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?

  • 1
    $\begingroup$ 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 $f_2'$? $\endgroup$
    – Giskard
    May 6, 2021 at 11:49
  • $\begingroup$ @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 $\epsilon = (bp)/(bp+a)$ $\endgroup$
    – WilliamT
    May 6, 2021 at 12:43
  • $\begingroup$ @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? $\endgroup$
    – WilliamT
    May 6, 2021 at 12:51

1 Answer 1


$$y'\frac{x}{y}= -\frac{f_1'(x,y)}{f_2'(x,y)}\frac{x}{y} = - \frac{y^2\exp(x+1/y)}{2y\exp(x+1/y)+y^2\exp(x+1/y)(-y^{-2})}\,\frac{x}{y}=\\=\frac{y^2}{1-2y}\,\frac{x}{y}=\frac{xy}{1-2y}.$$

  • $\begingroup$ thanks, I made mistake when I was calculating $d/dy$ of $e^{1/y}$ but I see that now thanks to you $\endgroup$
    – WilliamT
    May 7, 2021 at 10:39

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