; (log|a|+ipi))
=e^xlog; e^ipi x = |a|^x;big(cos(pi x)+isin(pi x)big)
$$
Now we are in a position to consider the derivative formula.
Use the product rule,
$$
fracddxbig(a^xbig) = |a|^x;big(-pisin(pi x)+i pi cos(pi x)big)
+|a|^x;log|a|;big(cos(pi x)+isin(pi x)big)
$$
and on the other hand
beginalign
a^x;mathrmLog;a &=
|a|^x;big(cos(pi x)+isin(pi x)big);big(log|a|+ipibig)
\
&=|a|^x;big(cos(pi x)+isin(pi x)big);big(log|a|big)
+|a|^x;big(cos(pi x)+isin(pi x)big);big(ipibig)
endalign
They agree.
answered yesterday
GEdgarGEdgar
63.3k268172
$endgroup$
add a comment |
$begingroup$
Of course $a^x$ where $a<0$ and $x$ is irrational is a complex number, not a real number. So let us define
$$
mathrmLog; a
$$
to be the principal value of the multivalued function $log a$; that is the solution $z$ of $e^z=a$ with argument in $(-pi,pi]$. Then define
$$
a^x = exp(x;mathrmLog;a)
$$
So, when $a$ is real and negative, we have $mathrmLog;a = log|a|+ipi$ and
$$
a^x = exp(x;mathrmLog;a) = exp(x; (log|a|+ipi))
=e^xlog; e^ipi x = |a|^x;big(cos(pi x)+isin(pi x)big)
$$
Now we are in a position to consider the derivative formula.
Use the product rule,
$$
fracddxbig(a^xbig) = |a|^x;big(-pisin(pi x)+i pi cos(pi x)big)
+|a|^x;log|a|;big(cos(pi x)+isin(pi x)big)
$$
and on the other hand
beginalign
a^x;mathrmLog;a &=
|a|^x;big(cos(pi x)+isin(pi x)big);big(log|a|+ipibig)
\
&=|a|^x;big(cos(pi x)+isin(pi x)big);big(log|a|big)
+|a|^x;big(cos(pi x)+isin(pi x)big);big(ipibig)
endalign
They agree.
answered yesterday
GEdgarGEdgar
63.3k268172
$endgroup$
add a comment |
$begingroup$
Of course $a^x$ where $a<0$ and $x$ is irrational is a complex number, not a real number. So let us define
$$
mathrmLog; a
$$
to be the principal value of the multivalued function $log a$; that is the solution $z$ of $e^z=a$ with argument in $(-pi,pi]$. Then define
$$
a^x = exp(x;mathrmLog;a)
$$
So, when $a$ is real and negative, we have $mathrmLog;a = log|a|+ipi$ and
$$
a^x = exp(x;mathrmLog;a) = exp(x; (log|a|+ipi))
=e^xlog; e^ipi x = |a|^x;big(cos(pi x)+isin(pi x)big)
$$
Now we are in a position to consider the derivative formula.
Use the product rule,
$$
fracddxbig(a^xbig) = |a|^x;big(-pisin(pi x)+i pi cos(pi x)big)
+|a|^x;log|a|;big(cos(pi x)+isin(pi x)big)
$$
and on the other hand
beginalign
a^x;mathrmLog;a &=
|a|^x;big(cos(pi x)+isin(pi x)big);big(log|a|+ipibig)
\
&=|a|^x;big(cos(pi x)+isin(pi x)big);big(log|a|big)
+|a|^x;big(cos(pi x)+isin(pi x)big);big(ipibig)
endalign
They agree.
answered yesterday
GEdgarGEdgar
63.3k268172
$endgroup$
Of course $a^x$ where $a<0$ and $x$ is irrational is a complex number, not a real number. So let us define
$$
mathrmLog; a
$$
to be the principal value of the multivalued function $log a$; that is the solution $z$ of $e^z=a$ with argument in $(-pi,pi]$. Then define
$$
a^x = exp(x;mathrmLog;a)
$$
So, when $a$ is real and negative, we have $mathrmLog;a = log|a|+ipi$ and
$$
a^x = exp(x;mathrmLog;a) = exp(x; (log|a|+ipi))
=e^xlog; e^ipi x = |a|^x;big(cos(pi x)+isin(pi x)big)
$$
Now we are in a position to consider the derivative formula.
Use the product rule,
$$
fracddxbig(a^xbig) = |a|^x;big(-pisin(pi x)+i pi cos(pi x)big)
+|a|^x;log|a|;big(cos(pi x)+isin(pi x)big)
$$
and on the other hand
beginalign
a^x;mathrmLog;a &=
|a|^x;big(cos(pi x)+isin(pi x)big);big(log|a|+ipibig)
\
&=|a|^x;big(cos(pi x)+isin(pi x)big);big(log|a|big)
+|a|^x;big(cos(pi x)+isin(pi x)big);big(ipibig)
endalign
They agree.
answered yesterday
GEdgarGEdgar
63.3k268172
edited yesterday
answered yesterday
GEdgarGEdgar
63.3k268172
answered yesterday
GEdgarGEdgar
63.3k268172
answered yesterday
GEdgarGEdgar
63.3k268172
63.3k268172
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6
$begingroup$
For $a<0$, there isn't even an obvious way to define that function, let alone compute its derivative.
$endgroup$
– Jack M
yesterday
1
$begingroup$
It would help a potential answer-writer give you a useful answer if you explained your motivation for asking this. The short answer is that (1) there is a way to make sense of the identity that makes it true (by regarding it as a complex-valued function), but (2) given what little context there is in the question it doesn't seem like that's what you want, and (3) even then there are issues about what one means by $a^x$ for $a$ not a positive real number.
$endgroup$
– Travis
yesterday
1
$begingroup$
"In this case the argument of logarithm is negative." Note that Wolfram Alpha already told you that $log(-10) = log(10) + mathrmi pi$.
$endgroup$
– L. F.
yesterday
1
$begingroup$
Thank you. Now, I explained my motivation - the question is edited.
$endgroup$
– Gatey
yesterday
1
$begingroup$
The edit fails to address all the concerns. For example, you seem to think that $C_1a^x+C_2(ka)^x$ has a sign, when either $a$ or $ka$ may be negative?
$endgroup$
– Jyrki Lahtonen
yesterday