Mixing
Implicit differentiation questions
$begingroup$
I want to solve $dy/dx$ for the following:
$x^2 + y^2 = R^2$ where $R$ is a constant.
I know to use implicit differentiation, though I have a question. When I derive $R^2$, do I obtain $2R$ or 0?
Additionally, deriving $y^2$ with respect to x yields $2y (dy/dx)$? This is different from a partial derivative?
Thanks!
derivatives
asked Jan 9 at 19:45
appleapple
445
$endgroup$
add a comment |
$begingroup$
I want to solve $dy/dx$ for the following:
$x^2 + y^2 = R^2$ where $R$ is a constant.
I know to use implicit differentiation, though I have a question. When I derive $R^2$, do I obtain $2R$ or 0?
Additionally, deriving $y^2$ with respect to x yields $2y (dy/dx)$? This is different from a partial derivative?
Thanks!
derivatives
asked Jan 9 at 19:45
appleapple
445
$endgroup$
$begingroup$
If you are looking for $frac{dy}{dx},$ then you need to derive the given equation in terms of $x.$ $R$ is a constant. Notice that $frac{d}{dx} c = 0.$
$endgroup$
– K. Jiang
Jan 9 at 19:47
$begingroup$
You are differentiating with respect to $x$, and since $R$ is independent of $x$, its derivative is zero. For the derivative of $y^2=f(x)^2$, use the chain rule; its derivative is $dfrac{d(y^2)}{df(x)}cdotdfrac{df(x)}x$
$endgroup$
– TheSimpliFire
Jan 9 at 19:49
add a comment |
$begingroup$
I want to solve $dy/dx$ for the following:
$x^2 + y^2 = R^2$ where $R$ is a constant.
I know to use implicit differentiation, though I have a question. When I derive $R^2$, do I obtain $2R$ or 0?
Additionally, deriving $y^2$ with respect to x yields $2y (dy/dx)$? This is different from a partial derivative?
Thanks!
derivatives
asked Jan 9 at 19:45
appleapple
445
$endgroup$
I want to solve $dy/dx$ for the following:
$x^2 + y^2 = R^2$ where $R$ is a constant.
I know to use implicit differentiation, though I have a question. When I derive $R^2$, do I obtain $2R$ or 0?
Additionally, deriving $y^2$ with respect to x yields $2y (dy/dx)$? This is different from a partial derivative?
Thanks!
derivatives
derivatives
asked Jan 9 at 19:45
appleapple
445
asked Jan 9 at 19:45
appleapple
445
asked Jan 9 at 19:45
appleapple
445
asked Jan 9 at 19:45
appleapple
445
asked Jan 9 at 19:45
appleapple
445
445
$begingroup$
If you are looking for $frac{dy}{dx},$ then you need to derive the given equation in terms of $x.$ $R$ is a constant. Notice that $frac{d}{dx} c = 0.$
$endgroup$
– K. Jiang
Jan 9 at 19:47
$begingroup$
You are differentiating with respect to $x$, and since $R$ is independent of $x$, its derivative is zero. For the derivative of $y^2=f(x)^2$, use the chain rule; its derivative is $dfrac{d(y^2)}{df(x)}cdotdfrac{df(x)}x$
$endgroup$
– TheSimpliFire
Jan 9 at 19:49
add a comment |
$begingroup$
If you are looking for $frac{dy}{dx},$ then you need to derive the given equation in terms of $x.$ $R$ is a constant. Notice that $frac{d}{dx} c = 0.$
$endgroup$
– K. Jiang
Jan 9 at 19:47
$begingroup$
You are differentiating with respect to $x$, and since $R$ is independent of $x$, its derivative is zero. For the derivative of $y^2=f(x)^2$, use the chain rule; its derivative is $dfrac{d(y^2)}{df(x)}cdotdfrac{df(x)}x$
$endgroup$
– TheSimpliFire
Jan 9 at 19:49
$begingroup$
If you are looking for $frac{dy}{dx},$ then you need to derive the given equation in terms of $x.$ $R$ is a constant. Notice that $frac{d}{dx} c = 0.$
$endgroup$
– K. Jiang
Jan 9 at 19:47
$begingroup$
If you are looking for $frac{dy}{dx},$ then you need to derive the given equation in terms of $x.$ $R$ is a constant. Notice that $frac{d}{dx} c = 0.$
$endgroup$
– K. Jiang
Jan 9 at 19:47
$begingroup$
You are differentiating with respect to $x$, and since $R$ is independent of $x$, its derivative is zero. For the derivative of $y^2=f(x)^2$, use the chain rule; its derivative is $dfrac{d(y^2)}{df(x)}cdotdfrac{df(x)}x$
$endgroup$
– TheSimpliFire
Jan 9 at 19:49
$begingroup$
You are differentiating with respect to $x$, and since $R$ is independent of $x$, its derivative is zero. For the derivative of $y^2=f(x)^2$, use the chain rule; its derivative is $dfrac{d(y^2)}{df(x)}cdotdfrac{df(x)}x$
$endgroup$
– TheSimpliFire
Jan 9 at 19:49
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
By the chaine rule you will get $$2x+2ycdot y'=0$$
answered Jan 9 at 19:47
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.9k42865
$endgroup$
add a comment |
$begingroup$
Suppose $R$ is a function of $x$ and $y$; then
$x^2 + y^2 = R^2(x, y); tag 1$
if we define
$F(x, y) = x^2 + y^2 - R^2(x, y), tag 2$
we may also write (1) as
$F(x, y) = x^2 + y^2 - R^2(x, y) = 0; tag 3$
by the implicit funtion theorem, this equation in fact may be seen as defining $y(x)$, a function of $x$, provided that
$dfrac{partial F(x, y)}{partial y} ne 0; tag 4$
we have
$dfrac{partial F(x, y)}{partial y} = 2y - 2R(x, y) dfrac{partial R(x, y)}{partial y} ne 0 tag 5$
provided
$y ne R(x, y) dfrac{partial R(x, y)}{partial y};tag 6$
under such circumstances, we may affirm $y(x)$ is uniquely determined as a differentiable function of $x$ in some neighborhood of any point $(x, y)$; then we have
$F(x, y) = x^2 + y^2(x) - R^2(x, y(x)) = 0; tag 7$
we may take the total derivative with respect to $x$ to obtain
$dfrac{dF(x, y)}{dx} = 2x + 2ydfrac{dy(x)}{dx} - 2R(x, y(x)) left ( dfrac{partial R(x, y(x))}{partial x} + dfrac{partial R(x, y(x))}{partial y} dfrac{dy(x)}{dx} right ) = 0; tag 8$
a little algebra allows us to isolate the terms containing $dy(x)/dx$:
$x + ydfrac{dy(x)}{dx} - R(x, y(x)) left ( dfrac{partial R(x, y(x))}{partial x} + dfrac{partial R(x, y(x))}{partial y} dfrac{dy(x)}{dx} right ) = 0; tag 9$
$ydfrac{dy(x)}{dx} - R(x, y(x)) dfrac{partial R(x, y(x))}{partial y} dfrac{dy(x)}{dx} = R(x, y(x))dfrac{partial R(x, y(x))}{partial x} - x; tag{10}$
$left ( y - R(x, y(x)) dfrac{partial R(x, y(x)}{partial y} right )dfrac{dy(x)}{dx} = R(x, y(x))dfrac{partial R(x, y(x))}{partial x} - x; tag{11}$
for the sake of compactess and brevity, we introduce the subscript notation
$R_x = dfrac{partial R}{partial x}, ; text{etc.}, tag{12}$
and write (11) in the form
$y'(x) = dfrac{RR_x - x}{y - RR_y} = -dfrac{x - RR_x}{y - RR_y}, tag{13}$
which gives a general expression for $y'(x)$; in the event that $R(x, y)$ is constant, we obtain
$y'(x) = -dfrac{x}{y}, tag{14}$
which the reader may recognize as the slope of the circle
$x^2 + y^2 = R^2 tag{15}$
at any point $(x, y)$ where $y ne 0$.
answered Jan 9 at 23:34
Robert LewisRobert Lewis
45.7k23065
$endgroup$
add a comment |
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$begingroup$
By the chaine rule you will get $$2x+2ycdot y'=0$$
answered Jan 9 at 19:47
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.9k42865
$endgroup$
add a comment |
$begingroup$
By the chaine rule you will get $$2x+2ycdot y'=0$$
answered Jan 9 at 19:47
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.9k42865
$endgroup$
add a comment |
$begingroup$
By the chaine rule you will get $$2x+2ycdot y'=0$$
answered Jan 9 at 19:47
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.9k42865
$endgroup$
By the chaine rule you will get $$2x+2ycdot y'=0$$
answered Jan 9 at 19:47
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.9k42865
answered Jan 9 at 19:47
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.9k42865
answered Jan 9 at 19:47
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.9k42865
answered Jan 9 at 19:47
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.9k42865
74.9k42865
add a comment |
add a comment |
$begingroup$
Suppose $R$ is a function of $x$ and $y$; then
$x^2 + y^2 = R^2(x, y); tag 1$
if we define
$F(x, y) = x^2 + y^2 - R^2(x, y), tag 2$
we may also write (1) as
$F(x, y) = x^2 + y^2 - R^2(x, y) = 0; tag 3$
by the implicit funtion theorem, this equation in fact may be seen as defining $y(x)$, a function of $x$, provided that
$dfrac{partial F(x, y)}{partial y} ne 0; tag 4$
we have
$dfrac{partial F(x, y)}{partial y} = 2y - 2R(x, y) dfrac{partial R(x, y)}{partial y} ne 0 tag 5$
provided
$y ne R(x, y) dfrac{partial R(x, y)}{partial y};tag 6$
under such circumstances, we may affirm $y(x)$ is uniquely determined as a differentiable function of $x$ in some neighborhood of any point $(x, y)$; then we have
$F(x, y) = x^2 + y^2(x) - R^2(x, y(x)) = 0; tag 7$
we may take the total derivative with respect to $x$ to obtain
$dfrac{dF(x, y)}{dx} = 2x + 2ydfrac{dy(x)}{dx} - 2R(x, y(x)) left ( dfrac{partial R(x, y(x))}{partial x} + dfrac{partial R(x, y(x))}{partial y} dfrac{dy(x)}{dx} right ) = 0; tag 8$
a little algebra allows us to isolate the terms containing $dy(x)/dx$:
$x + ydfrac{dy(x)}{dx} - R(x, y(x)) left ( dfrac{partial R(x, y(x))}{partial x} + dfrac{partial R(x, y(x))}{partial y} dfrac{dy(x)}{dx} right ) = 0; tag 9$
$ydfrac{dy(x)}{dx} - R(x, y(x)) dfrac{partial R(x, y(x))}{partial y} dfrac{dy(x)}{dx} = R(x, y(x))dfrac{partial R(x, y(x))}{partial x} - x; tag{10}$
$left ( y - R(x, y(x)) dfrac{partial R(x, y(x)}{partial y} right )dfrac{dy(x)}{dx} = R(x, y(x))dfrac{partial R(x, y(x))}{partial x} - x; tag{11}$
for the sake of compactess and brevity, we introduce the subscript notation
$R_x = dfrac{partial R}{partial x}, ; text{etc.}, tag{12}$
and write (11) in the form
$y'(x) = dfrac{RR_x - x}{y - RR_y} = -dfrac{x - RR_x}{y - RR_y}, tag{13}$
which gives a general expression for $y'(x)$; in the event that $R(x, y)$ is constant, we obtain
$y'(x) = -dfrac{x}{y}, tag{14}$
which the reader may recognize as the slope of the circle
$x^2 + y^2 = R^2 tag{15}$
at any point $(x, y)$ where $y ne 0$.
answered Jan 9 at 23:34
Robert LewisRobert Lewis
45.7k23065
$endgroup$
add a comment |
$begingroup$
Suppose $R$ is a function of $x$ and $y$; then
$x^2 + y^2 = R^2(x, y); tag 1$
if we define
$F(x, y) = x^2 + y^2 - R^2(x, y), tag 2$
we may also write (1) as
$F(x, y) = x^2 + y^2 - R^2(x, y) = 0; tag 3$
by the implicit funtion theorem, this equation in fact may be seen as defining $y(x)$, a function of $x$, provided that
$dfrac{partial F(x, y)}{partial y} ne 0; tag 4$
we have
$dfrac{partial F(x, y)}{partial y} = 2y - 2R(x, y) dfrac{partial R(x, y)}{partial y} ne 0 tag 5$
provided
$y ne R(x, y) dfrac{partial R(x, y)}{partial y};tag 6$
under such circumstances, we may affirm $y(x)$ is uniquely determined as a differentiable function of $x$ in some neighborhood of any point $(x, y)$; then we have
$F(x, y) = x^2 + y^2(x) - R^2(x, y(x)) = 0; tag 7$
we may take the total derivative with respect to $x$ to obtain
$dfrac{dF(x, y)}{dx} = 2x + 2ydfrac{dy(x)}{dx} - 2R(x, y(x)) left ( dfrac{partial R(x, y(x))}{partial x} + dfrac{partial R(x, y(x))}{partial y} dfrac{dy(x)}{dx} right ) = 0; tag 8$
a little algebra allows us to isolate the terms containing $dy(x)/dx$:
$x + ydfrac{dy(x)}{dx} - R(x, y(x)) left ( dfrac{partial R(x, y(x))}{partial x} + dfrac{partial R(x, y(x))}{partial y} dfrac{dy(x)}{dx} right ) = 0; tag 9$
$ydfrac{dy(x)}{dx} - R(x, y(x)) dfrac{partial R(x, y(x))}{partial y} dfrac{dy(x)}{dx} = R(x, y(x))dfrac{partial R(x, y(x))}{partial x} - x; tag{10}$
$left ( y - R(x, y(x)) dfrac{partial R(x, y(x)}{partial y} right )dfrac{dy(x)}{dx} = R(x, y(x))dfrac{partial R(x, y(x))}{partial x} - x; tag{11}$
for the sake of compactess and brevity, we introduce the subscript notation
$R_x = dfrac{partial R}{partial x}, ; text{etc.}, tag{12}$
and write (11) in the form
$y'(x) = dfrac{RR_x - x}{y - RR_y} = -dfrac{x - RR_x}{y - RR_y}, tag{13}$
which gives a general expression for $y'(x)$; in the event that $R(x, y)$ is constant, we obtain
$y'(x) = -dfrac{x}{y}, tag{14}$
which the reader may recognize as the slope of the circle
$x^2 + y^2 = R^2 tag{15}$
at any point $(x, y)$ where $y ne 0$.
answered Jan 9 at 23:34
Robert LewisRobert Lewis
45.7k23065
$endgroup$
add a comment |
$begingroup$
Suppose $R$ is a function of $x$ and $y$; then
$x^2 + y^2 = R^2(x, y); tag 1$
if we define
$F(x, y) = x^2 + y^2 - R^2(x, y), tag 2$
we may also write (1) as
$F(x, y) = x^2 + y^2 - R^2(x, y) = 0; tag 3$
by the implicit funtion theorem, this equation in fact may be seen as defining $y(x)$, a function of $x$, provided that
$dfrac{partial F(x, y)}{partial y} ne 0; tag 4$
we have
$dfrac{partial F(x, y)}{partial y} = 2y - 2R(x, y) dfrac{partial R(x, y)}{partial y} ne 0 tag 5$
provided
$y ne R(x, y) dfrac{partial R(x, y)}{partial y};tag 6$
under such circumstances, we may affirm $y(x)$ is uniquely determined as a differentiable function of $x$ in some neighborhood of any point $(x, y)$; then we have
$F(x, y) = x^2 + y^2(x) - R^2(x, y(x)) = 0; tag 7$
we may take the total derivative with respect to $x$ to obtain
$dfrac{dF(x, y)}{dx} = 2x + 2ydfrac{dy(x)}{dx} - 2R(x, y(x)) left ( dfrac{partial R(x, y(x))}{partial x} + dfrac{partial R(x, y(x))}{partial y} dfrac{dy(x)}{dx} right ) = 0; tag 8$
a little algebra allows us to isolate the terms containing $dy(x)/dx$:
$x + ydfrac{dy(x)}{dx} - R(x, y(x)) left ( dfrac{partial R(x, y(x))}{partial x} + dfrac{partial R(x, y(x))}{partial y} dfrac{dy(x)}{dx} right ) = 0; tag 9$
$ydfrac{dy(x)}{dx} - R(x, y(x)) dfrac{partial R(x, y(x))}{partial y} dfrac{dy(x)}{dx} = R(x, y(x))dfrac{partial R(x, y(x))}{partial x} - x; tag{10}$
$left ( y - R(x, y(x)) dfrac{partial R(x, y(x)}{partial y} right )dfrac{dy(x)}{dx} = R(x, y(x))dfrac{partial R(x, y(x))}{partial x} - x; tag{11}$
for the sake of compactess and brevity, we introduce the subscript notation
$R_x = dfrac{partial R}{partial x}, ; text{etc.}, tag{12}$
and write (11) in the form
$y'(x) = dfrac{RR_x - x}{y - RR_y} = -dfrac{x - RR_x}{y - RR_y}, tag{13}$
which gives a general expression for $y'(x)$; in the event that $R(x, y)$ is constant, we obtain
$y'(x) = -dfrac{x}{y}, tag{14}$
which the reader may recognize as the slope of the circle
$x^2 + y^2 = R^2 tag{15}$
at any point $(x, y)$ where $y ne 0$.
answered Jan 9 at 23:34
Robert LewisRobert Lewis
45.7k23065
$endgroup$
Suppose $R$ is a function of $x$ and $y$; then
$x^2 + y^2 = R^2(x, y); tag 1$
if we define
$F(x, y) = x^2 + y^2 - R^2(x, y), tag 2$
we may also write (1) as
$F(x, y) = x^2 + y^2 - R^2(x, y) = 0; tag 3$
by the implicit funtion theorem, this equation in fact may be seen as defining $y(x)$, a function of $x$, provided that
$dfrac{partial F(x, y)}{partial y} ne 0; tag 4$
we have
$dfrac{partial F(x, y)}{partial y} = 2y - 2R(x, y) dfrac{partial R(x, y)}{partial y} ne 0 tag 5$
provided
$y ne R(x, y) dfrac{partial R(x, y)}{partial y};tag 6$
under such circumstances, we may affirm $y(x)$ is uniquely determined as a differentiable function of $x$ in some neighborhood of any point $(x, y)$; then we have
$F(x, y) = x^2 + y^2(x) - R^2(x, y(x)) = 0; tag 7$
we may take the total derivative with respect to $x$ to obtain
$dfrac{dF(x, y)}{dx} = 2x + 2ydfrac{dy(x)}{dx} - 2R(x, y(x)) left ( dfrac{partial R(x, y(x))}{partial x} + dfrac{partial R(x, y(x))}{partial y} dfrac{dy(x)}{dx} right ) = 0; tag 8$
a little algebra allows us to isolate the terms containing $dy(x)/dx$:
$x + ydfrac{dy(x)}{dx} - R(x, y(x)) left ( dfrac{partial R(x, y(x))}{partial x} + dfrac{partial R(x, y(x))}{partial y} dfrac{dy(x)}{dx} right ) = 0; tag 9$
$ydfrac{dy(x)}{dx} - R(x, y(x)) dfrac{partial R(x, y(x))}{partial y} dfrac{dy(x)}{dx} = R(x, y(x))dfrac{partial R(x, y(x))}{partial x} - x; tag{10}$
$left ( y - R(x, y(x)) dfrac{partial R(x, y(x)}{partial y} right )dfrac{dy(x)}{dx} = R(x, y(x))dfrac{partial R(x, y(x))}{partial x} - x; tag{11}$
for the sake of compactess and brevity, we introduce the subscript notation
$R_x = dfrac{partial R}{partial x}, ; text{etc.}, tag{12}$
and write (11) in the form
$y'(x) = dfrac{RR_x - x}{y - RR_y} = -dfrac{x - RR_x}{y - RR_y}, tag{13}$
which gives a general expression for $y'(x)$; in the event that $R(x, y)$ is constant, we obtain
$y'(x) = -dfrac{x}{y}, tag{14}$
which the reader may recognize as the slope of the circle
$x^2 + y^2 = R^2 tag{15}$
at any point $(x, y)$ where $y ne 0$.
answered Jan 9 at 23:34
Robert LewisRobert Lewis
45.7k23065
answered Jan 9 at 23:34
Robert LewisRobert Lewis
45.7k23065
answered Jan 9 at 23:34
Robert LewisRobert Lewis
45.7k23065
answered Jan 9 at 23:34
Robert LewisRobert Lewis
45.7k23065
45.7k23065
add a comment |
add a comment |
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$begingroup$
If you are looking for $frac{dy}{dx},$ then you need to derive the given equation in terms of $x.$ $R$ is a constant. Notice that $frac{d}{dx} c = 0.$
$endgroup$
– K. Jiang
Jan 9 at 19:47
$begingroup$
You are differentiating with respect to $x$, and since $R$ is independent of $x$, its derivative is zero. For the derivative of $y^2=f(x)^2$, use the chain rule; its derivative is $dfrac{d(y^2)}{df(x)}cdotdfrac{df(x)}x$
$endgroup$
– TheSimpliFire
Jan 9 at 19:49