<p>To derive the formula for the elasticity of substition I consider a function with two arguments <span class="math-container">$f(x_1,x_2)$</span>. I then consider a level curve <span class="math-container">$f(x_1,x_2) = c$</span> assumed to define <span class="math-container">$x_2$</span> as a function of <span class="math-container">$x_1$</span>. From this it follows that</p>
<p><span class="math-container">$$\frac{\partial }{\partial x_1} f(x_1,x_2(x_1)) = c \Leftrightarrow \\[8pt]
f_1(x_1,x_2) + f_2(x_1,x_2) \frac{\partial x_2}{\partial x_1} = 0 \Leftrightarrow \\[8pt] \frac{\partial x_2}{\partial x_1} = - \frac{f_1(x_1,x_2) }{f_2(x_1,x_2) }$$</span></p>
<p>giving the slope of the level curve in terms of <span class="math-container">$f_i := \partial f/\partial x_i$</span> for <span class="math-container">$i=1,2$</span>. Given the definition of MRS I conclude that</p>
<p><span class="math-container">$$(1) \ \ MRS(x_1,x_2) := \frac{f_1(x_1,x_2) }{f_2(x_1,x_2) } = -\frac{\partial x_2}{\partial x_1}.$$</span></p>
<p>Then define the ratio function</p>
<p><span class="math-container">$$r(x_1) = x_2(x_1)/x_1$$</span></p>
<p>again using that <span class="math-container">$x_2$</span> is a function of <span class="math-container">$x_1$</span>. This function can be inverted to get <span class="math-container">$x_1 = r^{-1}(r)$</span> and using again that <span class="math-container">$x_2$</span> is a function of <span class="math-container">$x_1$</span> the MRS can be written as</p>
<p><span class="math-container">$$MRS(r) := \frac{f_1(r^{-1}(r),x_2(r^{-1}(r)))}{f_2(r^{-1}(r),x_2(r^{-1}(r)))} $$</span></p>
<p>and differentiated with respect to <span class="math-container">$r$</span> to get</p>
<p><span class="math-container">$$(2)\ \ \frac{\partial MRS(r)}{\partial r} = \left[\frac{\left(f_{11} + f_{12}\frac{\partial x_2}{\partial x_1}\right) f_2 - \left(f_{21} + f_{22}\frac{\partial x_2}{\partial x_1}\right)f_1}{f_2^2} \right]\frac{\partial r^{-1}(r)}{\partial r},$$</span></p>
<p>where the expression in square brackets is <span class="math-container">$\partial MRS(x_1,x_2(x_1))/\partial x_1$</span>.</p>
<p>I then use that <span class="math-container">$r^{-1}(r) = x_1$</span> implying that <span class="math-container">$\partial r^{-1}(r)/\partial r = 1/\partial r(x_1)/\partial x_1$</span> and from the definition of <span class="math-container">$r(x_1)$</span> it follows that</p>
<p><span class="math-container">$$(3) \ \ \frac{\partial r(x_1)}{\partial x_1} = \frac{\frac{\partial x_2}{\partial x_1} x_1 - x_2}{x_1^2} = \frac{}{}-\frac{f_1}{f_2}\frac{1}{x_1} - \frac{x_2}{x^2_1} = -f_1 r\left[ \frac{1}{f_2x_2} + \frac{1}{f_1x_1}\right],$$</span></p>
<p>using that <span class="math-container">$r =x_1/x_1$</span> and that <span class="math-container">$\frac{\partial x_2}{\partial x_1}  = - f_1/f_2$</span>. Inserting the reciprocal value of (3) into (2) it follows that</p>
<p><span class="math-container">$$(4)\ \ \frac{\partial MRS(r)}{\partial r} = -\left[\frac{\left(f_{11} + f_{12}\frac{\partial x_2}{\partial x_1}\right) f_2 - \left(f_{21} + f_{22}\frac{\partial x_2}{\partial x_1}\right)f_1}{f_2^2} \right] \left[ \frac{1}{f_1 r\left[ \frac{1}{f_2x_2} + \frac{1}{f_1x_1}\right]}\right].$$</span></p>
<p>What remains to be done is inserting this expression into the definition of the elasticity of substitution and cleaning it up. The Elasticity of substition is defined as the elasticity of <span class="math-container">$r = x_2/x_1$</span> as a function of <span class="math-container">$MRS$</span>:</p>
<p><span class="math-container">$$(5) \ \  \sigma := \frac{MRS}{r} \frac{\partial r}{\partial MRS} = \frac{1}{\frac{r}{MRS} \frac{\partial MRS(r)}{\partial r}}$$</span></p>
<p>being the reciprocal of the elasticity of MRS with respect to <span class="math-container">$r=x_2/x_1$</span>. Inserting (4) into (5) it therefore follows that</p>
<p><span class="math-container">$$1/\sigma =- \left[\frac{r}{f_1/f_2}\right]\left[\frac{\left(f_{11} + f_{12}\frac{\partial x_2}{\partial x_1}\right) f_2 - \left(f_{21} + f_{22}\frac{\partial x_2}{\partial x_1}\right)f_1}{f_2^2} \right] \left[ \frac{1}{f_1 r\left[ \frac{1}{f_2x_2} + \frac{1}{f_1x_1}\right]}\right],$$</span></p>
<p>which is reduced by removing the factor <span class="math-container">$r$</span> and using again that <span class="math-container">$\partial x_2/\partial x_2 = -f_1/f_2$</span> to get</p>
<p><span class="math-container">$$1/\sigma =- \left[\frac{1}{f_1/f_2}\right]\left[\frac{\left(f_{11} - f_{12}\frac{f_1}{f_2}\right) f_2 - \left(f_{21} - f_{22}\frac{f_1}{f_2}\right)f_1}{f_2^2} \right] \left[ \frac{1}{f_1 \left[ \frac{1}{f_2x_2} + \frac{1}{f_1x_1}\right]}\right] \\[8pt]
= -\left[\frac{\left(\frac{f_{11}f_2}{f_1} - f_{12}\right) f_2 - \left(f_{21} \frac{f_2}{f_1}- f_{22}\right)f_1}{f_2^2} \right] \left[ \frac{1}{f_1 \left[ \frac{1}{f_2x_2} + \frac{1}{f_1x_1}\right]}\right]\\[8pt]
= -\left[\frac{\frac{f_{11}f^2_2}{f_1} - f_{12} f_2 - f_{21} f_2+ f_{22}f_1}{f_2^2} \right] \left[ \frac{1}{f_1 \left[ \frac{1}{f_2x_2} + \frac{1}{f_1x_1}\right]}\right]\\[8pt]
= -\left[\frac{f_{11}}{f^2_1} - \frac{f_{12}}{f_1f_2}  - \frac{f_{21}}{f_1f_2} + \frac{f_{22}}{f_2^2} \right] \left[ \frac{1}{\left[ \frac{1}{f_2x_2} + \frac{1}{f_1x_1}\right]}\right].$$</span></p>
<p>Taking the reciprocal and using that <span class="math-container">$f_{12} = f_{21}$</span> it follows that</p>
<p><span class="math-container">$$ \sigma = - \frac{\frac{1}{f_2x_2} + \frac{1}{f_1x_1}}{\frac{f_{11}}{f^2_1} - \frac{2f_{12}}{f_1f_2}   + \frac{f_{22}}{f_2^2}} ,$$</span></p>
<p>which is the formula found in for example: Daniel McFadden <em>Constant Elasticity of Substitution Production Functions</em>
The Review of Economic Studies , Jun., 1963, Vol. 30, No. 2 (Jun., 1963), pp. 73-
83 <strong>[stated as equation (1)]</strong> or in</p>
<p>Knut Sydsæter, Arne Strøm, Peter Berck <em>Economists’ Mathematical
Manual</em> Fourth Edition  <strong>[stated as formula 5.21]</strong>.</p>
<p>I leave it for you to show that this formula is equivalent to the one stated in the question (I however prefer this one because the way the indeces are written makes it easier to remember).</p>

CC BY-SA 4.0
2021-04-06T17:21:52.833Z
See this post for an example of application of the formula: https://economics.stackexchange.com/questions/43270/how-to-compute-elasticity-of-substituion-in-general
Jesper Hybel
CC BY-SA 4.0
2021-04-06T17:17:13.560Z
2021-04-06T17:17:13.560Z
Jesper Hybel
<p>I would like to derive the elasticity of substitution. I'm aware that such a thread with a very straightforward explanation already exists, but my case is slightly different and I'm not sure how to differentiate the functions involved.</p>

<p>But let me first introduce you to my problem. I have to derive the elasticity of substitution 
$$\sigma = \frac{f_1 f_2(f_1z_1+f_2z_2)}{z_1z_2[2f_{1,2}f_1f_2-f_{11}(f_2)^2-f_{22}(f_1)^2]}$$
where I have the following production function $y^0 - f(z_1,rz_1)=0$ for which $r$ is defined the following: $r=\frac{z_2}{z_1}$.</p>

<p>As I have already mentioned this problem is already perfectly solved in this thread: <a href="https://economics.stackexchange.com/questions/5517/how-does-one-derive-the-elasticity-of-substitution">How does one derive the elasticity of substitution?</a> </p>

<p>But now comes the clue in my specification, where I do not now how to proceed: I have to express $z_1$ as function of $r$, such that $z_1=g(r)$. The marginal rate of substitution can thus be expressed as follows $$MRTS_{21}=\frac{f_1(g(r),rg(r))}{f_2(g(r),rg(r))}$$
How can I derive the differential of $f(g(r),rg(r))$, which is the first step in the derivation of the elasticity of substitution. I somehow should use the theorem of implicit functions, but I'm not sure how..</p>

<p>I hope someone can explain this differential to me!</p>

<p>Cheers</p>

<microeconomics><elasticity-of-substitution>
How does one derive the elasticity of substitution with implicit functions?
Louki
# How does one derive the elasticity of substitution with implicit functions?

(Asked by: username_0 on 2016-11-03)

I would like to derive the elasticity of substitution. I'm aware that such a thread with a very straightforward explanation already exists, but my case is slightly different and I'm not sure how to differentiate the functions involved.

But let me first introduce you to my problem. I have to derive the elasticity of substitution
$$\sigma = \frac{f\_1 f\_2(f\_1z\_1+f\_2z\_2)}{z\_1z\_2[2f\_{1,2}f\_1f\_2-f\_{11}(f\_2)^2-f\_{22}(f\_1)^2]}$$
where I have the following production function $y^0 - f(z\_1,rz\_1)=0$ for which $r$ is defined the following: $r=\frac{z\_2}{z\_1}$.

As I have already mentioned this problem is already perfectly solved in this thread: [How does one derive the elasticity of substitution?](https://economics.stackexchange.com/questions/5517/how-does-one-derive-the-elasticity-of-substitution)

But now comes the clue in my specification, where I do not now how to proceed: I have to express $z\_1$ as function of $r$, such that $z\_1=g(r)$. The marginal rate of substitution can thus be expressed as follows $$MRTS\_{21}=\frac{f\_1(g(r),rg(r))}{f\_2(g(r),rg(r))}$$
How can I derive the differential of $f(g(r),rg(r))$, which is the first step in the derivation of the elasticity of substitution. I somehow should use the theorem of implicit functions, but I'm not sure how..

I hope someone can explain this differential to me!

Cheers

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


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## 1 answers

### Answer by username_1 on 2021-04-06. Score: 4

To derive the formula for the elasticity of substition I consider a function with two arguments $f(x\_1,x\_2)$. I then consider a level curve $f(x\_1,x\_2) = c$ assumed to define $x\_2$ as a function of $x\_1$. From this it follows that

$$\frac{\partial }{\partial x\_1} f(x\_1,x\_2(x\_1)) = c \Leftrightarrow \\[8pt]
f\_1(x\_1,x\_2) + f\_2(x\_1,x\_2) \frac{\partial x\_2}{\partial x\_1} = 0 \Leftrightarrow \\[8pt] \frac{\partial x\_2}{\partial x\_1} = - \frac{f\_1(x\_1,x\_2) }{f\_2(x\_1,x\_2) }$$

giving the slope of the level curve in terms of $f\_i := \partial f/\partial x\_i$ for $i=1,2$. Given the definition of MRS I conclude that

$$(1) \ \ MRS(x\_1,x\_2) := \frac{f\_1(x\_1,x\_2) }{f\_2(x\_1,x\_2) } = -\frac{\partial x\_2}{\partial x\_1}.$$

Then define the ratio function

$$r(x\_1) = x\_2(x\_1)/x\_1$$

again using that $x\_2$ is a function of $x\_1$. This function can be inverted to get $x\_1 = r^{-1}(r)$ and using again that $x\_2$ is a function of $x\_1$ the MRS can be written as

$$MRS(r) := \frac{f\_1(r^{-1}(r),x\_2(r^{-1}(r)))}{f\_2(r^{-1}(r),x\_2(r^{-1}(r)))} $$

and differentiated with respect to $r$ to get

$$(2)\ \ \frac{\partial MRS(r)}{\partial r} = \left[\frac{\left(f\_{11} + f\_{12}\frac{\partial x\_2}{\partial x\_1}\right) f\_2 - \left(f\_{21} + f\_{22}\frac{\partial x\_2}{\partial x\_1}\right)f\_1}{f\_2^2} \right]\frac{\partial r^{-1}(r)}{\partial r},$$

where the expression in square brackets is $\partial MRS(x\_1,x\_2(x\_1))/\partial x\_1$.

I then use that $r^{-1}(r) = x\_1$ implying that $\partial r^{-1}(r)/\partial r = 1/\partial r(x\_1)/\partial x\_1$ and from the definition of $r(x\_1)$ it follows that

$$(3) \ \ \frac{\partial r(x\_1)}{\partial x\_1} = \frac{\frac{\partial x\_2}{\partial x\_1} x\_1 - x\_2}{x\_1^2} = \frac{}{}-\frac{f\_1}{f\_2}\frac{1}{x\_1} - \frac{x\_2}{x^2\_1} = -f\_1 r\left[ \frac{1}{f\_2x\_2} + \frac{1}{f\_1x\_1}\right],$$

using that $r =x\_1/x\_1$ and that $\frac{\partial x\_2}{\partial x\_1} = - f\_1/f\_2$. Inserting the reciprocal value of (3) into (2) it follows that

$$(4)\ \ \frac{\partial MRS(r)}{\partial r} = -\left[\frac{\left(f\_{11} + f\_{12}\frac{\partial x\_2}{\partial x\_1}\right) f\_2 - \left(f\_{21} + f\_{22}\frac{\partial x\_2}{\partial x\_1}\right)f\_1}{f\_2^2} \right] \left[ \frac{1}{f\_1 r\left[ \frac{1}{f\_2x\_2} + \frac{1}{f\_1x\_1}\right]}\right].$$

What remains to be done is inserting this expression into the definition of the elasticity of substitution and cleaning it up. The Elasticity of substition is defined as the elasticity of $r = x\_2/x\_1$ as a function of $MRS$:

$$(5) \ \ \sigma := \frac{MRS}{r} \frac{\partial r}{\partial MRS} = \frac{1}{\frac{r}{MRS} \frac{\partial MRS(r)}{\partial r}}$$

being the reciprocal of the elasticity of MRS with respect to $r=x\_2/x\_1$. Inserting (4) into (5) it therefore follows that

$$1/\sigma =- \left[\frac{r}{f\_1/f\_2}\right]\left[\frac{\left(f\_{11} + f\_{12}\frac{\partial x\_2}{\partial x\_1}\right) f\_2 - \left(f\_{21} + f\_{22}\frac{\partial x\_2}{\partial x\_1}\right)f\_1}{f\_2^2} \right] \left[ \frac{1}{f\_1 r\left[ \frac{1}{f\_2x\_2} + \frac{1}{f\_1x\_1}\right]}\right],$$

which is reduced by removing the factor $r$ and using again that $\partial x\_2/\partial x\_2 = -f\_1/f\_2$ to get

$$1/\sigma =- \left[\frac{1}{f\_1/f\_2}\right]\left[\frac{\left(f\_{11} - f\_{12}\frac{f\_1}{f\_2}\right) f\_2 - \left(f\_{21} - f\_{22}\frac{f\_1}{f\_2}\right)f\_1}{f\_2^2} \right] \left[ \frac{1}{f\_1 \left[ \frac{1}{f\_2x\_2} + \frac{1}{f\_1x\_1}\right]}\right] \\[8pt]
= -\left[\frac{\left(\frac{f\_{11}f\_2}{f\_1} - f\_{12}\right) f\_2 - \left(f\_{21} \frac{f\_2}{f\_1}- f\_{22}\right)f\_1}{f\_2^2} \right] \left[ \frac{1}{f\_1 \left[ \frac{1}{f\_2x\_2} + \frac{1}{f\_1x\_1}\right]}\right]\\[8pt]
= -\left[\frac{\frac{f\_{11}f^2\_2}{f\_1} - f\_{12} f\_2 - f\_{21} f\_2+ f\_{22}f\_1}{f\_2^2} \right] \left[ \frac{1}{f\_1 \left[ \frac{1}{f\_2x\_2} + \frac{1}{f\_1x\_1}\right]}\right]\\[8pt]
= -\left[\frac{f\_{11}}{f^2\_1} - \frac{f\_{12}}{f\_1f\_2} - \frac{f\_{21}}{f\_1f\_2} + \frac{f\_{22}}{f\_2^2} \right] \left[ \frac{1}{\left[ \frac{1}{f\_2x\_2} + \frac{1}{f\_1x\_1}\right]}\right].$$

Taking the reciprocal and using that $f\_{12} = f\_{21}$ it follows that

$$ \sigma = - \frac{\frac{1}{f\_2x\_2} + \frac{1}{f\_1x\_1}}{\frac{f\_{11}}{f^2\_1} - \frac{2f\_{12}}{f\_1f\_2} + \frac{f\_{22}}{f\_2^2}} ,$$

which is the formula found in for example: Daniel McFadden *Constant Elasticity of Substitution Production Functions*
The Review of Economic Studies , Jun., 1963, Vol. 30, No. 2 (Jun., 1963), pp. 73-
83 **[stated as equation (1)]** or in

Knut Sydsæter, Arne Strøm, Peter Berck *Economists’ Mathematical
Manual* Fourth Edition **[stated as formula 5.21]**.

I leave it for you to show that this formula is equivalent to the one stated in the question (I however prefer this one because the way the indeces are written makes it easier to remember).

> **Comments:**
> **username_1**: See this post for an example of application of the formula: https://economics.stackexchange.com/questions/43270/how-to-compute-elasticity-of-substituion-in-general  

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