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.
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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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