Differentiation of a function wrt different variables
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Suppose a function
"f (xy)", then does it equate to "f ' (xy) . x" when partial differentiating wrt y, and f '(xy).y when partial differentiating wrt x?
Cause I don't get it why differentiating f with x and y would both yield f ', any help?
calculus multivariable-calculus
asked Nov 18 at 5:08
Shahbaz Khan
31
add a comment |
up vote
0
down vote
favorite
Suppose a function
"f (xy)", then does it equate to "f ' (xy) . x" when partial differentiating wrt y, and f '(xy).y when partial differentiating wrt x?
Cause I don't get it why differentiating f with x and y would both yield f ', any help?
calculus multivariable-calculus
asked Nov 18 at 5:08
Shahbaz Khan
31
Careful: if this is a function of two variables you need the product rule - and saying $f'$ here makes no sense
– Sean Roberson
Nov 18 at 5:18
@SeanRoberson Exactly! I know right, I just saw it in my book, where it was mentioning how to form Partial Differential Equations, by eliminating arbitrary functions!
– Shahbaz Khan
Nov 18 at 5:33
As you have written it, $f$ appears to be a function of a single variable, say $t mapsto f(t)$, which you are evaluating at $t=xy$. This is a composition of functions, hence chain rule applies. So, for example, $$frac{mathrm{d}}{mathrm{d}x} f(xy) = f'(xy) frac{mathrm{d}}{mathrm{dx}}(xy) = f'(xy) y. $$
– Xander Henderson
Nov 18 at 6:04
@XanderHenderson I already know that much but $frac{mathrm{d}}{mathrm{d}y} f(xy) = f'(xy) frac{mathrm{d}}{mathrm{dy}}(xy) = f'(xy) x$. Is this f ' the same as the one we get when differentiating with x?
– Shahbaz Khan
Nov 18 at 13:54
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose a function
"f (xy)", then does it equate to "f ' (xy) . x" when partial differentiating wrt y, and f '(xy).y when partial differentiating wrt x?
Cause I don't get it why differentiating f with x and y would both yield f ', any help?
calculus multivariable-calculus
asked Nov 18 at 5:08
Shahbaz Khan
31
Suppose a function
"f (xy)", then does it equate to "f ' (xy) . x" when partial differentiating wrt y, and f '(xy).y when partial differentiating wrt x?
Cause I don't get it why differentiating f with x and y would both yield f ', any help?
calculus multivariable-calculus
calculus multivariable-calculus
asked Nov 18 at 5:08
Shahbaz Khan
31
asked Nov 18 at 5:08
Shahbaz Khan
31
asked Nov 18 at 5:08
Shahbaz Khan
31
asked Nov 18 at 5:08
Shahbaz Khan
31
asked Nov 18 at 5:08
Shahbaz Khan
31
31
Careful: if this is a function of two variables you need the product rule - and saying $f'$ here makes no sense
– Sean Roberson
Nov 18 at 5:18
@SeanRoberson Exactly! I know right, I just saw it in my book, where it was mentioning how to form Partial Differential Equations, by eliminating arbitrary functions!
– Shahbaz Khan
Nov 18 at 5:33
As you have written it, $f$ appears to be a function of a single variable, say $t mapsto f(t)$, which you are evaluating at $t=xy$. This is a composition of functions, hence chain rule applies. So, for example, $$frac{mathrm{d}}{mathrm{d}x} f(xy) = f'(xy) frac{mathrm{d}}{mathrm{dx}}(xy) = f'(xy) y. $$
– Xander Henderson
Nov 18 at 6:04
@XanderHenderson I already know that much but $frac{mathrm{d}}{mathrm{d}y} f(xy) = f'(xy) frac{mathrm{d}}{mathrm{dy}}(xy) = f'(xy) x$. Is this f ' the same as the one we get when differentiating with x?
– Shahbaz Khan
Nov 18 at 13:54
add a comment |
Careful: if this is a function of two variables you need the product rule - and saying $f'$ here makes no sense
– Sean Roberson
Nov 18 at 5:18
@SeanRoberson Exactly! I know right, I just saw it in my book, where it was mentioning how to form Partial Differential Equations, by eliminating arbitrary functions!
– Shahbaz Khan
Nov 18 at 5:33
As you have written it, $f$ appears to be a function of a single variable, say $t mapsto f(t)$, which you are evaluating at $t=xy$. This is a composition of functions, hence chain rule applies. So, for example, $$frac{mathrm{d}}{mathrm{d}x} f(xy) = f'(xy) frac{mathrm{d}}{mathrm{dx}}(xy) = f'(xy) y. $$
– Xander Henderson
Nov 18 at 6:04
@XanderHenderson I already know that much but $frac{mathrm{d}}{mathrm{d}y} f(xy) = f'(xy) frac{mathrm{d}}{mathrm{dy}}(xy) = f'(xy) x$. Is this f ' the same as the one we get when differentiating with x?
– Shahbaz Khan
Nov 18 at 13:54
Careful: if this is a function of two variables you need the product rule - and saying $f'$ here makes no sense
– Sean Roberson
Nov 18 at 5:18
Careful: if this is a function of two variables you need the product rule - and saying $f'$ here makes no sense
– Sean Roberson
Nov 18 at 5:18
@SeanRoberson Exactly! I know right, I just saw it in my book, where it was mentioning how to form Partial Differential Equations, by eliminating arbitrary functions!
– Shahbaz Khan
Nov 18 at 5:33
@SeanRoberson Exactly! I know right, I just saw it in my book, where it was mentioning how to form Partial Differential Equations, by eliminating arbitrary functions!
– Shahbaz Khan
Nov 18 at 5:33
As you have written it, $f$ appears to be a function of a single variable, say $t mapsto f(t)$, which you are evaluating at $t=xy$. This is a composition of functions, hence chain rule applies. So, for example, $$frac{mathrm{d}}{mathrm{d}x} f(xy) = f'(xy) frac{mathrm{d}}{mathrm{dx}}(xy) = f'(xy) y. $$
– Xander Henderson
Nov 18 at 6:04
As you have written it, $f$ appears to be a function of a single variable, say $t mapsto f(t)$, which you are evaluating at $t=xy$. This is a composition of functions, hence chain rule applies. So, for example, $$frac{mathrm{d}}{mathrm{d}x} f(xy) = f'(xy) frac{mathrm{d}}{mathrm{dx}}(xy) = f'(xy) y. $$
– Xander Henderson
Nov 18 at 6:04
@XanderHenderson I already know that much but $frac{mathrm{d}}{mathrm{d}y} f(xy) = f'(xy) frac{mathrm{d}}{mathrm{dy}}(xy) = f'(xy) x$. Is this f ' the same as the one we get when differentiating with x?
– Shahbaz Khan
Nov 18 at 13:54
@XanderHenderson I already know that much but $frac{mathrm{d}}{mathrm{d}y} f(xy) = f'(xy) frac{mathrm{d}}{mathrm{dy}}(xy) = f'(xy) x$. Is this f ' the same as the one we get when differentiating with x?
– Shahbaz Khan
Nov 18 at 13:54
add a comment |
1 Answer
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oldest
votes
up vote
0
down vote
accepted
Based on the comments, my guess is that the Newton notation is getting in the way a little. It might be more clear what is going on if we are a little more explicit with the functional composition, and if we use the Leibniz notation instead.
Suppose that
$$ f : mathbb{R} to mathbb{R}$$
is a real-valued function of a single real variable, and let
$$ u : mathbb{R}^2 to mathbb{R} $$
be a real-valued function of two real variables defined by the formula
$$ u = u(x,y) = xy.$$
Then the function $g = fcirc u$ is a real-valued function of two real variables. The partial derivatives of $g$ can be found via the chain rule:
$$ g_x
= frac{mathrm{d}(fcirc u)}{mathrm{d}x}
= frac{mathrm{d}f}{mathrm{d}u} cdot frac{mathrm{d}u}{mathrm{d}x}
= frac{mathrm{d}f}{mathrm{d}u} cdot y,$$
where $frac{mathrm{d}f}{mathrm{d}u}$ is the derivative of the single variable function $f$. This derivative is not taken with respect to either $x$ or $y$, but instead with respect to the variable of $f$, which we have labeled $u$. Note that as $f$ is a function of a single variable $u$, it makes sense to write $f'(u)$ instead of $frac{mathrm{d}f}{mathrm{d}u}(u)$.
Perhaps an example would help: suppose that $f(u) = cos(u)$, and take $u = xy$. Then
$$ frac{mathrm{d}f}{mathrm{d}x}
= frac{mathrm{df}}{mathrm{d}u} cdot frac{mathrm{d}u}{mathrm{d}x}. tag{1}$$
But
$$ frac{mathrm{d}f}{mathrm{d}u} = frac{mathrm{d}}{mathrm{d}u} cos(u)
= -sin(u). $$
Then, as $u = xy$, it follows that $frac{mathrm{d}f}{mathrm{d}u} = -sin(xy)$. Also note that $frac{mathrm{d}u}{mathrm{d}x} = y$. Substituting these into (1), we conclude that
$$frac{mathrm{d}f}{mathrm{d}x} = -sin(xy) cdot y. $$
answered Nov 18 at 15:44
Xander Henderson
13.9k93552
Yeah, thank you, that makes sense. I really hate books that don’t lay out a clear notation convention that they’re using.
– Shahbaz Khan
yesterday
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Based on the comments, my guess is that the Newton notation is getting in the way a little. It might be more clear what is going on if we are a little more explicit with the functional composition, and if we use the Leibniz notation instead.
Suppose that
$$ f : mathbb{R} to mathbb{R}$$
is a real-valued function of a single real variable, and let
$$ u : mathbb{R}^2 to mathbb{R} $$
be a real-valued function of two real variables defined by the formula
$$ u = u(x,y) = xy.$$
Then the function $g = fcirc u$ is a real-valued function of two real variables. The partial derivatives of $g$ can be found via the chain rule:
$$ g_x
= frac{mathrm{d}(fcirc u)}{mathrm{d}x}
= frac{mathrm{d}f}{mathrm{d}u} cdot frac{mathrm{d}u}{mathrm{d}x}
= frac{mathrm{d}f}{mathrm{d}u} cdot y,$$
where $frac{mathrm{d}f}{mathrm{d}u}$ is the derivative of the single variable function $f$. This derivative is not taken with respect to either $x$ or $y$, but instead with respect to the variable of $f$, which we have labeled $u$. Note that as $f$ is a function of a single variable $u$, it makes sense to write $f'(u)$ instead of $frac{mathrm{d}f}{mathrm{d}u}(u)$.
Perhaps an example would help: suppose that $f(u) = cos(u)$, and take $u = xy$. Then
$$ frac{mathrm{d}f}{mathrm{d}x}
= frac{mathrm{df}}{mathrm{d}u} cdot frac{mathrm{d}u}{mathrm{d}x}. tag{1}$$
But
$$ frac{mathrm{d}f}{mathrm{d}u} = frac{mathrm{d}}{mathrm{d}u} cos(u)
= -sin(u). $$
Then, as $u = xy$, it follows that $frac{mathrm{d}f}{mathrm{d}u} = -sin(xy)$. Also note that $frac{mathrm{d}u}{mathrm{d}x} = y$. Substituting these into (1), we conclude that
$$frac{mathrm{d}f}{mathrm{d}x} = -sin(xy) cdot y. $$
answered Nov 18 at 15:44
Xander Henderson
13.9k93552
Yeah, thank you, that makes sense. I really hate books that don’t lay out a clear notation convention that they’re using.
– Shahbaz Khan
yesterday
add a comment |
up vote
0
down vote
accepted
Based on the comments, my guess is that the Newton notation is getting in the way a little. It might be more clear what is going on if we are a little more explicit with the functional composition, and if we use the Leibniz notation instead.
Suppose that
$$ f : mathbb{R} to mathbb{R}$$
is a real-valued function of a single real variable, and let
$$ u : mathbb{R}^2 to mathbb{R} $$
be a real-valued function of two real variables defined by the formula
$$ u = u(x,y) = xy.$$
Then the function $g = fcirc u$ is a real-valued function of two real variables. The partial derivatives of $g$ can be found via the chain rule:
$$ g_x
= frac{mathrm{d}(fcirc u)}{mathrm{d}x}
= frac{mathrm{d}f}{mathrm{d}u} cdot frac{mathrm{d}u}{mathrm{d}x}
= frac{mathrm{d}f}{mathrm{d}u} cdot y,$$
where $frac{mathrm{d}f}{mathrm{d}u}$ is the derivative of the single variable function $f$. This derivative is not taken with respect to either $x$ or $y$, but instead with respect to the variable of $f$, which we have labeled $u$. Note that as $f$ is a function of a single variable $u$, it makes sense to write $f'(u)$ instead of $frac{mathrm{d}f}{mathrm{d}u}(u)$.
Perhaps an example would help: suppose that $f(u) = cos(u)$, and take $u = xy$. Then
$$ frac{mathrm{d}f}{mathrm{d}x}
= frac{mathrm{df}}{mathrm{d}u} cdot frac{mathrm{d}u}{mathrm{d}x}. tag{1}$$
But
$$ frac{mathrm{d}f}{mathrm{d}u} = frac{mathrm{d}}{mathrm{d}u} cos(u)
= -sin(u). $$
Then, as $u = xy$, it follows that $frac{mathrm{d}f}{mathrm{d}u} = -sin(xy)$. Also note that $frac{mathrm{d}u}{mathrm{d}x} = y$. Substituting these into (1), we conclude that
$$frac{mathrm{d}f}{mathrm{d}x} = -sin(xy) cdot y. $$
answered Nov 18 at 15:44
Xander Henderson
13.9k93552
Yeah, thank you, that makes sense. I really hate books that don’t lay out a clear notation convention that they’re using.
– Shahbaz Khan
yesterday
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Based on the comments, my guess is that the Newton notation is getting in the way a little. It might be more clear what is going on if we are a little more explicit with the functional composition, and if we use the Leibniz notation instead.
Suppose that
$$ f : mathbb{R} to mathbb{R}$$
is a real-valued function of a single real variable, and let
$$ u : mathbb{R}^2 to mathbb{R} $$
be a real-valued function of two real variables defined by the formula
$$ u = u(x,y) = xy.$$
Then the function $g = fcirc u$ is a real-valued function of two real variables. The partial derivatives of $g$ can be found via the chain rule:
$$ g_x
= frac{mathrm{d}(fcirc u)}{mathrm{d}x}
= frac{mathrm{d}f}{mathrm{d}u} cdot frac{mathrm{d}u}{mathrm{d}x}
= frac{mathrm{d}f}{mathrm{d}u} cdot y,$$
where $frac{mathrm{d}f}{mathrm{d}u}$ is the derivative of the single variable function $f$. This derivative is not taken with respect to either $x$ or $y$, but instead with respect to the variable of $f$, which we have labeled $u$. Note that as $f$ is a function of a single variable $u$, it makes sense to write $f'(u)$ instead of $frac{mathrm{d}f}{mathrm{d}u}(u)$.
Perhaps an example would help: suppose that $f(u) = cos(u)$, and take $u = xy$. Then
$$ frac{mathrm{d}f}{mathrm{d}x}
= frac{mathrm{df}}{mathrm{d}u} cdot frac{mathrm{d}u}{mathrm{d}x}. tag{1}$$
But
$$ frac{mathrm{d}f}{mathrm{d}u} = frac{mathrm{d}}{mathrm{d}u} cos(u)
= -sin(u). $$
Then, as $u = xy$, it follows that $frac{mathrm{d}f}{mathrm{d}u} = -sin(xy)$. Also note that $frac{mathrm{d}u}{mathrm{d}x} = y$. Substituting these into (1), we conclude that
$$frac{mathrm{d}f}{mathrm{d}x} = -sin(xy) cdot y. $$
answered Nov 18 at 15:44
Xander Henderson
13.9k93552
Based on the comments, my guess is that the Newton notation is getting in the way a little. It might be more clear what is going on if we are a little more explicit with the functional composition, and if we use the Leibniz notation instead.
Suppose that
$$ f : mathbb{R} to mathbb{R}$$
is a real-valued function of a single real variable, and let
$$ u : mathbb{R}^2 to mathbb{R} $$
be a real-valued function of two real variables defined by the formula
$$ u = u(x,y) = xy.$$
Then the function $g = fcirc u$ is a real-valued function of two real variables. The partial derivatives of $g$ can be found via the chain rule:
$$ g_x
= frac{mathrm{d}(fcirc u)}{mathrm{d}x}
= frac{mathrm{d}f}{mathrm{d}u} cdot frac{mathrm{d}u}{mathrm{d}x}
= frac{mathrm{d}f}{mathrm{d}u} cdot y,$$
where $frac{mathrm{d}f}{mathrm{d}u}$ is the derivative of the single variable function $f$. This derivative is not taken with respect to either $x$ or $y$, but instead with respect to the variable of $f$, which we have labeled $u$. Note that as $f$ is a function of a single variable $u$, it makes sense to write $f'(u)$ instead of $frac{mathrm{d}f}{mathrm{d}u}(u)$.
Perhaps an example would help: suppose that $f(u) = cos(u)$, and take $u = xy$. Then
$$ frac{mathrm{d}f}{mathrm{d}x}
= frac{mathrm{df}}{mathrm{d}u} cdot frac{mathrm{d}u}{mathrm{d}x}. tag{1}$$
But
$$ frac{mathrm{d}f}{mathrm{d}u} = frac{mathrm{d}}{mathrm{d}u} cos(u)
= -sin(u). $$
Then, as $u = xy$, it follows that $frac{mathrm{d}f}{mathrm{d}u} = -sin(xy)$. Also note that $frac{mathrm{d}u}{mathrm{d}x} = y$. Substituting these into (1), we conclude that
$$frac{mathrm{d}f}{mathrm{d}x} = -sin(xy) cdot y. $$
answered Nov 18 at 15:44
Xander Henderson
13.9k93552
answered Nov 18 at 15:44
Xander Henderson
13.9k93552
answered Nov 18 at 15:44
Xander Henderson
13.9k93552
answered Nov 18 at 15:44
Xander Henderson
13.9k93552
13.9k93552
Yeah, thank you, that makes sense. I really hate books that don’t lay out a clear notation convention that they’re using.
– Shahbaz Khan
yesterday
add a comment |
Yeah, thank you, that makes sense. I really hate books that don’t lay out a clear notation convention that they’re using.
– Shahbaz Khan
yesterday
Yeah, thank you, that makes sense. I really hate books that don’t lay out a clear notation convention that they’re using.
– Shahbaz Khan
yesterday
Yeah, thank you, that makes sense. I really hate books that don’t lay out a clear notation convention that they’re using.
– Shahbaz Khan
yesterday
add a comment |
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Careful: if this is a function of two variables you need the product rule - and saying $f'$ here makes no sense
– Sean Roberson
Nov 18 at 5:18
@SeanRoberson Exactly! I know right, I just saw it in my book, where it was mentioning how to form Partial Differential Equations, by eliminating arbitrary functions!
– Shahbaz Khan
Nov 18 at 5:33
As you have written it, $f$ appears to be a function of a single variable, say $t mapsto f(t)$, which you are evaluating at $t=xy$. This is a composition of functions, hence chain rule applies. So, for example, $$frac{mathrm{d}}{mathrm{d}x} f(xy) = f'(xy) frac{mathrm{d}}{mathrm{dx}}(xy) = f'(xy) y. $$
– Xander Henderson
Nov 18 at 6:04
@XanderHenderson I already know that much but $frac{mathrm{d}}{mathrm{d}y} f(xy) = f'(xy) frac{mathrm{d}}{mathrm{dy}}(xy) = f'(xy) x$. Is this f ' the same as the one we get when differentiating with x?
– Shahbaz Khan
Nov 18 at 13:54