<p><span class="math-container">$$y'\frac{x}{y}= -\frac{f_1'(x,y)}{f_2'(x,y)}\frac{x}{y} = - \frac{y^2\exp(x+1/y)}{2y\exp(x+1/y)+y^2\exp(x+1/y)(-y^{-2})}\,\frac{x}{y}=\\=\frac{y^2}{1-2y}\,\frac{x}{y}=\frac{xy}{1-2y}.$$</span></p>

CC BY-SA 4.0
2021-05-07T10:39:03.483Z
thanks, I made mistake when I was calculating $d/dy$ of $e^{1/y}$ but I see that now thanks to you
WilliamT
CC BY-SA 4.0
2021-05-06T14:03:51.317Z
2021-05-06T14:03:51.317Z
VARulle
<p>I want to calculate elasticity of y with respect to x  for <span class="math-container">$$y^2e^{x+\frac{1}{y}}=3$$</span></p>
<p>My attempt:</p>
<p>I calculate <span class="math-container">$y'$</span> using:</p>
<p><span class="math-container">$$y'= -\frac{f_1'(x,y)}{f_2'(x,y)} = - \frac{y}{2-y}$$</span></p>
<p>I calculated this using implicit differentiation that can be used where <span class="math-container">$f(x,y)=c$</span>, in my case <span class="math-container">$f(x,y)= y^2e^{x+\frac{1}{y}}$</span> and <span class="math-container">$c=3$</span>.</p>
<p>Then elasticity should be just</p>
<p><span class="math-container">$$y'\frac{x}{y} = -\frac{x}{2-y}$$</span></p>
<p>But I was told by professor that is a wrong answer  and that correct answer is: <span class="math-container">$\frac{xy}{1-2y}$</span>.</p>
<p>I don't understand where I have made a mistake and how come its  <span class="math-container">$\frac{xy}{1-2y}$</span>. Please can someone help me to understand it?</p>

CC BY-SA 4.0
2021-05-06T11:49:21.633Z
What do you mean by elasticity when you are talking about an equation? Functions have elasticities at points, but you are defining an entire curve. Also, can you elaborate on how you arrive at the values for $f_1'$ and $f_2'$?
Giskard
CC BY-SA 4.0
2021-05-06T12:43:30.163Z
@Giskard hi, I added explanation. Also, what do you mean by points? In class we are often asked to find elasticity where result is equation like find elasticity of Q with respect to p $Q= bp+a$ is $\epsilon = (bp)/(bp+a)$
WilliamT
CC BY-SA 4.0
2021-05-06T12:51:12.217Z
@Giskard I mean I know you can calculate elasticity at some point. So in last comment if b=1 and a=1 and p=2 elasticity will be 2, but isn't the function that tells us elasticity at any p elasticity?
WilliamT
<mathematical-economics><elasticity>
Calculating elasticity for $y^2e^{x+\frac{1}{y}}=3$
WilliamT
# Calculating elasticity for $y^2e^{x+\frac{1}{y}}=3$

(Asked by: username_0 on 2021-05-06)

I want to calculate elasticity of y with respect to x for $$y^2e^{x+\frac{1}{y}}=3$$

My attempt:

I calculate $y'$ using:

$$y'= -\frac{f\_1'(x,y)}{f\_2'(x,y)} = - \frac{y}{2-y}$$

I calculated this using implicit differentiation that can be used where $f(x,y)=c$, in my case $f(x,y)= y^2e^{x+\frac{1}{y}}$ and $c=3$.

Then elasticity should be just

$$y'\frac{x}{y} = -\frac{x}{2-y}$$

But I was told by professor that is a wrong answer and that correct answer is: $\frac{xy}{1-2y}$.

I don't understand where I have made a mistake and how come its $\frac{xy}{1-2y}$. Please can someone help me to understand it?

**Tags:** mathematical-economics, elasticity
> **Comments:**
> **username_1**: What do you mean by elasticity when you are talking about an equation? Functions have elasticities at points, but you are defining an entire curve. Also, can you elaborate on how you arrive at the values for $f_1'$ and $f_2'$?  
> **username_0**: @Giskard hi, I added explanation. Also, what do you mean by points? In class we are often asked to find elasticity where result is equation like find elasticity of Q with respect to p $Q= bp+a$ is $\epsilon = (bp)/(bp+a)$  
> **username_0**: @Giskard I mean I know you can calculate elasticity at some point. So in last comment if b=1 and a=1 and p=2 elasticity will be 2, but isn't the function that tells us elasticity at any p elasticity?  

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## 1 answers

### Answer by username_2 on 2021-05-06. Score: 5 | Accepted answer

$$y'\frac{x}{y}= -\frac{f\_1'(x,y)}{f\_2'(x,y)}\frac{x}{y} = - \frac{y^2\exp(x+1/y)}{2y\exp(x+1/y)+y^2\exp(x+1/y)(-y^{-2})}\,\frac{x}{y}=\\=\frac{y^2}{1-2y}\,\frac{x}{y}=\frac{xy}{1-2y}.$$

> **Comments:**
> **username_0**: thanks, I made mistake when I was calculating $d/dy$ of $e^{1/y}$ but I see that now thanks to you  

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