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Derivative of $G(r,phi)=f(rcos(phi),rsin(phi))$

Calculating partial derivative, polar and cartesian coordinatesDetermine $f'(12)$ when $f(x)=xcos(12/x)$. Chain RuleDerivative of $ h(t)= sin (cos^-1t$)?Chain rule version for partiel derivative?derivative of $f(rcosphi,rsinphi)=r^acos(aphi)$Partial derivative of $f(x,y) = (x/y) cos (1/y)$Determining $fracpartialpartial beta fracsin(alpha)cos(beta) + cos(alpha)sin(beta)sin(alpha)$Partial derivative of a function w.r.t an argument that occurs multiple timesLet $f: mathbbR^3to mathbbR^3$ be given by $f(rho, phi, theta) = (rhocostheta sin phi, rho sin theta sin phi, rho cos phi).$Partial derivation of $g:mathbbR^2 to mathbbR$ with $g(x,y) := (sin(xy))^2$

2

1

$begingroup$

Let $(x,y)=(rcos(phi),rsin(phi))$, $r>0$ and $f:mathbbR^2tomathbbR$ a $C^2-$function and $G(r,phi)=f(rcos(phi),rsin(phi))$. I want to know how to calculate the derivatives $fracpartial G(r,phi)partial r$, $fracpartial G(r,phi)partial phi$ and $fracpartial f(x,y)partial x$.

For example, I tried to calculate this $fracpartial G(r,phi)partial r$ derivation with the chain rule: $fracpartial G(r,phi)partial r=f_r'(rcos(phi),rsin(phi))cdot (cos(phi),sin(phi))$, but I think it isn't correct. Could you help me? Regards.

real-analysis derivatives

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edited Mar 10 at 18:57

Rócherz

2,9262821

asked Jun 10 '15 at 18:24

timbatimba

312

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add a comment | 

2

1

$begingroup$

Let $(x,y)=(rcos(phi),rsin(phi))$, $r>0$ and $f:mathbbR^2tomathbbR$ a $C^2-$function and $G(r,phi)=f(rcos(phi),rsin(phi))$. I want to know how to calculate the derivatives $fracpartial G(r,phi)partial r$, $fracpartial G(r,phi)partial phi$ and $fracpartial f(x,y)partial x$.

For example, I tried to calculate this $fracpartial G(r,phi)partial r$ derivation with the chain rule: $fracpartial G(r,phi)partial r=f_r'(rcos(phi),rsin(phi))cdot (cos(phi),sin(phi))$, but I think it isn't correct. Could you help me? Regards.

real-analysis derivatives

share|cite|improve this question

edited Mar 10 at 18:57

Rócherz

2,9262821

asked Jun 10 '15 at 18:24

timbatimba

312

$endgroup$

add a comment | 

$begingroup$

Let $(x,y)=(rcos(phi),rsin(phi))$, $r>0$ and $f:mathbbR^2tomathbbR$ a $C^2-$function and $G(r,phi)=f(rcos(phi),rsin(phi))$. I want to know how to calculate the derivatives $fracpartial G(r,phi)partial r$, $fracpartial G(r,phi)partial phi$ and $fracpartial f(x,y)partial x$.

For example, I tried to calculate this $fracpartial G(r,phi)partial r$ derivation with the chain rule: $fracpartial G(r,phi)partial r=f_r'(rcos(phi),rsin(phi))cdot (cos(phi),sin(phi))$, but I think it isn't correct. Could you help me? Regards.

real-analysis derivatives

share|cite|improve this question

edited Mar 10 at 18:57

Rócherz

2,9262821

asked Jun 10 '15 at 18:24

timbatimba

312

$endgroup$

share|cite|improve this question

edited Mar 10 at 18:57

Rócherz

2,9262821

asked Jun 10 '15 at 18:24

timbatimba

312

share|cite|improve this question

edited Mar 10 at 18:57

Rócherz

2,9262821

asked Jun 10 '15 at 18:24

timbatimba

312

edited Mar 10 at 18:57

Rócherz

2,9262821

edited Mar 10 at 18:57

Rócherz

2,9262821

asked Jun 10 '15 at 18:24

timbatimba

312

asked Jun 10 '15 at 18:24

timbatimba

312

2 Answers
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$begingroup$

By chain rule
$$fracpartial G(r,phi)partial r=f_xcdot cosphi+f_ycdotsinphi$$

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answered Jun 10 '15 at 18:29

KittyLKittyL

13.8k31534

$endgroup$

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0

$begingroup$

We have

$$G(r,phi)=f(rcos phi, rsin phi)$$

Using the chain rule, we have

$$fracpartial Gpartial r=f_1(rcos phi, rsin phi)cos phi+f_2(rcos phi, rsin phi)sin phi$$

and

$$fracpartial Gpartial phi=-f_1(rcos phi, rsin phi)rsin phi+f_2(rcos phi, rsin phi)rcos phi$$

where $f_1$ and $f_2$ designate partial derivatives of $f$ with respect to the first and second arguments, respectively.

share|cite|improve this answer

answered Jun 10 '15 at 18:29

Mark ViolaMark Viola

133k1278176

$endgroup$

add a comment | 

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0

$begingroup$

By chain rule
$$fracpartial G(r,phi)partial r=f_xcdot cosphi+f_ycdotsinphi$$

share|cite|improve this answer

answered Jun 10 '15 at 18:29

KittyLKittyL

13.8k31534

$endgroup$

add a comment | 

0

$begingroup$

By chain rule
$$fracpartial G(r,phi)partial r=f_xcdot cosphi+f_ycdotsinphi$$

share|cite|improve this answer

answered Jun 10 '15 at 18:29

KittyLKittyL

13.8k31534

$endgroup$

add a comment | 

$begingroup$

By chain rule
$$fracpartial G(r,phi)partial r=f_xcdot cosphi+f_ycdotsinphi$$

share|cite|improve this answer

answered Jun 10 '15 at 18:29

KittyLKittyL

13.8k31534

$endgroup$

share|cite|improve this answer

answered Jun 10 '15 at 18:29

KittyLKittyL

13.8k31534

answered Jun 10 '15 at 18:29

KittyLKittyL

13.8k31534

answered Jun 10 '15 at 18:29

KittyLKittyL

13.8k31534

0

$begingroup$

We have

$$G(r,phi)=f(rcos phi, rsin phi)$$

Using the chain rule, we have

$$fracpartial Gpartial r=f_1(rcos phi, rsin phi)cos phi+f_2(rcos phi, rsin phi)sin phi$$

and

$$fracpartial Gpartial phi=-f_1(rcos phi, rsin phi)rsin phi+f_2(rcos phi, rsin phi)rcos phi$$

where $f_1$ and $f_2$ designate partial derivatives of $f$ with respect to the first and second arguments, respectively.

share|cite|improve this answer

answered Jun 10 '15 at 18:29

Mark ViolaMark Viola

133k1278176

$endgroup$

add a comment | 

0

$begingroup$

We have

$$G(r,phi)=f(rcos phi, rsin phi)$$

Using the chain rule, we have

$$fracpartial Gpartial r=f_1(rcos phi, rsin phi)cos phi+f_2(rcos phi, rsin phi)sin phi$$

and

$$fracpartial Gpartial phi=-f_1(rcos phi, rsin phi)rsin phi+f_2(rcos phi, rsin phi)rcos phi$$

where $f_1$ and $f_2$ designate partial derivatives of $f$ with respect to the first and second arguments, respectively.

share|cite|improve this answer

answered Jun 10 '15 at 18:29

Mark ViolaMark Viola

133k1278176

$endgroup$

add a comment | 

$begingroup$

We have

$$G(r,phi)=f(rcos phi, rsin phi)$$

Using the chain rule, we have

$$fracpartial Gpartial r=f_1(rcos phi, rsin phi)cos phi+f_2(rcos phi, rsin phi)sin phi$$

and

$$fracpartial Gpartial phi=-f_1(rcos phi, rsin phi)rsin phi+f_2(rcos phi, rsin phi)rcos phi$$

where $f_1$ and $f_2$ designate partial derivatives of $f$ with respect to the first and second arguments, respectively.

share|cite|improve this answer

answered Jun 10 '15 at 18:29

Mark ViolaMark Viola

133k1278176

$endgroup$

share|cite|improve this answer

answered Jun 10 '15 at 18:29

Mark ViolaMark Viola

133k1278176

answered Jun 10 '15 at 18:29

Mark ViolaMark Viola

133k1278176

answered Jun 10 '15 at 18:29

Mark ViolaMark Viola

133k1278176