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Calculating the second order partial derivatives of $z(x,y.)$. [closed]

1

$begingroup$

If we know that $z_x^' = fracyzz^2 - xy ,$, how can I calculate the second order partial derivative with respect to x, knowing that the final answer should be (as given at the back of the book) $z_xx^'' = frac2xy^3z(z^2 - xy)^3 ,$. could anyone explain to me how to do it? I do not know from where the power 3 in the denominator came.

real-analysis multivariable-calculus implicit-differentiation implicit-function-theorem implicit-function

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edited Mar 26 at 10:14

hopefully

asked Mar 26 at 10:06

hopefullyhopefully

190215

$endgroup$

closed as off-topic by user21820, Shailesh, José Carlos Santos, vonbrand, Xander Henderson Apr 1 at 22:15

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Shailesh, José Carlos Santos, Xander Henderson

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is a bit strange; you want to find "the second order partial derivative with respect to $y$" but you refer to an answer for $z_xx^''$, which looks like a second order derivative with respect to $x$...? Do you have $z(x,y)$ as well? Please include all relevant information.
  $endgroup$
  – StackTD
  Mar 26 at 10:12
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am sorry I will fix this mistake @StackTD
  $endgroup$
  – hopefully
  Mar 26 at 10:14
  

  
  
  
  

add a comment | 

1

$begingroup$

If we know that $z_x^' = fracyzz^2 - xy ,$, how can I calculate the second order partial derivative with respect to x, knowing that the final answer should be (as given at the back of the book) $z_xx^'' = frac2xy^3z(z^2 - xy)^3 ,$. could anyone explain to me how to do it? I do not know from where the power 3 in the denominator came.

real-analysis multivariable-calculus implicit-differentiation implicit-function-theorem implicit-function

share|cite|improve this question

edited Mar 26 at 10:14

hopefully

asked Mar 26 at 10:06

hopefullyhopefully

190215

$endgroup$

closed as off-topic by user21820, Shailesh, José Carlos Santos, vonbrand, Xander Henderson Apr 1 at 22:15

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Shailesh, José Carlos Santos, Xander Henderson

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is a bit strange; you want to find "the second order partial derivative with respect to $y$" but you refer to an answer for $z_xx^''$, which looks like a second order derivative with respect to $x$...? Do you have $z(x,y)$ as well? Please include all relevant information.
  $endgroup$
  – StackTD
  Mar 26 at 10:12
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am sorry I will fix this mistake @StackTD
  $endgroup$
  – hopefully
  Mar 26 at 10:14
  

  
  
  
  

add a comment | 

$begingroup$

If we know that $z_x^' = fracyzz^2 - xy ,$, how can I calculate the second order partial derivative with respect to x, knowing that the final answer should be (as given at the back of the book) $z_xx^'' = frac2xy^3z(z^2 - xy)^3 ,$. could anyone explain to me how to do it? I do not know from where the power 3 in the denominator came.

real-analysis multivariable-calculus implicit-differentiation implicit-function-theorem implicit-function

share|cite|improve this question

edited Mar 26 at 10:14

hopefully

asked Mar 26 at 10:06

hopefullyhopefully

190215

$endgroup$

share|cite|improve this question

edited Mar 26 at 10:14

hopefully

asked Mar 26 at 10:06

hopefullyhopefully

190215

share|cite|improve this question

edited Mar 26 at 10:14

hopefully

asked Mar 26 at 10:06

hopefullyhopefully

190215

asked Mar 26 at 10:06

hopefullyhopefully

190215

asked Mar 26 at 10:06

hopefullyhopefully

190215

1 Answer
1

active

oldest

votes

1

$begingroup$

If we know that $z_x^' = fracyzz^2 - xy $, how can I calculate the second order partial derivative with respect to x

Take the partial derivative with respect to $x$: use the quotient rule and don't forget that $z$ is a function of $x$ (and $y$). This will then contain the first partial derivative $z'_x$ again, but you can substitute that with:
$$z_x^' = fracyzz^2 - xy tag$*$$$
and then simplify.

---

Here's a start:
$$beginalignz_xx^''=fracpartialpartial xz_x^'
& =fracpartialpartial xleft(fracyzz^2 - xy right) \[6pt]
& =fracleft(z^2 - xyright)colorredfracpartialpartial xleft(yzright)-left(yzright)colorbluefracpartialpartial xleft(z^2 - xyright)left(z^2 - xyright)^2 \
& = ldots
endalign$$
Now $colorredfracpartialpartial xleft(yzright)=yz_x^'$ and use $(*)$ for $z_x^'$ and don't forget the chain rule for $colorbluefracpartialpartial xleft(z^2 - xyright)$ and you'll be able to use $(*)$ once again to eliminate $z_x^'$.

share|cite|improve this answer

edited Mar 26 at 10:26

answered Mar 26 at 10:19

StackTDStackTD

24.3k2254

$endgroup$

add a comment | 

1

$begingroup$

If we know that $z_x^' = fracyzz^2 - xy $, how can I calculate the second order partial derivative with respect to x

Take the partial derivative with respect to $x$: use the quotient rule and don't forget that $z$ is a function of $x$ (and $y$). This will then contain the first partial derivative $z'_x$ again, but you can substitute that with:
$$z_x^' = fracyzz^2 - xy tag$*$$$
and then simplify.

---

Here's a start:
$$beginalignz_xx^''=fracpartialpartial xz_x^'
& =fracpartialpartial xleft(fracyzz^2 - xy right) \[6pt]
& =fracleft(z^2 - xyright)colorredfracpartialpartial xleft(yzright)-left(yzright)colorbluefracpartialpartial xleft(z^2 - xyright)left(z^2 - xyright)^2 \
& = ldots
endalign$$
Now $colorredfracpartialpartial xleft(yzright)=yz_x^'$ and use $(*)$ for $z_x^'$ and don't forget the chain rule for $colorbluefracpartialpartial xleft(z^2 - xyright)$ and you'll be able to use $(*)$ once again to eliminate $z_x^'$.

share|cite|improve this answer

edited Mar 26 at 10:26

answered Mar 26 at 10:19

StackTDStackTD

24.3k2254

$endgroup$

add a comment | 

1

$begingroup$

If we know that $z_x^' = fracyzz^2 - xy $, how can I calculate the second order partial derivative with respect to x

Take the partial derivative with respect to $x$: use the quotient rule and don't forget that $z$ is a function of $x$ (and $y$). This will then contain the first partial derivative $z'_x$ again, but you can substitute that with:
$$z_x^' = fracyzz^2 - xy tag$*$$$
and then simplify.

---

Here's a start:
$$beginalignz_xx^''=fracpartialpartial xz_x^'
& =fracpartialpartial xleft(fracyzz^2 - xy right) \[6pt]
& =fracleft(z^2 - xyright)colorredfracpartialpartial xleft(yzright)-left(yzright)colorbluefracpartialpartial xleft(z^2 - xyright)left(z^2 - xyright)^2 \
& = ldots
endalign$$
Now $colorredfracpartialpartial xleft(yzright)=yz_x^'$ and use $(*)$ for $z_x^'$ and don't forget the chain rule for $colorbluefracpartialpartial xleft(z^2 - xyright)$ and you'll be able to use $(*)$ once again to eliminate $z_x^'$.

share|cite|improve this answer

edited Mar 26 at 10:26

answered Mar 26 at 10:19

StackTDStackTD

24.3k2254

$endgroup$

add a comment | 

$begingroup$

If we know that $z_x^' = fracyzz^2 - xy $, how can I calculate the second order partial derivative with respect to x

Take the partial derivative with respect to $x$: use the quotient rule and don't forget that $z$ is a function of $x$ (and $y$). This will then contain the first partial derivative $z'_x$ again, but you can substitute that with:
$$z_x^' = fracyzz^2 - xy tag$*$$$
and then simplify.

---

Here's a start:
$$beginalignz_xx^''=fracpartialpartial xz_x^'
& =fracpartialpartial xleft(fracyzz^2 - xy right) \[6pt]
& =fracleft(z^2 - xyright)colorredfracpartialpartial xleft(yzright)-left(yzright)colorbluefracpartialpartial xleft(z^2 - xyright)left(z^2 - xyright)^2 \
& = ldots
endalign$$
Now $colorredfracpartialpartial xleft(yzright)=yz_x^'$ and use $(*)$ for $z_x^'$ and don't forget the chain rule for $colorbluefracpartialpartial xleft(z^2 - xyright)$ and you'll be able to use $(*)$ once again to eliminate $z_x^'$.

share|cite|improve this answer

edited Mar 26 at 10:26

answered Mar 26 at 10:19

StackTDStackTD

24.3k2254

$endgroup$

share|cite|improve this answer

edited Mar 26 at 10:26

answered Mar 26 at 10:19

StackTDStackTD

24.3k2254

answered Mar 26 at 10:19

StackTDStackTD

24.3k2254

answered Mar 26 at 10:19

StackTDStackTD

24.3k2254