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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 $11x^5 + x^4y + xy^5=18$
$begingroup$
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
asked Mar 15 at 16:57
LuminousNutriaLuminousNutria
45512
$endgroup$
|
show 4 more comments
$begingroup$
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
asked Mar 15 at 16:57
LuminousNutriaLuminousNutria
45512
$endgroup$
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
|
show 4 more comments
$begingroup$
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
asked Mar 15 at 16:57
LuminousNutriaLuminousNutria
45512
$endgroup$
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
asked Mar 15 at 16:57
LuminousNutriaLuminousNutria
45512
asked Mar 15 at 16:57
LuminousNutriaLuminousNutria
45512
edited Mar 15 at 17:55
LuminousNutria
asked Mar 15 at 16:57
LuminousNutriaLuminousNutria
45512
asked Mar 15 at 16:57
LuminousNutriaLuminousNutria
45512
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
|
show 4 more comments
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
|
show 4 more comments
3 Answers
3
active
oldest
votes
$begingroup$
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 $$fracxy'-yx^2=frac20-y'20x-y implies y'=frac20x^2+20xy-y^221x^2-xy$$
answered Mar 15 at 17:05
Yadati KiranYadati Kiran
2,1101622
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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:
$$y'=-fracF_xF_y=-frace^y/xcdot (-y/x^2)-2e^y/xcdot (1/x)+1=frac(2x-y)y+2x^2(2x-y)x+x^2=frac2x^2+2xy-y^23x^2-xy.$$
answered Mar 15 at 17:24
farruhotafarruhota
21.5k2842
$endgroup$
$begingroup$
What's wrong with leaving $e^y/x$?
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– Tomislav Ostojich
Mar 15 at 19:26
2
$begingroup$
Nothing, it is simpler, isn't it?
$endgroup$
– farruhota
Mar 15 at 19:43
add a comment |
$begingroup$
$e^y/x cdot fracx-yx^2 = 20 - fracdydx$. 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.
answered Mar 15 at 17:21
Michael RybkinMichael Rybkin
3,954421
$endgroup$
add a comment |
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3 Answers
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3 Answers
3
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
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 $$fracxy'-yx^2=frac20-y'20x-y implies y'=frac20x^2+20xy-y^221x^2-xy$$
answered Mar 15 at 17:05
Yadati KiranYadati Kiran
2,1101622
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $$fracxy'-yx^2=frac20-y'20x-y implies y'=frac20x^2+20xy-y^221x^2-xy$$
answered Mar 15 at 17:05
Yadati KiranYadati Kiran
2,1101622
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $$fracxy'-yx^2=frac20-y'20x-y implies y'=frac20x^2+20xy-y^221x^2-xy$$
answered Mar 15 at 17:05
Yadati KiranYadati Kiran
2,1101622
$endgroup$
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 $$fracxy'-yx^2=frac20-y'20x-y implies y'=frac20x^2+20xy-y^221x^2-xy$$
answered Mar 15 at 17:05
Yadati KiranYadati Kiran
2,1101622
edited Mar 15 at 17:46
answered Mar 15 at 17:05
Yadati KiranYadati Kiran
2,1101622
answered Mar 15 at 17:05
Yadati KiranYadati Kiran
2,1101622
answered Mar 15 at 17:05
Yadati KiranYadati Kiran
2,1101622
2,1101622
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How do you get rid of the second $fracdydx$ or $y'$ in the line $dfracddx(e^y/x)=e^y/xcdot dfracddxleft(dfracyxright)=e^y/xcdotdfracxfracdydx-yfracdxdxx^2=e^y/xcdotleft(dfracxy'-yx^2right)$?
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– LuminousNutria
Mar 15 at 17:33
1
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@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 $dfracddxleft(e^y/x = 20x-yright)$.
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– Yadati Kiran
Mar 15 at 17:38
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Oh, I see! I misread your math, I thought $fracdxdx$ was $fracdydx$
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– LuminousNutria
Mar 15 at 17:45
add a comment |
$begingroup$
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
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Oh, I see! I misread your math, I thought $fracdxdx$ was $fracdydx$
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– 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
add a comment |
$begingroup$
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:
$$y'=-fracF_xF_y=-frace^y/xcdot (-y/x^2)-2e^y/xcdot (1/x)+1=frac(2x-y)y+2x^2(2x-y)x+x^2=frac2x^2+2xy-y^23x^2-xy.$$
answered Mar 15 at 17:24
farruhotafarruhota
21.5k2842
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$begingroup$
What's wrong with leaving $e^y/x$?
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– Tomislav Ostojich
Mar 15 at 19:26
2
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Nothing, it is simpler, isn't it?
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– farruhota
Mar 15 at 19:43
add a comment |
$begingroup$
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:
$$y'=-fracF_xF_y=-frace^y/xcdot (-y/x^2)-2e^y/xcdot (1/x)+1=frac(2x-y)y+2x^2(2x-y)x+x^2=frac2x^2+2xy-y^23x^2-xy.$$
answered Mar 15 at 17:24
farruhotafarruhota
21.5k2842
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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:
$$y'=-fracF_xF_y=-frace^y/xcdot (-y/x^2)-2e^y/xcdot (1/x)+1=frac(2x-y)y+2x^2(2x-y)x+x^2=frac2x^2+2xy-y^23x^2-xy.$$
answered Mar 15 at 17:24
farruhotafarruhota
21.5k2842
$endgroup$
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:
$$y'=-fracF_xF_y=-frace^y/xcdot (-y/x^2)-2e^y/xcdot (1/x)+1=frac(2x-y)y+2x^2(2x-y)x+x^2=frac2x^2+2xy-y^23x^2-xy.$$
answered Mar 15 at 17:24
farruhotafarruhota
21.5k2842
answered Mar 15 at 17:24
farruhotafarruhota
21.5k2842
answered Mar 15 at 17:24
farruhotafarruhota
21.5k2842
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
$e^y/x cdot fracx-yx^2 = 20 - fracdydx$. 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.
answered Mar 15 at 17:21
Michael RybkinMichael Rybkin
3,954421
$endgroup$
add a comment |
$begingroup$
$e^y/x cdot fracx-yx^2 = 20 - fracdydx$. 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.
answered Mar 15 at 17:21
Michael RybkinMichael Rybkin
3,954421
$endgroup$
add a comment |
$begingroup$
$e^y/x cdot fracx-yx^2 = 20 - fracdydx$. 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.
answered Mar 15 at 17:21
Michael RybkinMichael Rybkin
3,954421
$endgroup$
$e^y/x cdot fracx-yx^2 = 20 - fracdydx$. 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.
answered Mar 15 at 17:21
Michael RybkinMichael Rybkin
3,954421
edited Mar 15 at 19:39
answered Mar 15 at 17:21
Michael RybkinMichael Rybkin
3,954421
answered Mar 15 at 17:21
Michael RybkinMichael Rybkin
3,954421
answered Mar 15 at 17:21
Michael RybkinMichael Rybkin
3,954421
3,954421
add a comment |
add a comment |
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$begingroup$
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 $y/x$.
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– TM Gallagher
Mar 15 at 17:02
1
$begingroup$
You make a mistake on d/dx(e^(y/x))
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– DragunityMAX
Mar 15 at 17:03
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@DragunityMAX I'm not sure how to handle derivating $e^x$ when x is not a lone variable.
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– LuminousNutria
Mar 15 at 17:09
1
$begingroup$
@LuminousNutria Exactly what you did. You just make a calculation mistake when applying quotient rule.
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– DragunityMAX
Mar 15 at 17:18
1
$begingroup$
@LuminousNutria Yadati Kiran gives the ans. Note that y is function of x.
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– DragunityMAX
Mar 15 at 17:22