How do you differentiate with respect to y?
$begingroup$
Find the gradient of
$$z=x^y$$
I understand how to get it with respect to $x$ since $y$ is treated as a constant. But when trying to solve it with respect to $y$, why is it incorrect to implicitly differentiate and use the product rule:
$$ln(z)=ycdot ln(x)$$
$$frac{(z_y)}{z}=Bigr(ycdot frac{1}{x}Bigr)+(1cdot ln(x))$$
$$z_y=zBigr(frac{y}{x}+ln(x)Bigr)$$
$$z_y=x^yBigr(frac{y}{x}+ln(x)Bigr)$$
derivatives
edited 4 hours ago
Xander Henderson
14.2k103554
asked 4 hours ago
Random StudentRandom Student
182
New contributor
Random Student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Find the gradient of
$$z=x^y$$
I understand how to get it with respect to $x$ since $y$ is treated as a constant. But when trying to solve it with respect to $y$, why is it incorrect to implicitly differentiate and use the product rule:
$$ln(z)=ycdot ln(x)$$
$$frac{(z_y)}{z}=Bigr(ycdot frac{1}{x}Bigr)+(1cdot ln(x))$$
$$z_y=zBigr(frac{y}{x}+ln(x)Bigr)$$
$$z_y=x^yBigr(frac{y}{x}+ln(x)Bigr)$$
derivatives
edited 4 hours ago
Xander Henderson
14.2k103554
asked 4 hours ago
Random StudentRandom Student
182
New contributor
Random Student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Find the gradient of
$$z=x^y$$
I understand how to get it with respect to $x$ since $y$ is treated as a constant. But when trying to solve it with respect to $y$, why is it incorrect to implicitly differentiate and use the product rule:
$$ln(z)=ycdot ln(x)$$
$$frac{(z_y)}{z}=Bigr(ycdot frac{1}{x}Bigr)+(1cdot ln(x))$$
$$z_y=zBigr(frac{y}{x}+ln(x)Bigr)$$
$$z_y=x^yBigr(frac{y}{x}+ln(x)Bigr)$$
derivatives
edited 4 hours ago
Xander Henderson
14.2k103554
asked 4 hours ago
Random StudentRandom Student
182
New contributor
Random Student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Find the gradient of
$$z=x^y$$
I understand how to get it with respect to $x$ since $y$ is treated as a constant. But when trying to solve it with respect to $y$, why is it incorrect to implicitly differentiate and use the product rule:
$$ln(z)=ycdot ln(x)$$
$$frac{(z_y)}{z}=Bigr(ycdot frac{1}{x}Bigr)+(1cdot ln(x))$$
$$z_y=zBigr(frac{y}{x}+ln(x)Bigr)$$
$$z_y=x^yBigr(frac{y}{x}+ln(x)Bigr)$$
derivatives
derivatives
edited 4 hours ago
Xander Henderson
14.2k103554
asked 4 hours ago
Random StudentRandom Student
182
New contributor
Random Student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 hours ago
Xander Henderson
14.2k103554
asked 4 hours ago
Random StudentRandom Student
182
New contributor
Random Student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 hours ago
Xander Henderson
14.2k103554
edited 4 hours ago
Xander Henderson
14.2k103554
edited 4 hours ago
Xander Henderson
14.2k103554
14.2k103554
asked 4 hours ago
Random StudentRandom Student
182
New contributor
Random Student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Random StudentRandom Student
182
asked 4 hours ago
Random StudentRandom Student
182
182
New contributor
Random Student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Random Student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Random Student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
In single-variable calculus, a first application of implicit differentiation is typically to find the derivative of $x mapsto a^x$, where $a > 0$. The typical argument is
begin{align*}
y = a^x
&implies log(y) = xlog(a) \
&implies frac{1}{y} y' = log(a) \
&implies y' = ylog(a) = a^x log(a).
end{align*}
In your problem, when you differentiate with respect to $y$, you need to regard $x$ as a constant (you should also probably assume that $x > 0$). You can then apply the single-variable result to get
$$ z_y
= frac{mathrm{d}}{mathrm{d}y} x^y
= x^ylog(x).
$$
answered 4 hours ago
Xander HendersonXander Henderson
14.2k103554
$endgroup$
add a comment |
$begingroup$
When take the derivative of $x^y$ with respect to $y$, are you treating $x$ as a constant or as a function of $y$?
If the former then (using your notation) you get $frac{z_y}{z}=1 times ln(x)$ and so $z_y = x^y ln(x)$
If the latter then you get $frac{z_y}{z}=y times frac1xtimes x_y+1 times ln(x)$ and so $z_y =y x^{y-1} x_y+ x^y ln(x)$. Then you might also want to say $z_x = y x^{y-1}+ x^y ln(x) y_x$ when taking the derivative of $x^y$ with respect to $x$
answered 4 hours ago
HenryHenry
98.7k476163
$endgroup$
add a comment |
$begingroup$
When we're differentiating with respect to $y$, $x$ should be treated as a constant. The $frac yx$ term in there that comes from differentiating $x$ shouldn't be there.
answered 4 hours ago
jmerryjmerry
3,592514
$endgroup$
add a comment |
$begingroup$
Note that $x $ is not a function of $y $. Hence $x_y=0$. So the correct one is:
$$ln(z)=ycdotln(x)$$
$$frac{(z_y)}{z}=1cdotln(x)$$
$$z_y=zln(x)$$
$$z_y=x^yln(x)$$
answered 4 hours ago
Thomas ShelbyThomas Shelby
2,119220
$endgroup$
add a comment |
$begingroup$
The variables $x$ and $y$ are to be considered independent when computing the gradient of $z$, that is,
$nabla z = (z_x, z_y), tag 1$
which is a vector. We have
$z = x^y, tag 2$
whence
$z_x = yx^{y - 1}, tag 3$
as our OP Random Student has noted. As for $z_y$, we have
$ln z = y ln x, tag 4$
whence
$dfrac{z_y}{z} = ln x, tag 5$
or
$z_y = z ln x = x^y ln x; tag 6$
thus,
$nabla z = (z_x, z_y) = (yx^{y - 1}, x^y ln x). tag 7$
In (4)-(6), we have implicitly differentated $z$ with respect to $y$ alone, holding $x$ constant.
answered 4 hours ago
Robert LewisRobert Lewis
44.5k22964
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In single-variable calculus, a first application of implicit differentiation is typically to find the derivative of $x mapsto a^x$, where $a > 0$. The typical argument is
begin{align*}
y = a^x
&implies log(y) = xlog(a) \
&implies frac{1}{y} y' = log(a) \
&implies y' = ylog(a) = a^x log(a).
end{align*}
In your problem, when you differentiate with respect to $y$, you need to regard $x$ as a constant (you should also probably assume that $x > 0$). You can then apply the single-variable result to get
$$ z_y
= frac{mathrm{d}}{mathrm{d}y} x^y
= x^ylog(x).
$$
answered 4 hours ago
Xander HendersonXander Henderson
14.2k103554
$endgroup$
add a comment |
$begingroup$
In single-variable calculus, a first application of implicit differentiation is typically to find the derivative of $x mapsto a^x$, where $a > 0$. The typical argument is
begin{align*}
y = a^x
&implies log(y) = xlog(a) \
&implies frac{1}{y} y' = log(a) \
&implies y' = ylog(a) = a^x log(a).
end{align*}
In your problem, when you differentiate with respect to $y$, you need to regard $x$ as a constant (you should also probably assume that $x > 0$). You can then apply the single-variable result to get
$$ z_y
= frac{mathrm{d}}{mathrm{d}y} x^y
= x^ylog(x).
$$
answered 4 hours ago
Xander HendersonXander Henderson
14.2k103554
$endgroup$
add a comment |
$begingroup$
In single-variable calculus, a first application of implicit differentiation is typically to find the derivative of $x mapsto a^x$, where $a > 0$. The typical argument is
begin{align*}
y = a^x
&implies log(y) = xlog(a) \
&implies frac{1}{y} y' = log(a) \
&implies y' = ylog(a) = a^x log(a).
end{align*}
In your problem, when you differentiate with respect to $y$, you need to regard $x$ as a constant (you should also probably assume that $x > 0$). You can then apply the single-variable result to get
$$ z_y
= frac{mathrm{d}}{mathrm{d}y} x^y
= x^ylog(x).
$$
answered 4 hours ago
Xander HendersonXander Henderson
14.2k103554
$endgroup$
In single-variable calculus, a first application of implicit differentiation is typically to find the derivative of $x mapsto a^x$, where $a > 0$. The typical argument is
begin{align*}
y = a^x
&implies log(y) = xlog(a) \
&implies frac{1}{y} y' = log(a) \
&implies y' = ylog(a) = a^x log(a).
end{align*}
In your problem, when you differentiate with respect to $y$, you need to regard $x$ as a constant (you should also probably assume that $x > 0$). You can then apply the single-variable result to get
$$ z_y
= frac{mathrm{d}}{mathrm{d}y} x^y
= x^ylog(x).
$$
answered 4 hours ago
Xander HendersonXander Henderson
14.2k103554
answered 4 hours ago
Xander HendersonXander Henderson
14.2k103554
answered 4 hours ago
Xander HendersonXander Henderson
14.2k103554
answered 4 hours ago
Xander HendersonXander Henderson
14.2k103554
14.2k103554
add a comment |
add a comment |
$begingroup$
When take the derivative of $x^y$ with respect to $y$, are you treating $x$ as a constant or as a function of $y$?
If the former then (using your notation) you get $frac{z_y}{z}=1 times ln(x)$ and so $z_y = x^y ln(x)$
If the latter then you get $frac{z_y}{z}=y times frac1xtimes x_y+1 times ln(x)$ and so $z_y =y x^{y-1} x_y+ x^y ln(x)$. Then you might also want to say $z_x = y x^{y-1}+ x^y ln(x) y_x$ when taking the derivative of $x^y$ with respect to $x$
answered 4 hours ago
HenryHenry
98.7k476163
$endgroup$
add a comment |
$begingroup$
When take the derivative of $x^y$ with respect to $y$, are you treating $x$ as a constant or as a function of $y$?
If the former then (using your notation) you get $frac{z_y}{z}=1 times ln(x)$ and so $z_y = x^y ln(x)$
If the latter then you get $frac{z_y}{z}=y times frac1xtimes x_y+1 times ln(x)$ and so $z_y =y x^{y-1} x_y+ x^y ln(x)$. Then you might also want to say $z_x = y x^{y-1}+ x^y ln(x) y_x$ when taking the derivative of $x^y$ with respect to $x$
answered 4 hours ago
HenryHenry
98.7k476163
$endgroup$
add a comment |
$begingroup$
When take the derivative of $x^y$ with respect to $y$, are you treating $x$ as a constant or as a function of $y$?
If the former then (using your notation) you get $frac{z_y}{z}=1 times ln(x)$ and so $z_y = x^y ln(x)$
If the latter then you get $frac{z_y}{z}=y times frac1xtimes x_y+1 times ln(x)$ and so $z_y =y x^{y-1} x_y+ x^y ln(x)$. Then you might also want to say $z_x = y x^{y-1}+ x^y ln(x) y_x$ when taking the derivative of $x^y$ with respect to $x$
answered 4 hours ago
HenryHenry
98.7k476163
$endgroup$
When take the derivative of $x^y$ with respect to $y$, are you treating $x$ as a constant or as a function of $y$?
If the former then (using your notation) you get $frac{z_y}{z}=1 times ln(x)$ and so $z_y = x^y ln(x)$
If the latter then you get $frac{z_y}{z}=y times frac1xtimes x_y+1 times ln(x)$ and so $z_y =y x^{y-1} x_y+ x^y ln(x)$. Then you might also want to say $z_x = y x^{y-1}+ x^y ln(x) y_x$ when taking the derivative of $x^y$ with respect to $x$
answered 4 hours ago
HenryHenry
98.7k476163
answered 4 hours ago
HenryHenry
98.7k476163
answered 4 hours ago
HenryHenry
98.7k476163
answered 4 hours ago
HenryHenry
98.7k476163
98.7k476163
add a comment |
add a comment |
$begingroup$
When we're differentiating with respect to $y$, $x$ should be treated as a constant. The $frac yx$ term in there that comes from differentiating $x$ shouldn't be there.
answered 4 hours ago
jmerryjmerry
3,592514
$endgroup$
add a comment |
$begingroup$
When we're differentiating with respect to $y$, $x$ should be treated as a constant. The $frac yx$ term in there that comes from differentiating $x$ shouldn't be there.
answered 4 hours ago
jmerryjmerry
3,592514
$endgroup$
add a comment |
$begingroup$
When we're differentiating with respect to $y$, $x$ should be treated as a constant. The $frac yx$ term in there that comes from differentiating $x$ shouldn't be there.
answered 4 hours ago
jmerryjmerry
3,592514
$endgroup$
When we're differentiating with respect to $y$, $x$ should be treated as a constant. The $frac yx$ term in there that comes from differentiating $x$ shouldn't be there.
answered 4 hours ago
jmerryjmerry
3,592514
answered 4 hours ago
jmerryjmerry
3,592514
answered 4 hours ago
jmerryjmerry
3,592514
answered 4 hours ago
jmerryjmerry
3,592514
3,592514
add a comment |
add a comment |
$begingroup$
Note that $x $ is not a function of $y $. Hence $x_y=0$. So the correct one is:
$$ln(z)=ycdotln(x)$$
$$frac{(z_y)}{z}=1cdotln(x)$$
$$z_y=zln(x)$$
$$z_y=x^yln(x)$$
answered 4 hours ago
Thomas ShelbyThomas Shelby
2,119220
$endgroup$
add a comment |
$begingroup$
Note that $x $ is not a function of $y $. Hence $x_y=0$. So the correct one is:
$$ln(z)=ycdotln(x)$$
$$frac{(z_y)}{z}=1cdotln(x)$$
$$z_y=zln(x)$$
$$z_y=x^yln(x)$$
answered 4 hours ago
Thomas ShelbyThomas Shelby
2,119220
$endgroup$
add a comment |
$begingroup$
Note that $x $ is not a function of $y $. Hence $x_y=0$. So the correct one is:
$$ln(z)=ycdotln(x)$$
$$frac{(z_y)}{z}=1cdotln(x)$$
$$z_y=zln(x)$$
$$z_y=x^yln(x)$$
answered 4 hours ago
Thomas ShelbyThomas Shelby
2,119220
$endgroup$
Note that $x $ is not a function of $y $. Hence $x_y=0$. So the correct one is:
$$ln(z)=ycdotln(x)$$
$$frac{(z_y)}{z}=1cdotln(x)$$
$$z_y=zln(x)$$
$$z_y=x^yln(x)$$
answered 4 hours ago
Thomas ShelbyThomas Shelby
2,119220
answered 4 hours ago
Thomas ShelbyThomas Shelby
2,119220
answered 4 hours ago
Thomas ShelbyThomas Shelby
2,119220
answered 4 hours ago
Thomas ShelbyThomas Shelby
2,119220
2,119220
add a comment |
add a comment |
$begingroup$
The variables $x$ and $y$ are to be considered independent when computing the gradient of $z$, that is,
$nabla z = (z_x, z_y), tag 1$
which is a vector. We have
$z = x^y, tag 2$
whence
$z_x = yx^{y - 1}, tag 3$
as our OP Random Student has noted. As for $z_y$, we have
$ln z = y ln x, tag 4$
whence
$dfrac{z_y}{z} = ln x, tag 5$
or
$z_y = z ln x = x^y ln x; tag 6$
thus,
$nabla z = (z_x, z_y) = (yx^{y - 1}, x^y ln x). tag 7$
In (4)-(6), we have implicitly differentated $z$ with respect to $y$ alone, holding $x$ constant.
answered 4 hours ago
Robert LewisRobert Lewis
44.5k22964
$endgroup$
add a comment |
$begingroup$
The variables $x$ and $y$ are to be considered independent when computing the gradient of $z$, that is,
$nabla z = (z_x, z_y), tag 1$
which is a vector. We have
$z = x^y, tag 2$
whence
$z_x = yx^{y - 1}, tag 3$
as our OP Random Student has noted. As for $z_y$, we have
$ln z = y ln x, tag 4$
whence
$dfrac{z_y}{z} = ln x, tag 5$
or
$z_y = z ln x = x^y ln x; tag 6$
thus,
$nabla z = (z_x, z_y) = (yx^{y - 1}, x^y ln x). tag 7$
In (4)-(6), we have implicitly differentated $z$ with respect to $y$ alone, holding $x$ constant.
answered 4 hours ago
Robert LewisRobert Lewis
44.5k22964
$endgroup$
add a comment |
$begingroup$
The variables $x$ and $y$ are to be considered independent when computing the gradient of $z$, that is,
$nabla z = (z_x, z_y), tag 1$
which is a vector. We have
$z = x^y, tag 2$
whence
$z_x = yx^{y - 1}, tag 3$
as our OP Random Student has noted. As for $z_y$, we have
$ln z = y ln x, tag 4$
whence
$dfrac{z_y}{z} = ln x, tag 5$
or
$z_y = z ln x = x^y ln x; tag 6$
thus,
$nabla z = (z_x, z_y) = (yx^{y - 1}, x^y ln x). tag 7$
In (4)-(6), we have implicitly differentated $z$ with respect to $y$ alone, holding $x$ constant.
answered 4 hours ago
Robert LewisRobert Lewis
44.5k22964
$endgroup$
The variables $x$ and $y$ are to be considered independent when computing the gradient of $z$, that is,
$nabla z = (z_x, z_y), tag 1$
which is a vector. We have
$z = x^y, tag 2$
whence
$z_x = yx^{y - 1}, tag 3$
as our OP Random Student has noted. As for $z_y$, we have
$ln z = y ln x, tag 4$
whence
$dfrac{z_y}{z} = ln x, tag 5$
or
$z_y = z ln x = x^y ln x; tag 6$
thus,
$nabla z = (z_x, z_y) = (yx^{y - 1}, x^y ln x). tag 7$
In (4)-(6), we have implicitly differentated $z$ with respect to $y$ alone, holding $x$ constant.
answered 4 hours ago
Robert LewisRobert Lewis
44.5k22964
edited 2 hours ago
answered 4 hours ago
Robert LewisRobert Lewis
44.5k22964
answered 4 hours ago
Robert LewisRobert Lewis
44.5k22964
answered 4 hours ago
Robert LewisRobert Lewis
44.5k22964
44.5k22964
add a comment |
add a comment |
Random Student is a new contributor. Be nice, and check out our Code of Conduct.
Random Student is a new contributor. Be nice, and check out our Code of Conduct.
Random Student is a new contributor. Be nice, and check out our Code of Conduct.
Random Student is a new contributor. Be nice, and check out our Code of Conduct.
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StackExchange.using('gps', function() { StackExchange.gps.track('embedded_signup_form.view', { location: 'question_page' }); });
$window.unbind('scroll', onScroll);
}
};
$window.on('scroll', onScroll);
});
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
var $window = $(window),
onScroll = function(e) {
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom)) {
StackExchange.using('gps', function() { StackExchange.gps.track('embedded_signup_form.view', { location: 'question_page' }); });
$window.unbind('scroll', onScroll);
}
};
$window.on('scroll', onScroll);
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
