<p>The formula you already have there is a general formula for elasticity of substitution, but I can see that it might be difficult to apply to your problem here given that <span class="math-container">$MU_{x_2}=1$</span>. There is also another way how formula for elasticity of substitution can be expressed. You can use 'partial derivative formula' (e.g. see Sydsaeter et al. EMEA pp 430) which is the inelegant 'monstrosity' given below. For elasticity of substitution between <span class="math-container">$x_2$</span> and <span class="math-container">$x_1$</span> it will be given by:</p>
<p><span class="math-container">$$\sigma_{x_2x_1} = \frac{−F_1^′F_2^′(x_1 F_1^′ + x_2 F_2^′)}{x_1x_2[(F_2^′)^2F_{11}^{''}
 − 2F_1^′F_2^′F_{12}^{''}+ (F_1^′)^2F_{22}^{''}]}, \text{ for }  F (x_1, x_2) = c$$</span></p>
<p>So in your case:</p>
<p><span class="math-container">$$\sigma_{x_2x_1} = \frac{−x_1 (x_1^2 + x_2)}{x_1x_2[1^2\cdot 1
]}= \frac{-x_1^2-x_2}{x_2}$$</span></p>
<p>More specifically for <span class="math-container">$(x_1, x_2) = (1,1)$</span>: <span class="math-container">$\sigma_{x_2x_1} = -2$</span></p>

CC BY-SA 4.0
2021-03-31T16:45:11.350Z
2021-03-31T17:08:45.780Z
2021-03-31T17:08:45.780Z
1muflon1
<p>We know that the elasticity of substitution is defined as
<span class="math-container">$$
e=\frac{d \ln(x_2/x_1)}{d \ln(MRS_{12})}=\frac{\frac{d(x_2/x_1)}{x_2/x_1}}{\frac{d(MU_1/MU_2)}{MU_1/MU_2}}
$$</span></p>
<p>When we compute ES for CES functions, MRS is function of <span class="math-container">$x_2/x_1$</span>, so we can let <span class="math-container">$z=x_2/x_1$</span> and do the math easily.</p>
<p>But for other utility/production functions, such as quasilinear <span class="math-container">$u(x_1,x_2)=2x_1^{0.5}+x_2$</span>, what am I supposed to do?</p>
<p>I believe ES will be depending on <span class="math-container">$(x_1,x_2)$</span>. So maybe the question can be asked as
&quot;What is ES of <span class="math-container">$u(x_1,x_2)=2x_1^{0.5}+x_2$</span> at <span class="math-container">$(x_1,x_2)=(1,1)$</span>?&quot;</p>

CC BY-SA 4.0
2021-04-06T17:21:13.280Z
See this post for derivation of formula given in answer: https://economics.stackexchange.com/questions/14074/how-does-one-derive-the-elasticity-of-substitution-with-implicit-functions/43355#43355
Jesper Hybel
<microeconomics><elasticity-of-substitution>
How to compute elasticity of substituion in general?
Andy Xu
# How to compute elasticity of substituion in general?

(Asked by: username_0 on 2021-03-31)

We know that the elasticity of substitution is defined as
$$
e=\frac{d \ln(x\_2/x\_1)}{d \ln(MRS\_{12})}=\frac{\frac{d(x\_2/x\_1)}{x\_2/x\_1}}{\frac{d(MU\_1/MU\_2)}{MU\_1/MU\_2}}
$$

When we compute ES for CES functions, MRS is function of $x\_2/x\_1$, so we can let $z=x\_2/x\_1$ and do the math easily.

But for other utility/production functions, such as quasilinear $u(x\_1,x\_2)=2x\_1^{0.5}+x\_2$, what am I supposed to do?

I believe ES will be depending on $(x\_1,x\_2)$. So maybe the question can be asked as
"What is ES of $u(x\_1,x\_2)=2x\_1^{0.5}+x\_2$ at $(x\_1,x\_2)=(1,1)$?"

**Tags:** microeconomics, elasticity-of-substitution

> **Comments:**
> **username_1**: See this post for derivation of formula given in answer: https://economics.stackexchange.com/questions/14074/how-does-one-derive-the-elasticity-of-substitution-with-implicit-functions/43355#43355  

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## 1 answers

### Answer by username_2 on 2021-03-31. Score: 1

The formula you already have there is a general formula for elasticity of substitution, but I can see that it might be difficult to apply to your problem here given that $MU\_{x\_2}=1$. There is also another way how formula for elasticity of substitution can be expressed. You can use 'partial derivative formula' (e.g. see Sydsaeter et al. EMEA pp 430) which is the inelegant 'monstrosity' given below. For elasticity of substitution between $x\_2$ and $x\_1$ it will be given by:

$$\sigma\_{x\_2x\_1} = \frac{−F\_1^′F\_2^′(x\_1 F\_1^′ + x\_2 F\_2^′)}{x\_1x\_2[(F\_2^′)^2F\_{11}^{''}
− 2F\_1^′F\_2^′F\_{12}^{''}+ (F\_1^′)^2F\_{22}^{''}]}, \text{ for } F (x\_1, x\_2) = c$$

So in your case:

$$\sigma\_{x\_2x\_1} = \frac{−x\_1 (x\_1^2 + x\_2)}{x\_1x\_2[1^2\cdot 1
]}= \frac{-x\_1^2-x\_2}{x\_2}$$

More specifically for $(x\_1, x\_2) = (1,1)$: $\sigma\_{x\_2x\_1} = -2$


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