How to Isolate all terms with $dy$ divided by $dx$ as a factor on one side of the equation

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
0
down vote

favorite












The question i cant figure out is: $$6x^2+8xy+4y^2+17y-6=0$$
I understand that you are supposed to take the derivative but after i do it says to Isolate all terms with $dy$ divided by $dx$ as a factor on one side of the equation. I cant understand why some terms are $dy/dx$ and other are not.




calculus derivatives implicit-differentiation




share|cite|improve this question





















  • forgot to add it but the question is asking me to differentiate implicity to find dy/dx
    – Keaton Schmidt
    Jul 31 at 5:36














up vote
0
down vote

favorite












The question i cant figure out is: $$6x^2+8xy+4y^2+17y-6=0$$
I understand that you are supposed to take the derivative but after i do it says to Isolate all terms with $dy$ divided by $dx$ as a factor on one side of the equation. I cant understand why some terms are $dy/dx$ and other are not.




calculus derivatives implicit-differentiation




share|cite|improve this question





















  • forgot to add it but the question is asking me to differentiate implicity to find dy/dx
    – Keaton Schmidt
    Jul 31 at 5:36












up vote
0
down vote

favorite









up vote
0
down vote

favorite











The question i cant figure out is: $$6x^2+8xy+4y^2+17y-6=0$$
I understand that you are supposed to take the derivative but after i do it says to Isolate all terms with $dy$ divided by $dx$ as a factor on one side of the equation. I cant understand why some terms are $dy/dx$ and other are not.




calculus derivatives implicit-differentiation




share|cite|improve this question













The question i cant figure out is: $$6x^2+8xy+4y^2+17y-6=0$$
I understand that you are supposed to take the derivative but after i do it says to Isolate all terms with $dy$ divided by $dx$ as a factor on one side of the equation. I cant understand why some terms are $dy/dx$ and other are not.





calculus derivatives implicit-differentiation





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 31 at 5:30









Piyush Divyanakar

3,258122




3,258122









asked Jul 31 at 5:28









Keaton Schmidt

1




1











  • forgot to add it but the question is asking me to differentiate implicity to find dy/dx
    – Keaton Schmidt
    Jul 31 at 5:36
















  • forgot to add it but the question is asking me to differentiate implicity to find dy/dx
    – Keaton Schmidt
    Jul 31 at 5:36















forgot to add it but the question is asking me to differentiate implicity to find dy/dx
– Keaton Schmidt
Jul 31 at 5:36




forgot to add it but the question is asking me to differentiate implicity to find dy/dx
– Keaton Schmidt
Jul 31 at 5:36










1 Answer
1






active

oldest

votes

















up vote
0
down vote













When you take the derivative all variables other than the one that you are taking the derivative with respect to are assumed to be functions of that variable. So $$fracddxx^2=2xquadtext and quad fracddxy^2=fracd(y^2)dxfracdydy=fracdydxfracd(y^2)dy=2yfracdydx$$



This is the result of chain rule for differentiation.
$$fracdzdx=fracdzdyfracdydx$$
So to answer your question the $dy/dx$ comes from the $y$ containing terms in the equation.






share|cite|improve this answer























  • so for instance in the problem i gave the derivative of 8xy is 8y+8x why is the 8y not a dy/dx
    – Keaton Schmidt
    Jul 31 at 5:39










  • $fracddx8xy=8y+8xfracdydx$, it is $8y$ because you are using the product rule and taking the derivative of the $x$. $fracddx8xy=8yfracdxdx+8xfracdydx$
    – Piyush Divyanakar
    Jul 31 at 5:40











Your Answer




StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);








 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2867689%2fhow-to-isolate-all-terms-with-dy-divided-by-dx-as-a-factor-on-one-side-of-th%23new-answer', 'question_page');

);

Post as a guest

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote













When you take the derivative all variables other than the one that you are taking the derivative with respect to are assumed to be functions of that variable. So $$fracddxx^2=2xquadtext and quad fracddxy^2=fracd(y^2)dxfracdydy=fracdydxfracd(y^2)dy=2yfracdydx$$



This is the result of chain rule for differentiation.
$$fracdzdx=fracdzdyfracdydx$$
So to answer your question the $dy/dx$ comes from the $y$ containing terms in the equation.






share|cite|improve this answer























  • so for instance in the problem i gave the derivative of 8xy is 8y+8x why is the 8y not a dy/dx
    – Keaton Schmidt
    Jul 31 at 5:39










  • $fracddx8xy=8y+8xfracdydx$, it is $8y$ because you are using the product rule and taking the derivative of the $x$. $fracddx8xy=8yfracdxdx+8xfracdydx$
    – Piyush Divyanakar
    Jul 31 at 5:40















up vote
0
down vote













When you take the derivative all variables other than the one that you are taking the derivative with respect to are assumed to be functions of that variable. So $$fracddxx^2=2xquadtext and quad fracddxy^2=fracd(y^2)dxfracdydy=fracdydxfracd(y^2)dy=2yfracdydx$$



This is the result of chain rule for differentiation.
$$fracdzdx=fracdzdyfracdydx$$
So to answer your question the $dy/dx$ comes from the $y$ containing terms in the equation.






share|cite|improve this answer























  • so for instance in the problem i gave the derivative of 8xy is 8y+8x why is the 8y not a dy/dx
    – Keaton Schmidt
    Jul 31 at 5:39










  • $fracddx8xy=8y+8xfracdydx$, it is $8y$ because you are using the product rule and taking the derivative of the $x$. $fracddx8xy=8yfracdxdx+8xfracdydx$
    – Piyush Divyanakar
    Jul 31 at 5:40













up vote
0
down vote










up vote
0
down vote









When you take the derivative all variables other than the one that you are taking the derivative with respect to are assumed to be functions of that variable. So $$fracddxx^2=2xquadtext and quad fracddxy^2=fracd(y^2)dxfracdydy=fracdydxfracd(y^2)dy=2yfracdydx$$



This is the result of chain rule for differentiation.
$$fracdzdx=fracdzdyfracdydx$$
So to answer your question the $dy/dx$ comes from the $y$ containing terms in the equation.






share|cite|improve this answer















When you take the derivative all variables other than the one that you are taking the derivative with respect to are assumed to be functions of that variable. So $$fracddxx^2=2xquadtext and quad fracddxy^2=fracd(y^2)dxfracdydy=fracdydxfracd(y^2)dy=2yfracdydx$$



This is the result of chain rule for differentiation.
$$fracdzdx=fracdzdyfracdydx$$
So to answer your question the $dy/dx$ comes from the $y$ containing terms in the equation.







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 31 at 5:39


























answered Jul 31 at 5:35









Piyush Divyanakar

3,258122




3,258122











  • so for instance in the problem i gave the derivative of 8xy is 8y+8x why is the 8y not a dy/dx
    – Keaton Schmidt
    Jul 31 at 5:39










  • $fracddx8xy=8y+8xfracdydx$, it is $8y$ because you are using the product rule and taking the derivative of the $x$. $fracddx8xy=8yfracdxdx+8xfracdydx$
    – Piyush Divyanakar
    Jul 31 at 5:40

















  • so for instance in the problem i gave the derivative of 8xy is 8y+8x why is the 8y not a dy/dx
    – Keaton Schmidt
    Jul 31 at 5:39










  • $fracddx8xy=8y+8xfracdydx$, it is $8y$ because you are using the product rule and taking the derivative of the $x$. $fracddx8xy=8yfracdxdx+8xfracdydx$
    – Piyush Divyanakar
    Jul 31 at 5:40
















so for instance in the problem i gave the derivative of 8xy is 8y+8x why is the 8y not a dy/dx
– Keaton Schmidt
Jul 31 at 5:39




so for instance in the problem i gave the derivative of 8xy is 8y+8x why is the 8y not a dy/dx
– Keaton Schmidt
Jul 31 at 5:39












$fracddx8xy=8y+8xfracdydx$, it is $8y$ because you are using the product rule and taking the derivative of the $x$. $fracddx8xy=8yfracdxdx+8xfracdydx$
– Piyush Divyanakar
Jul 31 at 5:40





$fracddx8xy=8y+8xfracdydx$, it is $8y$ because you are using the product rule and taking the derivative of the $x$. $fracddx8xy=8yfracdxdx+8xfracdydx$
– Piyush Divyanakar
Jul 31 at 5:40













 

draft saved


draft discarded


























 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2867689%2fhow-to-isolate-all-terms-with-dy-divided-by-dx-as-a-factor-on-one-side-of-th%23new-answer', 'question_page');

);

Post as a guest
































































Comments

Post a Comment

Popular posts from this blog

What is the equation of a 3D cone with generalised tilt?

Image
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an...
Read more

Color the edges and diagonals of a regular polygon

Image
Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
Read more

Relationship between determinant of matrix and determinant of adjoint?

Image
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
Read more