<p>For a function <span class="math-container">$f$</span> of <span class="math-container">$2$</span> variables <span class="math-container">$x,y$</span> its first order differential is</p>
<p><span class="math-container">$df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy$</span></p>
<p>Note this can be seen as a linear operator <span class="math-container">$d$</span> applied to the function <span class="math-container">$f$</span></p>
<p><span class="math-container">$d(f) =(\frac{\partial}{\partial x} dx + \frac{\partial}{\partial y} dy) f$</span></p>
<p>Here the operator <span class="math-container">$d$</span> is</p>
<p><span class="math-container">$d = \frac{\partial}{\partial x} dx + \frac{\partial}{\partial y} dy$</span></p>
<p><span class="math-container">$d^2 = (\frac{\partial}{\partial x} dx + \frac{\partial}{\partial y} dy)^2$</span></p>
<p><span class="math-container">$ d^2 = (\frac{\partial}{\partial x} dx + \frac{\partial}{\partial y} dy)(\frac{\partial}{\partial x} dx + \frac{\partial}{\partial y} dy)$</span></p>
<p><span class="math-container">$ d^2 = \frac{\partial^2}{\partial x^2} dx^2 + \frac{\partial^2}{\partial y^2} dy^2 + 2 \frac{\partial^2}{\partial x \partial y} dx dy$</span></p>
<p>Therefore, the second order differential of <span class="math-container">$f$</span> is:</p>
<p><span class="math-container">$d^2 f = \frac{\partial^2 f}{\partial x^2} dx^2 + \frac{\partial^2 f}{\partial y^2} dy^2 + 2 \frac{\partial^2 f}{\partial x \partial y} dx dy$</span></p>
<p>Now for <span class="math-container">$3$</span> variables <span class="math-container">$x,y,z$</span></p>
<p><span class="math-container">$df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz$</span></p>
<p>Note this can be seen as a linear operator <span class="math-container">$d$</span> applied to the function <span class="math-container">$f$</span></p>
<p><span class="math-container">$d(f) =(\frac{\partial}{\partial x} dx + \frac{\partial}{\partial y} dy + \frac{\partial}{\partial z} dz) f$</span></p>
<p>Here the operator <span class="math-container">$d$</span> is</p>
<p><span class="math-container">$d = \frac{\partial}{\partial x} dx + \frac{\partial}{\partial y} dy + \frac{\partial}{\partial z} dz$</span></p>
<p><span class="math-container">$d^2 = (\frac{\partial}{\partial x} dx + \frac{\partial}{\partial y} dy + \frac{\partial}{\partial z} dz)^2$</span></p>
<p><span class="math-container">$ d^2 = (\frac{\partial}{\partial x} dx + \frac{\partial}{\partial y} dy  + \frac{\partial}{\partial z} dz)^2)(\frac{\partial}{\partial x} dx + \frac{\partial}{\partial y} dy  + \frac{\partial}{\partial z} dz)^2)$</span></p>
<p><span class="math-container">$ d^2 = \frac{\partial^2}{\partial x^2} dx^2 + \frac{\partial^2}{\partial y^2} dy^2 + \frac{\partial f^2}{\partial z^2} dz^2 + 2 \frac{\partial^2}{\partial x \partial y} dx dy + 2 \frac{\partial^2}{\partial x \partial z} dx dz + 2 \frac{\partial^2}{\partial y \partial z} dy dz$</span></p>
<p>Therefore, the second order differential of <span class="math-container">$f$</span> is:</p>
<p><span class="math-container">$d^2 f = \frac{\partial^2 f}{\partial x^2} dx^2 + \frac{\partial^2 f}{\partial y^2} dy^2 + \frac{\partial f^2}{\partial z^2} dz^2 + 2 \frac{\partial^2 f}{\partial x \partial y} dx dy + 2 \frac{\partial^2 f}{\partial x \partial z} dx dz + 2 \frac{\partial^2 f}{\partial y \partial z} dy dz$</span></p>
<p>Similarly, for <span class="math-container">$n$</span> variables <span class="math-container">$x_1,\dots,x_n$</span> it’s</p>
<p><span class="math-container">$d^2 f = \sum_{i = 1}^{n} \frac{\partial^2 f}{\partial {x_i}^2} d{x_i}^2 + 2 \sum_{i = 1}^{n} \sum_{j = 1, j &gt; i}^{n} \frac{\partial^2 f}{\partial x_i \partial x_j} dx_i dx_j$</span></p>

CC BY-SA 4.0
2023-04-16T14:10:27.163Z
Thanks, but I also saw a formula for the second order total differential which only multiplied by 2 on one of the last three terms.
sffffii
CC BY-SA 4.0
2023-04-15T21:02:42.003Z
2023-04-15T21:44:20.033Z
2023-04-15T21:44:20.033Z
Nicolas Torres
<p>As the title says, how would you find a second order total differential for a three variable function. For instance:
<a href="https://i.sstatic.net/kw5p1.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/kw5p1.png" alt="enter image description here" /></a></p>

<mathematical-economics>
Second order total differential for three variables
sffffii
# Second order total differential for three variables

(Asked by: username_0 on 2023-04-15)

As the title says, how would you find a second order total differential for a three variable function. For instance:
[![enter image description here](https://i.sstatic.net/kw5p1.png)](https://i.sstatic.net/kw5p1.png)

**Tags:** mathematical-economics


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## 1 answers

### Answer by username_1 on 2023-04-15. Score: 1

For a function $f$ of $2$ variables $x,y$ its first order differential is

$df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy$

Note this can be seen as a linear operator $d$ applied to the function $f$

$d(f) =(\frac{\partial}{\partial x} dx + \frac{\partial}{\partial y} dy) f$

Here the operator $d$ is

$d = \frac{\partial}{\partial x} dx + \frac{\partial}{\partial y} dy$

$d^2 = (\frac{\partial}{\partial x} dx + \frac{\partial}{\partial y} dy)^2$

$ d^2 = (\frac{\partial}{\partial x} dx + \frac{\partial}{\partial y} dy)(\frac{\partial}{\partial x} dx + \frac{\partial}{\partial y} dy)$

$ d^2 = \frac{\partial^2}{\partial x^2} dx^2 + \frac{\partial^2}{\partial y^2} dy^2 + 2 \frac{\partial^2}{\partial x \partial y} dx dy$

Therefore, the second order differential of $f$ is:

$d^2 f = \frac{\partial^2 f}{\partial x^2} dx^2 + \frac{\partial^2 f}{\partial y^2} dy^2 + 2 \frac{\partial^2 f}{\partial x \partial y} dx dy$

Now for $3$ variables $x,y,z$

$df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz$

Note this can be seen as a linear operator $d$ applied to the function $f$

$d(f) =(\frac{\partial}{\partial x} dx + \frac{\partial}{\partial y} dy + \frac{\partial}{\partial z} dz) f$

Here the operator $d$ is

$d = \frac{\partial}{\partial x} dx + \frac{\partial}{\partial y} dy + \frac{\partial}{\partial z} dz$

$d^2 = (\frac{\partial}{\partial x} dx + \frac{\partial}{\partial y} dy + \frac{\partial}{\partial z} dz)^2$

$ d^2 = (\frac{\partial}{\partial x} dx + \frac{\partial}{\partial y} dy + \frac{\partial}{\partial z} dz)^2)(\frac{\partial}{\partial x} dx + \frac{\partial}{\partial y} dy + \frac{\partial}{\partial z} dz)^2)$

$ d^2 = \frac{\partial^2}{\partial x^2} dx^2 + \frac{\partial^2}{\partial y^2} dy^2 + \frac{\partial f^2}{\partial z^2} dz^2 + 2 \frac{\partial^2}{\partial x \partial y} dx dy + 2 \frac{\partial^2}{\partial x \partial z} dx dz + 2 \frac{\partial^2}{\partial y \partial z} dy dz$

Therefore, the second order differential of $f$ is:

$d^2 f = \frac{\partial^2 f}{\partial x^2} dx^2 + \frac{\partial^2 f}{\partial y^2} dy^2 + \frac{\partial f^2}{\partial z^2} dz^2 + 2 \frac{\partial^2 f}{\partial x \partial y} dx dy + 2 \frac{\partial^2 f}{\partial x \partial z} dx dz + 2 \frac{\partial^2 f}{\partial y \partial z} dy dz$

Similarly, for $n$ variables $x\_1,\dots,x\_n$ it’s

$d^2 f = \sum\_{i = 1}^{n} \frac{\partial^2 f}{\partial {x\_i}^2} d{x\_i}^2 + 2 \sum\_{i = 1}^{n} \sum\_{j = 1, j > i}^{n} \frac{\partial^2 f}{\partial x\_i \partial x\_j} dx\_i dx\_j$

> **Comments:**
> **username_0**: Thanks, but I also saw a formula for the second order total differential which only multiplied by 2 on one of the last three terms.  

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