How is it that the following is reduced on the left hand side to d/dx(x^-2y)?Is the following legal? $dfracdd thetacos(ntheta)=dfracd cos nthetad cos thetadfracd cos thetad theta$.Strange things happening with derivativesLeft Hand Derivative DefinitionHow do I solve $fracddtxleft ( t right )=-5xleft ( t right )cosleft ( t right )$?existence of right/left hand derivative of convex function in the interior.Differentiability: What if Left-hand and Right-hand Limit are Equal in $x$ but differ from $f(x)$?About the Left Hand Derivative = Right Hand Derivative condition of differentiabilityWhat is the explanation of taking $dx$ to right hand side if $fracddx$ is an operator?How to use derivatives to prove that $f(x)=2cos^2left(fracpi 4-fracx2right)-sin left(xright)=1$?Proving a left-hand limit exists
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How is it that the following is reduced on the left hand side to d/dx(x^-2y)?
Is the following legal? $dfracdd thetacos(ntheta)=dfracd cos nthetad cos thetadfracd cos thetad theta$.Strange things happening with derivativesLeft Hand Derivative DefinitionHow do I solve $fracddtxleft ( t right )=-5xleft ( t right )cosleft ( t right )$?existence of right/left hand derivative of convex function in the interior.Differentiability: What if Left-hand and Right-hand Limit are Equal in $x$ but differ from $f(x)$?About the Left Hand Derivative = Right Hand Derivative condition of differentiabilityWhat is the explanation of taking $dx$ to right hand side if $fracddx$ is an operator?How to use derivatives to prove that $f(x)=2cos^2left(fracpi 4-fracx2right)-sin left(xright)=1$?Proving a left-hand limit exists
$begingroup$
How is it that the following is reduced to $fracddx(x^-2y)$ on the left hand side?
Is this an intermediary step? $fracddx(x^-2y - 2x^-3y)$
$$ x^-2fracdydx -2x^-3y = cos x$$
$$ fracddxleft(x^-2y right) = cos x$$
derivatives
edited 20 hours ago
John_dydx
3,7211924
asked 21 hours ago
GarySharpeGarySharpe
1033
New contributor
GarySharpe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
How is it that the following is reduced to $fracddx(x^-2y)$ on the left hand side?
Is this an intermediary step? $fracddx(x^-2y - 2x^-3y)$
$$ x^-2fracdydx -2x^-3y = cos x$$
$$ fracddxleft(x^-2y right) = cos x$$
derivatives
edited 20 hours ago
John_dydx
3,7211924
asked 21 hours ago
GarySharpeGarySharpe
1033
New contributor
GarySharpe 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 |
$begingroup$
How is it that the following is reduced to $fracddx(x^-2y)$ on the left hand side?
Is this an intermediary step? $fracddx(x^-2y - 2x^-3y)$
$$ x^-2fracdydx -2x^-3y = cos x$$
$$ fracddxleft(x^-2y right) = cos x$$
derivatives
edited 20 hours ago
John_dydx
3,7211924
asked 21 hours ago
GarySharpeGarySharpe
1033
New contributor
GarySharpe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
How is it that the following is reduced to $fracddx(x^-2y)$ on the left hand side?
Is this an intermediary step? $fracddx(x^-2y - 2x^-3y)$
$$ x^-2fracdydx -2x^-3y = cos x$$
$$ fracddxleft(x^-2y right) = cos x$$
derivatives
derivatives
edited 20 hours ago
John_dydx
3,7211924
asked 21 hours ago
GarySharpeGarySharpe
1033
New contributor
GarySharpe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 20 hours ago
John_dydx
3,7211924
asked 21 hours ago
GarySharpeGarySharpe
1033
New contributor
GarySharpe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 20 hours ago
John_dydx
3,7211924
edited 20 hours ago
John_dydx
3,7211924
edited 20 hours ago
John_dydx
3,7211924
3,7211924
asked 21 hours ago
GarySharpeGarySharpe
1033
New contributor
GarySharpe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 21 hours ago
GarySharpeGarySharpe
1033
asked 21 hours ago
GarySharpeGarySharpe
1033
1033
New contributor
GarySharpe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
GarySharpe 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 |
2 Answers
2
active
oldest
votes
$begingroup$
It's the product rule:
$$
fracddx(uv)=ufracdvdx+fracdudxv
$$
Applied to $u=x^-2, v=y$.
answered 21 hours ago
ArthurArthur
120k7120204
$endgroup$
add a comment |
$begingroup$
It is simply observation regarding product rule. We have d(xy)=xdy+ ydx.
So wherever you see a form xdy+ydx you can write it as d(xy). $int (xdy+ydx)$ reduces to xy using same thing. So in this example if you observe,you have product of two functions$ x^-2$ and$ y$ in the same form ,so it can be reduced to $d(x^-2y)$
answered 21 hours ago
TojrahTojrah
2536
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
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votes
$begingroup$
It's the product rule:
$$
fracddx(uv)=ufracdvdx+fracdudxv
$$
Applied to $u=x^-2, v=y$.
answered 21 hours ago
ArthurArthur
120k7120204
$endgroup$
add a comment |
$begingroup$
It's the product rule:
$$
fracddx(uv)=ufracdvdx+fracdudxv
$$
Applied to $u=x^-2, v=y$.
answered 21 hours ago
ArthurArthur
120k7120204
$endgroup$
add a comment |
$begingroup$
It's the product rule:
$$
fracddx(uv)=ufracdvdx+fracdudxv
$$
Applied to $u=x^-2, v=y$.
answered 21 hours ago
ArthurArthur
120k7120204
$endgroup$
It's the product rule:
$$
fracddx(uv)=ufracdvdx+fracdudxv
$$
Applied to $u=x^-2, v=y$.
answered 21 hours ago
ArthurArthur
120k7120204
answered 21 hours ago
ArthurArthur
120k7120204
answered 21 hours ago
ArthurArthur
120k7120204
answered 21 hours ago
ArthurArthur
120k7120204
120k7120204
add a comment |
add a comment |
$begingroup$
It is simply observation regarding product rule. We have d(xy)=xdy+ ydx.
So wherever you see a form xdy+ydx you can write it as d(xy). $int (xdy+ydx)$ reduces to xy using same thing. So in this example if you observe,you have product of two functions$ x^-2$ and$ y$ in the same form ,so it can be reduced to $d(x^-2y)$
answered 21 hours ago
TojrahTojrah
2536
$endgroup$
add a comment |
$begingroup$
It is simply observation regarding product rule. We have d(xy)=xdy+ ydx.
So wherever you see a form xdy+ydx you can write it as d(xy). $int (xdy+ydx)$ reduces to xy using same thing. So in this example if you observe,you have product of two functions$ x^-2$ and$ y$ in the same form ,so it can be reduced to $d(x^-2y)$
answered 21 hours ago
TojrahTojrah
2536
$endgroup$
add a comment |
$begingroup$
It is simply observation regarding product rule. We have d(xy)=xdy+ ydx.
So wherever you see a form xdy+ydx you can write it as d(xy). $int (xdy+ydx)$ reduces to xy using same thing. So in this example if you observe,you have product of two functions$ x^-2$ and$ y$ in the same form ,so it can be reduced to $d(x^-2y)$
answered 21 hours ago
TojrahTojrah
2536
$endgroup$
It is simply observation regarding product rule. We have d(xy)=xdy+ ydx.
So wherever you see a form xdy+ydx you can write it as d(xy). $int (xdy+ydx)$ reduces to xy using same thing. So in this example if you observe,you have product of two functions$ x^-2$ and$ y$ in the same form ,so it can be reduced to $d(x^-2y)$
answered 21 hours ago
TojrahTojrah
2536
answered 21 hours ago
TojrahTojrah
2536
answered 21 hours ago
TojrahTojrah
2536
answered 21 hours ago
TojrahTojrah
2536
2536
add a comment |
add a comment |
GarySharpe is a new contributor. Be nice, and check out our Code of Conduct.
GarySharpe is a new contributor. Be nice, and check out our Code of Conduct.
GarySharpe is a new contributor. Be nice, and check out our Code of Conduct.
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