Mixing
Why there should be a negative sign in the chain rule?
$begingroup$
Suppose I have a function $$f(x,y)=0$$ and suppose I want to find $frac{dx}{dy}$ then I think I can get it by doing $$frac{dx}{dy}=frac{df}{dy}frac{dx}{df}.$$ But I do not understand why do we need to put a negative sign in front of $frac{df}{dy}frac{dx}{df}$. Please explain. Thanks in advance.
calculus derivatives partial-derivative
asked Jan 30 at 17:11
Frank MosesFrank Moses
1,199419
$endgroup$
add a comment |
$begingroup$
Suppose I have a function $$f(x,y)=0$$ and suppose I want to find $frac{dx}{dy}$ then I think I can get it by doing $$frac{dx}{dy}=frac{df}{dy}frac{dx}{df}.$$ But I do not understand why do we need to put a negative sign in front of $frac{df}{dy}frac{dx}{df}$. Please explain. Thanks in advance.
calculus derivatives partial-derivative
asked Jan 30 at 17:11
Frank MosesFrank Moses
1,199419
$endgroup$
$begingroup$
$f$ is a function of two variables. It does not make sense to write $frac{df}{dy}$.
$endgroup$
– user587192
Jan 30 at 17:21
$begingroup$
dx/df doesn't really make sense here, because x is not an implicit function of f. The sort of chain rule you wrote is really just for 1 variable. Yves' answer shows the correct, multivariable chain rule.
$endgroup$
– Nathaniel Mayer
Jan 30 at 17:26
$begingroup$
I think that your surprise comes from a confusion between $f(x,y)=0$ (a plane curve) and $z=varphi(x,y)$ (a surface).
$endgroup$
– Jean Marie
Jan 30 at 18:36
add a comment |
$begingroup$
Suppose I have a function $$f(x,y)=0$$ and suppose I want to find $frac{dx}{dy}$ then I think I can get it by doing $$frac{dx}{dy}=frac{df}{dy}frac{dx}{df}.$$ But I do not understand why do we need to put a negative sign in front of $frac{df}{dy}frac{dx}{df}$. Please explain. Thanks in advance.
calculus derivatives partial-derivative
asked Jan 30 at 17:11
Frank MosesFrank Moses
1,199419
$endgroup$
Suppose I have a function $$f(x,y)=0$$ and suppose I want to find $frac{dx}{dy}$ then I think I can get it by doing $$frac{dx}{dy}=frac{df}{dy}frac{dx}{df}.$$ But I do not understand why do we need to put a negative sign in front of $frac{df}{dy}frac{dx}{df}$. Please explain. Thanks in advance.
calculus derivatives partial-derivative
calculus derivatives partial-derivative
asked Jan 30 at 17:11
Frank MosesFrank Moses
1,199419
asked Jan 30 at 17:11
Frank MosesFrank Moses
1,199419
asked Jan 30 at 17:11
Frank MosesFrank Moses
1,199419
asked Jan 30 at 17:11
Frank MosesFrank Moses
1,199419
asked Jan 30 at 17:11
Frank MosesFrank Moses
1,199419
1,199419
$begingroup$
$f$ is a function of two variables. It does not make sense to write $frac{df}{dy}$.
$endgroup$
– user587192
Jan 30 at 17:21
$begingroup$
dx/df doesn't really make sense here, because x is not an implicit function of f. The sort of chain rule you wrote is really just for 1 variable. Yves' answer shows the correct, multivariable chain rule.
$endgroup$
– Nathaniel Mayer
Jan 30 at 17:26
$begingroup$
I think that your surprise comes from a confusion between $f(x,y)=0$ (a plane curve) and $z=varphi(x,y)$ (a surface).
$endgroup$
– Jean Marie
Jan 30 at 18:36
add a comment |
$begingroup$
$f$ is a function of two variables. It does not make sense to write $frac{df}{dy}$.
$endgroup$
– user587192
Jan 30 at 17:21
$begingroup$
dx/df doesn't really make sense here, because x is not an implicit function of f. The sort of chain rule you wrote is really just for 1 variable. Yves' answer shows the correct, multivariable chain rule.
$endgroup$
– Nathaniel Mayer
Jan 30 at 17:26
$begingroup$
I think that your surprise comes from a confusion between $f(x,y)=0$ (a plane curve) and $z=varphi(x,y)$ (a surface).
$endgroup$
– Jean Marie
Jan 30 at 18:36
$begingroup$
$f$ is a function of two variables. It does not make sense to write $frac{df}{dy}$.
$endgroup$
– user587192
Jan 30 at 17:21
$begingroup$
$f$ is a function of two variables. It does not make sense to write $frac{df}{dy}$.
$endgroup$
– user587192
Jan 30 at 17:21
$begingroup$
dx/df doesn't really make sense here, because x is not an implicit function of f. The sort of chain rule you wrote is really just for 1 variable. Yves' answer shows the correct, multivariable chain rule.
$endgroup$
– Nathaniel Mayer
Jan 30 at 17:26
$begingroup$
dx/df doesn't really make sense here, because x is not an implicit function of f. The sort of chain rule you wrote is really just for 1 variable. Yves' answer shows the correct, multivariable chain rule.
$endgroup$
– Nathaniel Mayer
Jan 30 at 17:26
$begingroup$
I think that your surprise comes from a confusion between $f(x,y)=0$ (a plane curve) and $z=varphi(x,y)$ (a surface).
$endgroup$
– Jean Marie
Jan 30 at 18:36
$begingroup$
I think that your surprise comes from a confusion between $f(x,y)=0$ (a plane curve) and $z=varphi(x,y)$ (a surface).
$endgroup$
– Jean Marie
Jan 30 at 18:36
add a comment |
2 Answers
2
active
oldest
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$begingroup$
The exact rule is
$$frac{partial f(x,y)}{partial x}dx+frac{partial f(x,y)}{partial y}dy=0impliesfrac{dy}{dx}=-dfrac{dfrac{partial f(x,y)}{partial x}}{dfrac{partial f(x,y)}{partial y}}.$$
E.g., with $$x^2+y^2-1=0$$
$$frac{dy}{dx}=-frac{2x}{2y}.$$
Compare to
$$y=sqrt{1-x^2}$$ then
$$frac{dy}{dx}=-frac{2x}{2sqrt{1-x^2}}$$ (and similarly with the negative branch).
answered Jan 30 at 17:15
Yves DaoustYves Daoust
132k676230
$endgroup$
add a comment |
$begingroup$
$f$ is a function of two variables. It does not make sense to write $frac{df}{dy}$.
If $f(x,y)=0$, the derivative of the implicit function $x(y)$ is given by:
$${displaystyle {frac {dx}{dy}}=-{frac {partial f/partial y}{partial f/partial x}}=-{frac {f_{y}}{f_{x}}},}
$$
where $f_x$ and $f_y$ indicate the partial derivatives of $f$ with respect to $x$ and $y$.
The above formula comes from using the generalized chain rule to obtain the total derivative—with respect to $y$—of both sides of $f(x, y) = 0$:
$${displaystyle {frac {partial f}{partial x}}{frac {dx}{dy}}+{frac {partial f}{partial y}}{frac {dy}{dy}}=0,}
$$
hence
$$
{displaystyle {frac {partial f}{partial y}}
+{frac {partial f}{partial x}}{frac {dx}{dy}}=0,} $$
which, when solved for $dx/dy$, gives the expression above.
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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$begingroup$
The exact rule is
$$frac{partial f(x,y)}{partial x}dx+frac{partial f(x,y)}{partial y}dy=0impliesfrac{dy}{dx}=-dfrac{dfrac{partial f(x,y)}{partial x}}{dfrac{partial f(x,y)}{partial y}}.$$
E.g., with $$x^2+y^2-1=0$$
$$frac{dy}{dx}=-frac{2x}{2y}.$$
Compare to
$$y=sqrt{1-x^2}$$ then
$$frac{dy}{dx}=-frac{2x}{2sqrt{1-x^2}}$$ (and similarly with the negative branch).
answered Jan 30 at 17:15
Yves DaoustYves Daoust
132k676230
$endgroup$
add a comment |
$begingroup$
The exact rule is
$$frac{partial f(x,y)}{partial x}dx+frac{partial f(x,y)}{partial y}dy=0impliesfrac{dy}{dx}=-dfrac{dfrac{partial f(x,y)}{partial x}}{dfrac{partial f(x,y)}{partial y}}.$$
E.g., with $$x^2+y^2-1=0$$
$$frac{dy}{dx}=-frac{2x}{2y}.$$
Compare to
$$y=sqrt{1-x^2}$$ then
$$frac{dy}{dx}=-frac{2x}{2sqrt{1-x^2}}$$ (and similarly with the negative branch).
answered Jan 30 at 17:15
Yves DaoustYves Daoust
132k676230
$endgroup$
add a comment |
$begingroup$
The exact rule is
$$frac{partial f(x,y)}{partial x}dx+frac{partial f(x,y)}{partial y}dy=0impliesfrac{dy}{dx}=-dfrac{dfrac{partial f(x,y)}{partial x}}{dfrac{partial f(x,y)}{partial y}}.$$
E.g., with $$x^2+y^2-1=0$$
$$frac{dy}{dx}=-frac{2x}{2y}.$$
Compare to
$$y=sqrt{1-x^2}$$ then
$$frac{dy}{dx}=-frac{2x}{2sqrt{1-x^2}}$$ (and similarly with the negative branch).
answered Jan 30 at 17:15
Yves DaoustYves Daoust
132k676230
$endgroup$
The exact rule is
$$frac{partial f(x,y)}{partial x}dx+frac{partial f(x,y)}{partial y}dy=0impliesfrac{dy}{dx}=-dfrac{dfrac{partial f(x,y)}{partial x}}{dfrac{partial f(x,y)}{partial y}}.$$
E.g., with $$x^2+y^2-1=0$$
$$frac{dy}{dx}=-frac{2x}{2y}.$$
Compare to
$$y=sqrt{1-x^2}$$ then
$$frac{dy}{dx}=-frac{2x}{2sqrt{1-x^2}}$$ (and similarly with the negative branch).
answered Jan 30 at 17:15
Yves DaoustYves Daoust
132k676230
edited Jan 30 at 17:20
answered Jan 30 at 17:15
Yves DaoustYves Daoust
132k676230
answered Jan 30 at 17:15
Yves DaoustYves Daoust
132k676230
answered Jan 30 at 17:15
Yves DaoustYves Daoust
132k676230
132k676230
add a comment |
add a comment |
$begingroup$
$f$ is a function of two variables. It does not make sense to write $frac{df}{dy}$.
If $f(x,y)=0$, the derivative of the implicit function $x(y)$ is given by:
$${displaystyle {frac {dx}{dy}}=-{frac {partial f/partial y}{partial f/partial x}}=-{frac {f_{y}}{f_{x}}},}
$$
where $f_x$ and $f_y$ indicate the partial derivatives of $f$ with respect to $x$ and $y$.
The above formula comes from using the generalized chain rule to obtain the total derivative—with respect to $y$—of both sides of $f(x, y) = 0$:
$${displaystyle {frac {partial f}{partial x}}{frac {dx}{dy}}+{frac {partial f}{partial y}}{frac {dy}{dy}}=0,}
$$
hence
$$
{displaystyle {frac {partial f}{partial y}}
+{frac {partial f}{partial x}}{frac {dx}{dy}}=0,} $$
which, when solved for $dx/dy$, gives the expression above.
$endgroup$
add a comment |
$begingroup$
$f$ is a function of two variables. It does not make sense to write $frac{df}{dy}$.
If $f(x,y)=0$, the derivative of the implicit function $x(y)$ is given by:
$${displaystyle {frac {dx}{dy}}=-{frac {partial f/partial y}{partial f/partial x}}=-{frac {f_{y}}{f_{x}}},}
$$
where $f_x$ and $f_y$ indicate the partial derivatives of $f$ with respect to $x$ and $y$.
The above formula comes from using the generalized chain rule to obtain the total derivative—with respect to $y$—of both sides of $f(x, y) = 0$:
$${displaystyle {frac {partial f}{partial x}}{frac {dx}{dy}}+{frac {partial f}{partial y}}{frac {dy}{dy}}=0,}
$$
hence
$$
{displaystyle {frac {partial f}{partial y}}
+{frac {partial f}{partial x}}{frac {dx}{dy}}=0,} $$
which, when solved for $dx/dy$, gives the expression above.
$endgroup$
add a comment |
$begingroup$
$f$ is a function of two variables. It does not make sense to write $frac{df}{dy}$.
If $f(x,y)=0$, the derivative of the implicit function $x(y)$ is given by:
$${displaystyle {frac {dx}{dy}}=-{frac {partial f/partial y}{partial f/partial x}}=-{frac {f_{y}}{f_{x}}},}
$$
where $f_x$ and $f_y$ indicate the partial derivatives of $f$ with respect to $x$ and $y$.
The above formula comes from using the generalized chain rule to obtain the total derivative—with respect to $y$—of both sides of $f(x, y) = 0$:
$${displaystyle {frac {partial f}{partial x}}{frac {dx}{dy}}+{frac {partial f}{partial y}}{frac {dy}{dy}}=0,}
$$
hence
$$
{displaystyle {frac {partial f}{partial y}}
+{frac {partial f}{partial x}}{frac {dx}{dy}}=0,} $$
which, when solved for $dx/dy$, gives the expression above.
$endgroup$
$f$ is a function of two variables. It does not make sense to write $frac{df}{dy}$.
If $f(x,y)=0$, the derivative of the implicit function $x(y)$ is given by:
$${displaystyle {frac {dx}{dy}}=-{frac {partial f/partial y}{partial f/partial x}}=-{frac {f_{y}}{f_{x}}},}
$$
where $f_x$ and $f_y$ indicate the partial derivatives of $f$ with respect to $x$ and $y$.
The above formula comes from using the generalized chain rule to obtain the total derivative—with respect to $y$—of both sides of $f(x, y) = 0$:
$${displaystyle {frac {partial f}{partial x}}{frac {dx}{dy}}+{frac {partial f}{partial y}}{frac {dy}{dy}}=0,}
$$
hence
$$
{displaystyle {frac {partial f}{partial y}}
+{frac {partial f}{partial x}}{frac {dx}{dy}}=0,} $$
which, when solved for $dx/dy$, gives the expression above.
answered Jan 30 at 17:30
user587192
add a comment |
add a comment |
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$begingroup$
$f$ is a function of two variables. It does not make sense to write $frac{df}{dy}$.
$endgroup$
– user587192
Jan 30 at 17:21
$begingroup$
dx/df doesn't really make sense here, because x is not an implicit function of f. The sort of chain rule you wrote is really just for 1 variable. Yves' answer shows the correct, multivariable chain rule.
$endgroup$
– Nathaniel Mayer
Jan 30 at 17:26
$begingroup$
I think that your surprise comes from a confusion between $f(x,y)=0$ (a plane curve) and $z=varphi(x,y)$ (a surface).
$endgroup$
– Jean Marie
Jan 30 at 18:36