Differentiate $e^y/x = 20x-y$Differentiating both sides of an equationFinding centre of ellipse using a tangent line?Trouble with Logarithmic DifferentiationHelp with Implicit Differentiation: Finding an equation for a tangent to a given point on a curveDifferentiate folowing expression, how much simplifying?Why can't one implicitly differentiate these two relations?Differentiate $y = 6 cdot 3^2x - 1$Implicit differentiation and exponentialProblem in trying to link chain rule and implicit differentiationDifferentiate $11x^5 + x^4y + xy^5=18$

The IT department bottlenecks progress. How should I handle this?

What does chmod -u do?

Is the U.S. Code copyrighted by the Government?

What is this called? Old film camera viewer?

When were female captains banned from Starfleet?

Why can Carol Danvers change her suit colours in the first place?

Non-trope happy ending?

Store Credit Card Information in Password Manager?

Must Legal Documents Be Siged In Standard Pen Colors?

Did Swami Prabhupada reject Advaita?

Is there a name for this algorithm to calculate the concentration of a mixture of two solutions containing the same solute?

Multiplicative persistence

Why did the EU agree to delay the Brexit deadline?

How do I find all files that end with a dot

Lowest total scrabble score

What is this cable/device?

What if a revenant (monster) gains fire resistance?

2.8 Why are collections grayed out? How can I open them?

Removing files under particular conditions (number of files, file age)

dpdt switch to spst switch

Why does the Sun have different day lengths, but not the gas giants?

What does "Scientists rise up against statistical significance" mean? (Comment in Nature)

Did arcade monitors have same pixel aspect ratio as TV sets?

How do I color the graph in datavisualization?



Differentiate $e^y/x = 20x-y$


Differentiating both sides of an equationFinding centre of ellipse using a tangent line?Trouble with Logarithmic DifferentiationHelp with Implicit Differentiation: Finding an equation for a tangent to a given point on a curveDifferentiate folowing expression, how much simplifying?Why can't one implicitly differentiate these two relations?Differentiate $y = 6 cdot 3^2x - 1$Implicit differentiation and exponentialProblem in trying to link chain rule and implicit differentiationDifferentiate $11x^5 + x^4y + xy^5=18$













3












$begingroup$


I am trying to use implicit differentiation to differentiate $e^y/x = 20x-y$. I get $frac202 e^y/x cdot fracx-yx^2$, but according to the math website I'm using, "WebWork", this is wrong.



I'm not sure how to handle it when an equation has two $fracdydx$ floating around, but I'm not sure this is the problem.



Here are my steps:



$$fracddx(e^y/x) = fracddx(20x-y)$$



Factoring both sides independently:



$$fracddx(e^y/x) = e^y/xcdotfracx-yx^2 fracdydx$$



$$fracddx(20x-y)=20-fracdydx$$



Finding $fracdydx$



$$e^y/x cdot fracx-yx^2 fracdydx = 20 - fracdydx$$



$$fracdydx+fracx-yx^2 fracdydx = frac20e^y/x$$



$$fracdydx + fracdydx = frac20e^y/xcdotfracx-yx^2$$



$$2fracdydx = frac20e^y/xcdotfracx-yx^2$$



$$fracdydx = frac202e^y/xcdotfracx-yx^2$$






calculus derivatives implicit-differentiation







share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    You want to factor $dy/dx$ out on the left hand side of your computation before dividing by $(x-y)/x^2$. Also, double check your computation of the derivative of $y/x$.
    $endgroup$
    – TM Gallagher
    Mar 15 at 17:02







  • 1




    $begingroup$
    You make a mistake on d/dx(e^(y/x))
    $endgroup$
    – DragunityMAX
    Mar 15 at 17:03










  • $begingroup$
    @DragunityMAX I'm not sure how to handle derivating $e^x$ when x is not a lone variable.
    $endgroup$
    – LuminousNutria
    Mar 15 at 17:09







  • 1




    $begingroup$
    @LuminousNutria Exactly what you did. You just make a calculation mistake when applying quotient rule.
    $endgroup$
    – DragunityMAX
    Mar 15 at 17:18






  • 1




    $begingroup$
    @LuminousNutria Yadati Kiran gives the ans. Note that y is function of x.
    $endgroup$
    – DragunityMAX
    Mar 15 at 17:22















3












$begingroup$


I am trying to use implicit differentiation to differentiate $e^y/x = 20x-y$. I get $frac202 e^y/x cdot fracx-yx^2$, but according to the math website I'm using, "WebWork", this is wrong.



I'm not sure how to handle it when an equation has two $fracdydx$ floating around, but I'm not sure this is the problem.



Here are my steps:



$$fracddx(e^y/x) = fracddx(20x-y)$$



Factoring both sides independently:



$$fracddx(e^y/x) = e^y/xcdotfracx-yx^2 fracdydx$$



$$fracddx(20x-y)=20-fracdydx$$



Finding $fracdydx$



$$e^y/x cdot fracx-yx^2 fracdydx = 20 - fracdydx$$



$$fracdydx+fracx-yx^2 fracdydx = frac20e^y/x$$



$$fracdydx + fracdydx = frac20e^y/xcdotfracx-yx^2$$



$$2fracdydx = frac20e^y/xcdotfracx-yx^2$$



$$fracdydx = frac202e^y/xcdotfracx-yx^2$$






calculus derivatives implicit-differentiation







share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    You want to factor $dy/dx$ out on the left hand side of your computation before dividing by $(x-y)/x^2$. Also, double check your computation of the derivative of $y/x$.
    $endgroup$
    – TM Gallagher
    Mar 15 at 17:02







  • 1




    $begingroup$
    You make a mistake on d/dx(e^(y/x))
    $endgroup$
    – DragunityMAX
    Mar 15 at 17:03










  • $begingroup$
    @DragunityMAX I'm not sure how to handle derivating $e^x$ when x is not a lone variable.
    $endgroup$
    – LuminousNutria
    Mar 15 at 17:09







  • 1




    $begingroup$
    @LuminousNutria Exactly what you did. You just make a calculation mistake when applying quotient rule.
    $endgroup$
    – DragunityMAX
    Mar 15 at 17:18






  • 1




    $begingroup$
    @LuminousNutria Yadati Kiran gives the ans. Note that y is function of x.
    $endgroup$
    – DragunityMAX
    Mar 15 at 17:22













3












3








3


0



$begingroup$


I am trying to use implicit differentiation to differentiate $e^y/x = 20x-y$. I get $frac202 e^y/x cdot fracx-yx^2$, but according to the math website I'm using, "WebWork", this is wrong.



I'm not sure how to handle it when an equation has two $fracdydx$ floating around, but I'm not sure this is the problem.



Here are my steps:



$$fracddx(e^y/x) = fracddx(20x-y)$$



Factoring both sides independently:



$$fracddx(e^y/x) = e^y/xcdotfracx-yx^2 fracdydx$$



$$fracddx(20x-y)=20-fracdydx$$



Finding $fracdydx$



$$e^y/x cdot fracx-yx^2 fracdydx = 20 - fracdydx$$



$$fracdydx+fracx-yx^2 fracdydx = frac20e^y/x$$



$$fracdydx + fracdydx = frac20e^y/xcdotfracx-yx^2$$



$$2fracdydx = frac20e^y/xcdotfracx-yx^2$$



$$fracdydx = frac202e^y/xcdotfracx-yx^2$$






calculus derivatives implicit-differentiation







share|cite|improve this question











$endgroup$




I am trying to use implicit differentiation to differentiate $e^y/x = 20x-y$. I get $frac202 e^y/x cdot fracx-yx^2$, but according to the math website I'm using, "WebWork", this is wrong.



I'm not sure how to handle it when an equation has two $fracdydx$ floating around, but I'm not sure this is the problem.



Here are my steps:



$$fracddx(e^y/x) = fracddx(20x-y)$$



Factoring both sides independently:



$$fracddx(e^y/x) = e^y/xcdotfracx-yx^2 fracdydx$$



$$fracddx(20x-y)=20-fracdydx$$



Finding $fracdydx$



$$e^y/x cdot fracx-yx^2 fracdydx = 20 - fracdydx$$



$$fracdydx+fracx-yx^2 fracdydx = frac20e^y/x$$



$$fracdydx + fracdydx = frac20e^y/xcdotfracx-yx^2$$



$$2fracdydx = frac20e^y/xcdotfracx-yx^2$$



$$fracdydx = frac202e^y/xcdotfracx-yx^2$$






calculus derivatives implicit-differentiation




calculus derivatives implicit-differentiation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 15 at 17:55







LuminousNutria

















asked Mar 15 at 16:57









LuminousNutriaLuminousNutria

45512




45512







  • 1




    $begingroup$
    You want to factor $dy/dx$ out on the left hand side of your computation before dividing by $(x-y)/x^2$. Also, double check your computation of the derivative of $y/x$.
    $endgroup$
    – TM Gallagher
    Mar 15 at 17:02







  • 1




    $begingroup$
    You make a mistake on d/dx(e^(y/x))
    $endgroup$
    – DragunityMAX
    Mar 15 at 17:03










  • $begingroup$
    @DragunityMAX I'm not sure how to handle derivating $e^x$ when x is not a lone variable.
    $endgroup$
    – LuminousNutria
    Mar 15 at 17:09







  • 1




    $begingroup$
    @LuminousNutria Exactly what you did. You just make a calculation mistake when applying quotient rule.
    $endgroup$
    – DragunityMAX
    Mar 15 at 17:18






  • 1




    $begingroup$
    @LuminousNutria Yadati Kiran gives the ans. Note that y is function of x.
    $endgroup$
    – DragunityMAX
    Mar 15 at 17:22












  • 1




    $begingroup$
    You want to factor $dy/dx$ out on the left hand side of your computation before dividing by $(x-y)/x^2$. Also, double check your computation of the derivative of $y/x$.
    $endgroup$
    – TM Gallagher
    Mar 15 at 17:02







  • 1




    $begingroup$
    You make a mistake on d/dx(e^(y/x))
    $endgroup$
    – DragunityMAX
    Mar 15 at 17:03










  • $begingroup$
    @DragunityMAX I'm not sure how to handle derivating $e^x$ when x is not a lone variable.
    $endgroup$
    – LuminousNutria
    Mar 15 at 17:09







  • 1




    $begingroup$
    @LuminousNutria Exactly what you did. You just make a calculation mistake when applying quotient rule.
    $endgroup$
    – DragunityMAX
    Mar 15 at 17:18






  • 1




    $begingroup$
    @LuminousNutria Yadati Kiran gives the ans. Note that y is function of x.
    $endgroup$
    – DragunityMAX
    Mar 15 at 17:22







1




1




$begingroup$
You want to factor $dy/dx$ out on the left hand side of your computation before dividing by $(x-y)/x^2$. Also, double check your computation of the derivative of $y/x$.
$endgroup$
– TM Gallagher
Mar 15 at 17:02





$begingroup$
You want to factor $dy/dx$ out on the left hand side of your computation before dividing by $(x-y)/x^2$. Also, double check your computation of the derivative of $y/x$.
$endgroup$
– TM Gallagher
Mar 15 at 17:02





1




1




$begingroup$
You make a mistake on d/dx(e^(y/x))
$endgroup$
– DragunityMAX
Mar 15 at 17:03




$begingroup$
You make a mistake on d/dx(e^(y/x))
$endgroup$
– DragunityMAX
Mar 15 at 17:03












$begingroup$
@DragunityMAX I'm not sure how to handle derivating $e^x$ when x is not a lone variable.
$endgroup$
– LuminousNutria
Mar 15 at 17:09





$begingroup$
@DragunityMAX I'm not sure how to handle derivating $e^x$ when x is not a lone variable.
$endgroup$
– LuminousNutria
Mar 15 at 17:09





1




1




$begingroup$
@LuminousNutria Exactly what you did. You just make a calculation mistake when applying quotient rule.
$endgroup$
– DragunityMAX
Mar 15 at 17:18




$begingroup$
@LuminousNutria Exactly what you did. You just make a calculation mistake when applying quotient rule.
$endgroup$
– DragunityMAX
Mar 15 at 17:18




1




1




$begingroup$
@LuminousNutria Yadati Kiran gives the ans. Note that y is function of x.
$endgroup$
– DragunityMAX
Mar 15 at 17:22




$begingroup$
@LuminousNutria Yadati Kiran gives the ans. Note that y is function of x.
$endgroup$
– DragunityMAX
Mar 15 at 17:22










3 Answers
3






active

oldest

votes


















1












$begingroup$

Consider representing $fracdydx=y'$ to make the equation clearer and easier to deal with terms.



Assuming your original equation is $e^y/x = 20x-y$.




$dfracddx(e^y/x) = e^y/xcdotdfracx-yx^2 dfracdydx.quad$ Incorrect! You have to consider $y$ as a function of $x$ as well.




$dfracddx(e^y/x)=e^y/xcdot dfracddxleft(dfracyxright)=e^y/xcdotdfracxdfracdydx-ydfracdxdxx^2=e^y/xcdotleft(dfracxy'-yx^2right)tag1$



So $dfracddxleft(e^y/x = 20x-yright)impliesdfracddxleft(e^y/x right)=dfracddxleft(20x-yright)$. Upon substituting $(1)$ we have,
$beginaligne^y/xcdotleft(dfracxy'-yx^2right)=20-y'&implies e^y/xxy'-e^y/xy=20x^2-x^2y'\&implies y'(x^2+xe^y/x)=20x^2+e^y/xyendalign$



I suppose you can take it from here.



Taking logarithms can be considered as well.$$fracyx=ln(20x-y)$$ Differentiating $$fracxy'-yx^2=frac20-y'20x-y implies y'=frac20x^2+20xy-y^221x^2-xy$$






share|cite|improve this answer











$endgroup$












  • $begingroup$
    How do you get rid of the second $fracdydx$ or $y'$ in the line $dfracddx(e^y/x)=e^y/xcdot dfracddxleft(dfracyxright)=e^y/xcdotdfracxfracdydx-yfracdxdxx^2=e^y/xcdotleft(dfracxy'-yx^2right)$?
    $endgroup$
    – LuminousNutria
    Mar 15 at 17:33







  • 1




    $begingroup$
    @LuminousNutria : I did not. I replaced $fracdydx$ by $y'$ and $fracdxdx$ by $1$. I got factor out $y'$ in the next line when I write $dfracddxleft(e^y/x = 20x-yright)$.
    $endgroup$
    – Yadati Kiran
    Mar 15 at 17:38











  • $begingroup$
    Oh, I see! I misread your math, I thought $fracdxdx$ was $fracdydx$
    $endgroup$
    – LuminousNutria
    Mar 15 at 17:45



















2












$begingroup$

You get:
$$(e^y/x)'_x=(2x-y)'_x Rightarrow \
e^y/xcdot (y/x)'_x=2-y'Rightarrow \
(2x-y)cdot fracy'x-yx^2=2-y' Rightarrow \
(2x^2-xy)y'-(2x-y)y=2x^2-x^2y'Rightarrow \
y'=frac2x^2+2xy-y^23x^2-xy.$$



If you are familiar with multivariable calculus, consider: $F(x,y)=e^y/x-2x+y=0$. Then:
$$y'=-fracF_xF_y=-frace^y/xcdot (-y/x^2)-2e^y/xcdot (1/x)+1=frac(2x-y)y+2x^2(2x-y)x+x^2=frac2x^2+2xy-y^23x^2-xy.$$






share|cite|improve this answer









$endgroup$












  • $begingroup$
    What's wrong with leaving $e^y/x$?
    $endgroup$
    – Tomislav Ostojich
    Mar 15 at 19:26






  • 2




    $begingroup$
    Nothing, it is simpler, isn't it?
    $endgroup$
    – farruhota
    Mar 15 at 19:43


















1












$begingroup$

$e^y/x cdot fracx-yx^2 = 20 - fracdydx$. You forgot that there was a $fracdydx$ term on the left which should have prevented you from simply adding those two fractions together unless you had slid it over into the numerator. But I don't even see it there. Compare your solution to mine:



$$
e^fracyx = 20x-y\
fracddxleft(e^fracyxright) = fracddx(20x-y)\
e^fracyxfracddxleft(fracyxright)=20-fracdydx\
e^fracyxleft(fracddxleft(frac1xright)y+frac1xfracdydxright)=20-fracdydx\
e^fracyxleft(-frac1x^2y+frac1xfracdydxright)=20-fracdydx\
-frac1x^2ye^fracyx+frac1xfracdydxe^fracyx=20-fracdydx\
frac1xfracdydxe^fracyx+fracdydx=20+frac1x^2ye^fracyx\
fracdydxleft(frac1xe^fracyx+1right)=20+frac1x^2ye^fracyx\
fracdydx=frac20+frac1x^2ye^fracyxfrac1xe^fracyx+1\
fracdydx=frac20+frac1x^2ye^fracyxfrac1xe^fracyx+1cdotfracx^2x^2\
fracdydx=frac20x^2+ye^fracyxxe^fracyx+x^2\
$$



Wolfram Alpha check.






share|cite|improve this answer











$endgroup$












    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',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader:
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    ,
    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%2f3149540%2fdifferentiate-ey-x-20x-y%23new-answer', 'question_page');

    );

    Post as a guest















    Required, but never shown

























    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Consider representing $fracdydx=y'$ to make the equation clearer and easier to deal with terms.



    Assuming your original equation is $e^y/x = 20x-y$.




    $dfracddx(e^y/x) = e^y/xcdotdfracx-yx^2 dfracdydx.quad$ Incorrect! You have to consider $y$ as a function of $x$ as well.




    $dfracddx(e^y/x)=e^y/xcdot dfracddxleft(dfracyxright)=e^y/xcdotdfracxdfracdydx-ydfracdxdxx^2=e^y/xcdotleft(dfracxy'-yx^2right)tag1$



    So $dfracddxleft(e^y/x = 20x-yright)impliesdfracddxleft(e^y/x right)=dfracddxleft(20x-yright)$. Upon substituting $(1)$ we have,
    $beginaligne^y/xcdotleft(dfracxy'-yx^2right)=20-y'&implies e^y/xxy'-e^y/xy=20x^2-x^2y'\&implies y'(x^2+xe^y/x)=20x^2+e^y/xyendalign$



    I suppose you can take it from here.



    Taking logarithms can be considered as well.$$fracyx=ln(20x-y)$$ Differentiating $$fracxy'-yx^2=frac20-y'20x-y implies y'=frac20x^2+20xy-y^221x^2-xy$$






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      How do you get rid of the second $fracdydx$ or $y'$ in the line $dfracddx(e^y/x)=e^y/xcdot dfracddxleft(dfracyxright)=e^y/xcdotdfracxfracdydx-yfracdxdxx^2=e^y/xcdotleft(dfracxy'-yx^2right)$?
      $endgroup$
      – LuminousNutria
      Mar 15 at 17:33







    • 1




      $begingroup$
      @LuminousNutria : I did not. I replaced $fracdydx$ by $y'$ and $fracdxdx$ by $1$. I got factor out $y'$ in the next line when I write $dfracddxleft(e^y/x = 20x-yright)$.
      $endgroup$
      – Yadati Kiran
      Mar 15 at 17:38











    • $begingroup$
      Oh, I see! I misread your math, I thought $fracdxdx$ was $fracdydx$
      $endgroup$
      – LuminousNutria
      Mar 15 at 17:45
















    1












    $begingroup$

    Consider representing $fracdydx=y'$ to make the equation clearer and easier to deal with terms.



    Assuming your original equation is $e^y/x = 20x-y$.




    $dfracddx(e^y/x) = e^y/xcdotdfracx-yx^2 dfracdydx.quad$ Incorrect! You have to consider $y$ as a function of $x$ as well.




    $dfracddx(e^y/x)=e^y/xcdot dfracddxleft(dfracyxright)=e^y/xcdotdfracxdfracdydx-ydfracdxdxx^2=e^y/xcdotleft(dfracxy'-yx^2right)tag1$



    So $dfracddxleft(e^y/x = 20x-yright)impliesdfracddxleft(e^y/x right)=dfracddxleft(20x-yright)$. Upon substituting $(1)$ we have,
    $beginaligne^y/xcdotleft(dfracxy'-yx^2right)=20-y'&implies e^y/xxy'-e^y/xy=20x^2-x^2y'\&implies y'(x^2+xe^y/x)=20x^2+e^y/xyendalign$



    I suppose you can take it from here.



    Taking logarithms can be considered as well.$$fracyx=ln(20x-y)$$ Differentiating $$fracxy'-yx^2=frac20-y'20x-y implies y'=frac20x^2+20xy-y^221x^2-xy$$






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      How do you get rid of the second $fracdydx$ or $y'$ in the line $dfracddx(e^y/x)=e^y/xcdot dfracddxleft(dfracyxright)=e^y/xcdotdfracxfracdydx-yfracdxdxx^2=e^y/xcdotleft(dfracxy'-yx^2right)$?
      $endgroup$
      – LuminousNutria
      Mar 15 at 17:33







    • 1




      $begingroup$
      @LuminousNutria : I did not. I replaced $fracdydx$ by $y'$ and $fracdxdx$ by $1$. I got factor out $y'$ in the next line when I write $dfracddxleft(e^y/x = 20x-yright)$.
      $endgroup$
      – Yadati Kiran
      Mar 15 at 17:38











    • $begingroup$
      Oh, I see! I misread your math, I thought $fracdxdx$ was $fracdydx$
      $endgroup$
      – LuminousNutria
      Mar 15 at 17:45














    1












    1








    1





    $begingroup$

    Consider representing $fracdydx=y'$ to make the equation clearer and easier to deal with terms.



    Assuming your original equation is $e^y/x = 20x-y$.




    $dfracddx(e^y/x) = e^y/xcdotdfracx-yx^2 dfracdydx.quad$ Incorrect! You have to consider $y$ as a function of $x$ as well.




    $dfracddx(e^y/x)=e^y/xcdot dfracddxleft(dfracyxright)=e^y/xcdotdfracxdfracdydx-ydfracdxdxx^2=e^y/xcdotleft(dfracxy'-yx^2right)tag1$



    So $dfracddxleft(e^y/x = 20x-yright)impliesdfracddxleft(e^y/x right)=dfracddxleft(20x-yright)$. Upon substituting $(1)$ we have,
    $beginaligne^y/xcdotleft(dfracxy'-yx^2right)=20-y'&implies e^y/xxy'-e^y/xy=20x^2-x^2y'\&implies y'(x^2+xe^y/x)=20x^2+e^y/xyendalign$



    I suppose you can take it from here.



    Taking logarithms can be considered as well.$$fracyx=ln(20x-y)$$ Differentiating $$fracxy'-yx^2=frac20-y'20x-y implies y'=frac20x^2+20xy-y^221x^2-xy$$






    share|cite|improve this answer











    $endgroup$



    Consider representing $fracdydx=y'$ to make the equation clearer and easier to deal with terms.



    Assuming your original equation is $e^y/x = 20x-y$.




    $dfracddx(e^y/x) = e^y/xcdotdfracx-yx^2 dfracdydx.quad$ Incorrect! You have to consider $y$ as a function of $x$ as well.




    $dfracddx(e^y/x)=e^y/xcdot dfracddxleft(dfracyxright)=e^y/xcdotdfracxdfracdydx-ydfracdxdxx^2=e^y/xcdotleft(dfracxy'-yx^2right)tag1$



    So $dfracddxleft(e^y/x = 20x-yright)impliesdfracddxleft(e^y/x right)=dfracddxleft(20x-yright)$. Upon substituting $(1)$ we have,
    $beginaligne^y/xcdotleft(dfracxy'-yx^2right)=20-y'&implies e^y/xxy'-e^y/xy=20x^2-x^2y'\&implies y'(x^2+xe^y/x)=20x^2+e^y/xyendalign$



    I suppose you can take it from here.



    Taking logarithms can be considered as well.$$fracyx=ln(20x-y)$$ Differentiating $$fracxy'-yx^2=frac20-y'20x-y implies y'=frac20x^2+20xy-y^221x^2-xy$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Mar 15 at 17:46

























    answered Mar 15 at 17:05









    Yadati KiranYadati Kiran

    2,1101622




    2,1101622











    • $begingroup$
      How do you get rid of the second $fracdydx$ or $y'$ in the line $dfracddx(e^y/x)=e^y/xcdot dfracddxleft(dfracyxright)=e^y/xcdotdfracxfracdydx-yfracdxdxx^2=e^y/xcdotleft(dfracxy'-yx^2right)$?
      $endgroup$
      – LuminousNutria
      Mar 15 at 17:33







    • 1




      $begingroup$
      @LuminousNutria : I did not. I replaced $fracdydx$ by $y'$ and $fracdxdx$ by $1$. I got factor out $y'$ in the next line when I write $dfracddxleft(e^y/x = 20x-yright)$.
      $endgroup$
      – Yadati Kiran
      Mar 15 at 17:38











    • $begingroup$
      Oh, I see! I misread your math, I thought $fracdxdx$ was $fracdydx$
      $endgroup$
      – LuminousNutria
      Mar 15 at 17:45

















    • $begingroup$
      How do you get rid of the second $fracdydx$ or $y'$ in the line $dfracddx(e^y/x)=e^y/xcdot dfracddxleft(dfracyxright)=e^y/xcdotdfracxfracdydx-yfracdxdxx^2=e^y/xcdotleft(dfracxy'-yx^2right)$?
      $endgroup$
      – LuminousNutria
      Mar 15 at 17:33







    • 1




      $begingroup$
      @LuminousNutria : I did not. I replaced $fracdydx$ by $y'$ and $fracdxdx$ by $1$. I got factor out $y'$ in the next line when I write $dfracddxleft(e^y/x = 20x-yright)$.
      $endgroup$
      – Yadati Kiran
      Mar 15 at 17:38











    • $begingroup$
      Oh, I see! I misread your math, I thought $fracdxdx$ was $fracdydx$
      $endgroup$
      – LuminousNutria
      Mar 15 at 17:45
















    $begingroup$
    How do you get rid of the second $fracdydx$ or $y'$ in the line $dfracddx(e^y/x)=e^y/xcdot dfracddxleft(dfracyxright)=e^y/xcdotdfracxfracdydx-yfracdxdxx^2=e^y/xcdotleft(dfracxy'-yx^2right)$?
    $endgroup$
    – LuminousNutria
    Mar 15 at 17:33





    $begingroup$
    How do you get rid of the second $fracdydx$ or $y'$ in the line $dfracddx(e^y/x)=e^y/xcdot dfracddxleft(dfracyxright)=e^y/xcdotdfracxfracdydx-yfracdxdxx^2=e^y/xcdotleft(dfracxy'-yx^2right)$?
    $endgroup$
    – LuminousNutria
    Mar 15 at 17:33





    1




    1




    $begingroup$
    @LuminousNutria : I did not. I replaced $fracdydx$ by $y'$ and $fracdxdx$ by $1$. I got factor out $y'$ in the next line when I write $dfracddxleft(e^y/x = 20x-yright)$.
    $endgroup$
    – Yadati Kiran
    Mar 15 at 17:38





    $begingroup$
    @LuminousNutria : I did not. I replaced $fracdydx$ by $y'$ and $fracdxdx$ by $1$. I got factor out $y'$ in the next line when I write $dfracddxleft(e^y/x = 20x-yright)$.
    $endgroup$
    – Yadati Kiran
    Mar 15 at 17:38













    $begingroup$
    Oh, I see! I misread your math, I thought $fracdxdx$ was $fracdydx$
    $endgroup$
    – LuminousNutria
    Mar 15 at 17:45





    $begingroup$
    Oh, I see! I misread your math, I thought $fracdxdx$ was $fracdydx$
    $endgroup$
    – LuminousNutria
    Mar 15 at 17:45












    2












    $begingroup$

    You get:
    $$(e^y/x)'_x=(2x-y)'_x Rightarrow \
    e^y/xcdot (y/x)'_x=2-y'Rightarrow \
    (2x-y)cdot fracy'x-yx^2=2-y' Rightarrow \
    (2x^2-xy)y'-(2x-y)y=2x^2-x^2y'Rightarrow \
    y'=frac2x^2+2xy-y^23x^2-xy.$$



    If you are familiar with multivariable calculus, consider: $F(x,y)=e^y/x-2x+y=0$. Then:
    $$y'=-fracF_xF_y=-frace^y/xcdot (-y/x^2)-2e^y/xcdot (1/x)+1=frac(2x-y)y+2x^2(2x-y)x+x^2=frac2x^2+2xy-y^23x^2-xy.$$






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      What's wrong with leaving $e^y/x$?
      $endgroup$
      – Tomislav Ostojich
      Mar 15 at 19:26






    • 2




      $begingroup$
      Nothing, it is simpler, isn't it?
      $endgroup$
      – farruhota
      Mar 15 at 19:43















    2












    $begingroup$

    You get:
    $$(e^y/x)'_x=(2x-y)'_x Rightarrow \
    e^y/xcdot (y/x)'_x=2-y'Rightarrow \
    (2x-y)cdot fracy'x-yx^2=2-y' Rightarrow \
    (2x^2-xy)y'-(2x-y)y=2x^2-x^2y'Rightarrow \
    y'=frac2x^2+2xy-y^23x^2-xy.$$



    If you are familiar with multivariable calculus, consider: $F(x,y)=e^y/x-2x+y=0$. Then:
    $$y'=-fracF_xF_y=-frace^y/xcdot (-y/x^2)-2e^y/xcdot (1/x)+1=frac(2x-y)y+2x^2(2x-y)x+x^2=frac2x^2+2xy-y^23x^2-xy.$$






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      What's wrong with leaving $e^y/x$?
      $endgroup$
      – Tomislav Ostojich
      Mar 15 at 19:26






    • 2




      $begingroup$
      Nothing, it is simpler, isn't it?
      $endgroup$
      – farruhota
      Mar 15 at 19:43













    2












    2








    2





    $begingroup$

    You get:
    $$(e^y/x)'_x=(2x-y)'_x Rightarrow \
    e^y/xcdot (y/x)'_x=2-y'Rightarrow \
    (2x-y)cdot fracy'x-yx^2=2-y' Rightarrow \
    (2x^2-xy)y'-(2x-y)y=2x^2-x^2y'Rightarrow \
    y'=frac2x^2+2xy-y^23x^2-xy.$$



    If you are familiar with multivariable calculus, consider: $F(x,y)=e^y/x-2x+y=0$. Then:
    $$y'=-fracF_xF_y=-frace^y/xcdot (-y/x^2)-2e^y/xcdot (1/x)+1=frac(2x-y)y+2x^2(2x-y)x+x^2=frac2x^2+2xy-y^23x^2-xy.$$






    share|cite|improve this answer









    $endgroup$



    You get:
    $$(e^y/x)'_x=(2x-y)'_x Rightarrow \
    e^y/xcdot (y/x)'_x=2-y'Rightarrow \
    (2x-y)cdot fracy'x-yx^2=2-y' Rightarrow \
    (2x^2-xy)y'-(2x-y)y=2x^2-x^2y'Rightarrow \
    y'=frac2x^2+2xy-y^23x^2-xy.$$



    If you are familiar with multivariable calculus, consider: $F(x,y)=e^y/x-2x+y=0$. Then:
    $$y'=-fracF_xF_y=-frace^y/xcdot (-y/x^2)-2e^y/xcdot (1/x)+1=frac(2x-y)y+2x^2(2x-y)x+x^2=frac2x^2+2xy-y^23x^2-xy.$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Mar 15 at 17:24









    farruhotafarruhota

    21.5k2842




    21.5k2842











    • $begingroup$
      What's wrong with leaving $e^y/x$?
      $endgroup$
      – Tomislav Ostojich
      Mar 15 at 19:26






    • 2




      $begingroup$
      Nothing, it is simpler, isn't it?
      $endgroup$
      – farruhota
      Mar 15 at 19:43
















    • $begingroup$
      What's wrong with leaving $e^y/x$?
      $endgroup$
      – Tomislav Ostojich
      Mar 15 at 19:26






    • 2




      $begingroup$
      Nothing, it is simpler, isn't it?
      $endgroup$
      – farruhota
      Mar 15 at 19:43















    $begingroup$
    What's wrong with leaving $e^y/x$?
    $endgroup$
    – Tomislav Ostojich
    Mar 15 at 19:26




    $begingroup$
    What's wrong with leaving $e^y/x$?
    $endgroup$
    – Tomislav Ostojich
    Mar 15 at 19:26




    2




    2




    $begingroup$
    Nothing, it is simpler, isn't it?
    $endgroup$
    – farruhota
    Mar 15 at 19:43




    $begingroup$
    Nothing, it is simpler, isn't it?
    $endgroup$
    – farruhota
    Mar 15 at 19:43











    1












    $begingroup$

    $e^y/x cdot fracx-yx^2 = 20 - fracdydx$. You forgot that there was a $fracdydx$ term on the left which should have prevented you from simply adding those two fractions together unless you had slid it over into the numerator. But I don't even see it there. Compare your solution to mine:



    $$
    e^fracyx = 20x-y\
    fracddxleft(e^fracyxright) = fracddx(20x-y)\
    e^fracyxfracddxleft(fracyxright)=20-fracdydx\
    e^fracyxleft(fracddxleft(frac1xright)y+frac1xfracdydxright)=20-fracdydx\
    e^fracyxleft(-frac1x^2y+frac1xfracdydxright)=20-fracdydx\
    -frac1x^2ye^fracyx+frac1xfracdydxe^fracyx=20-fracdydx\
    frac1xfracdydxe^fracyx+fracdydx=20+frac1x^2ye^fracyx\
    fracdydxleft(frac1xe^fracyx+1right)=20+frac1x^2ye^fracyx\
    fracdydx=frac20+frac1x^2ye^fracyxfrac1xe^fracyx+1\
    fracdydx=frac20+frac1x^2ye^fracyxfrac1xe^fracyx+1cdotfracx^2x^2\
    fracdydx=frac20x^2+ye^fracyxxe^fracyx+x^2\
    $$



    Wolfram Alpha check.






    share|cite|improve this answer











    $endgroup$

















      1












      $begingroup$

      $e^y/x cdot fracx-yx^2 = 20 - fracdydx$. You forgot that there was a $fracdydx$ term on the left which should have prevented you from simply adding those two fractions together unless you had slid it over into the numerator. But I don't even see it there. Compare your solution to mine:



      $$
      e^fracyx = 20x-y\
      fracddxleft(e^fracyxright) = fracddx(20x-y)\
      e^fracyxfracddxleft(fracyxright)=20-fracdydx\
      e^fracyxleft(fracddxleft(frac1xright)y+frac1xfracdydxright)=20-fracdydx\
      e^fracyxleft(-frac1x^2y+frac1xfracdydxright)=20-fracdydx\
      -frac1x^2ye^fracyx+frac1xfracdydxe^fracyx=20-fracdydx\
      frac1xfracdydxe^fracyx+fracdydx=20+frac1x^2ye^fracyx\
      fracdydxleft(frac1xe^fracyx+1right)=20+frac1x^2ye^fracyx\
      fracdydx=frac20+frac1x^2ye^fracyxfrac1xe^fracyx+1\
      fracdydx=frac20+frac1x^2ye^fracyxfrac1xe^fracyx+1cdotfracx^2x^2\
      fracdydx=frac20x^2+ye^fracyxxe^fracyx+x^2\
      $$



      Wolfram Alpha check.






      share|cite|improve this answer











      $endgroup$















        1












        1








        1





        $begingroup$

        $e^y/x cdot fracx-yx^2 = 20 - fracdydx$. You forgot that there was a $fracdydx$ term on the left which should have prevented you from simply adding those two fractions together unless you had slid it over into the numerator. But I don't even see it there. Compare your solution to mine:



        $$
        e^fracyx = 20x-y\
        fracddxleft(e^fracyxright) = fracddx(20x-y)\
        e^fracyxfracddxleft(fracyxright)=20-fracdydx\
        e^fracyxleft(fracddxleft(frac1xright)y+frac1xfracdydxright)=20-fracdydx\
        e^fracyxleft(-frac1x^2y+frac1xfracdydxright)=20-fracdydx\
        -frac1x^2ye^fracyx+frac1xfracdydxe^fracyx=20-fracdydx\
        frac1xfracdydxe^fracyx+fracdydx=20+frac1x^2ye^fracyx\
        fracdydxleft(frac1xe^fracyx+1right)=20+frac1x^2ye^fracyx\
        fracdydx=frac20+frac1x^2ye^fracyxfrac1xe^fracyx+1\
        fracdydx=frac20+frac1x^2ye^fracyxfrac1xe^fracyx+1cdotfracx^2x^2\
        fracdydx=frac20x^2+ye^fracyxxe^fracyx+x^2\
        $$



        Wolfram Alpha check.






        share|cite|improve this answer











        $endgroup$



        $e^y/x cdot fracx-yx^2 = 20 - fracdydx$. You forgot that there was a $fracdydx$ term on the left which should have prevented you from simply adding those two fractions together unless you had slid it over into the numerator. But I don't even see it there. Compare your solution to mine:



        $$
        e^fracyx = 20x-y\
        fracddxleft(e^fracyxright) = fracddx(20x-y)\
        e^fracyxfracddxleft(fracyxright)=20-fracdydx\
        e^fracyxleft(fracddxleft(frac1xright)y+frac1xfracdydxright)=20-fracdydx\
        e^fracyxleft(-frac1x^2y+frac1xfracdydxright)=20-fracdydx\
        -frac1x^2ye^fracyx+frac1xfracdydxe^fracyx=20-fracdydx\
        frac1xfracdydxe^fracyx+fracdydx=20+frac1x^2ye^fracyx\
        fracdydxleft(frac1xe^fracyx+1right)=20+frac1x^2ye^fracyx\
        fracdydx=frac20+frac1x^2ye^fracyxfrac1xe^fracyx+1\
        fracdydx=frac20+frac1x^2ye^fracyxfrac1xe^fracyx+1cdotfracx^2x^2\
        fracdydx=frac20x^2+ye^fracyxxe^fracyx+x^2\
        $$



        Wolfram Alpha check.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Mar 15 at 19:39

























        answered Mar 15 at 17:21









        Michael RybkinMichael Rybkin

        3,954421




        3,954421



























            draft saved

            draft discarded
















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …


            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.

            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3149540%2fdifferentiate-ey-x-20x-y%23new-answer', 'question_page');

            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            How should I support this large drywall patch? Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?How do I cover large gaps in drywall?How do I keep drywall around a patch from crumbling?Can I glue a second layer of drywall?How to patch long strip on drywall?Large drywall patch: how to avoid bulging seams?Drywall Mesh Patch vs. Bulge? To remove or not to remove?How to fix this drywall job?Prep drywall before backsplashWhat's the best way to fix this horrible drywall patch job?Drywall patching using 3M Patch Plus Primer

            Image
            How do living politicians protect their readily obtainable signatures from misuse? Is it possible to force a specific program to remain in memory after closing it? File name problem(?) How to pronounce 伝統色 How does the math work when buying airline miles? Sum letters are not two different An adverb for when you're not exaggerating What does it mean that physics no longer uses mechanical models to describe phenomena? Take 2! Is this homebrew Lady of Pain warlock patron balanced? How many time has Arya actually used Needle? A letter with no particular backstory Converted a Scalar function to a TVF function for parallel execution-Still running in Serial mode Generate an RGB colour grid Co-worker has annoying ringtone Strange behavior of Object.defineProperty() in JavaScript "Lost his faith in humanity in the trenches of Verdun" — last line of an SF story What is this round thing on the pantry door in The Shining What would you ca...
            Read more

            random experiment with two different functions on unit interval Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Random variable and probability space notionsRandom Walk with EdgesFinding functions where the increase over a random interval is Poisson distributedNumber of days until dayCan an observed event in fact be of zero probability?Unit random processmodels of coins and uniform distributionHow to get the number of successes given $n$ trials , probability $P$ and a random variable $X$Absorbing Markov chain in a computer. Is “almost every” turned into always convergence in computer executions?Stopped random walk is not uniformly integrable

            Image
            What would you call this weird metallic apparatus that allows you to lift people? Is it possible to give , in economics, an example of a relation ( set of ordered pairs) that is not a function? Maximum summed subsequences with non-adjacent items How many time has Arya actually used Needle? Is there public access to the Meteor Crater in Arizona? Belief In God or Knowledge Of God. Which is better? What is the meaning of 'breadth' in breadth first search? Why is there Net Work Done on a Pressure/Volume Cycle? How much damage would a cupful of neutron star matter do to the Earth? Sum letters are not two different Why does it sometimes sound good to play a grace note as a lead in to a note in a melody? What initially awakened the Balrog? How to pronounce 伝統色 QGIS virtual layer functionality does not seem to support memory layers AppleTVs create a chatty alternate WiFi network How to write capital alpha? How often does castling occur in gra...
            Read more

            Lowndes Grove History Architecture References Navigation menu32°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661132°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661178002500"National Register Information System"Historic houses of South Carolina"Lowndes Grove""+32° 48' 6.00", −79° 57' 58.00""Lowndes Grove, Charleston County (260 St. Margaret St., Charleston)""Lowndes Grove"The Charleston ExpositionIt Happened in South Carolina"Lowndes Grove (House), Saint Margaret Street & Sixth Avenue, Charleston, Charleston County, SC(Photographs)"Plantations of the Carolina Low Countrye

            Image
            Contributing propertyKeeper of the RegisterHistoric districtHistory of the National Register of Historic PlacesNational Park ServiceProperty typesCharlestonColumbiaGreenvilleNorth CharlestonRock Hill Houses on the National Register of Historic Places in South CarolinaGeorgian architecture in South CarolinaHouses completed in 1786Plantation houses in South CarolinaNational Register of Historic Places in Charleston, South CarolinaHouses in Charleston, South Carolina plantation houseAshley RiverCharlestonNational Register of Historic PlacesRevolutionary WarWilliam LowndesU.S. CongressSouth Carolina Inter-State and West Indian ExpositionGeorgianDoricbalustradespedimentoculusentablaturemodilliondentillightship roofterra cottatilescorbelledcornicesRegency stylespiral staircaseskylight Lowndes Grove U.S. National Register of Historic Places Lowndes Grove in 1940 Show map of South Carolina Show map of the United States Location 260 St. Margaret St., Charleston, South Carolina Coordinates 32...
            Read more