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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

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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











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    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



























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