<p>You wrote:</p>
<blockquote>
<p>Unless our Hicksian demand was linear, <strong>it would be inappropriate to
actually calculate the SE for large changes in P using:</strong></p>
<p><span class="math-container">$\frac{\partial h}{\partial p_{x}}∆P_x$</span></p>
</blockquote>
<p>You are right, a differential of a function of one variable <span class="math-container">$f(x)$</span> is a good approximation of the increase of the function <strong>only for small changes of <span class="math-container">$x$</span>.</strong></p>
<p>To summarize the reason why this approximation is not good for large changes of <span class="math-container">$x$</span>, I synthesize below, in the picture, the geometric meaning of the differential of a function <span class="math-container">$y=f(x)$</span> of <strong>one</strong> single variable.</p>
<p><a href="https://i.sstatic.net/o0aeJ.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/o0aeJ.png" alt="enter image description here" /></a></p>
<p>As I think you already know, the derivative of a function <span class="math-container">$f: \mathbb {R} \rightarrow \mathbb {R}$</span> in a point <span class="math-container">$x_0$</span> represents geometrically <strong>the slope of the tangent line</strong> to the function at <span class="math-container">$x_0$</span> (the trigonometric tangent of the angle <span class="math-container">$\alpha$</span> in the picture above).</p>
<p>The <strong>differential <span class="math-container">$dy$</span> of <span class="math-container">$f(x)$</span></strong> in <span class="math-container">$x_0$</span> is, by definition:</p>
<p><span class="math-container">$$dy= f'(x_0) \Delta x \;\;\;\;\;\;\;(1)$$</span></p>
<p>where <span class="math-container">$f'(x_0)$</span> is the derivative of <span class="math-container">$f$</span> at <span class="math-container">$x_0$</span>.</p>
<p>When <span class="math-container">$x$</span> increase from <span class="math-container">$x_0$</span> to <span class="math-container">$x_0+\Delta x$</span>, the function increases from <span class="math-container">$f(x_0)$</span> to <span class="math-container">$f(x_0+\Delta x)$</span>, and its increase is <span class="math-container">$\Delta Y$</span>, which in the picture is equal to the segment <span class="math-container">$\bar {BS}$</span>.</p>
<p>The differential <span class="math-container">$dy$</span> in the picture is equal (according to <span class="math-container">$(1)$</span>) to the segment <span class="math-container">$TS$</span>, that is the increase of the <strong>ordinate of the tangent line</strong> following the increase <span class="math-container">$\Delta x$</span>.</p>
<p>The segment <span class="math-container">$\bar {BT}= \bar {BS} -\bar {TS}$</span> represents the <strong>'error'</strong> we make when approximating the increase of the function <span class="math-container">$\Delta Y$</span> with the differential <span class="math-container">$dy$</span>.</p>
<p>As can be seen in the picture, this error becomes larger and larger as <span class="math-container">$\Delta x$</span> becomes  larger, and smaller as <span class="math-container">$\Delta x$</span> becomes smaller, and tends to <span class="math-container">$0$</span> as <span class="math-container">$\Delta x \rightarrow 0$</span>.</p>
<p>Of course this is a geometric, informal argument, but what we said can be showed formally and rigorously.</p>
<p><span class="math-container">$$\;$$</span></p>
<p><strong>OBSERVATION.</strong> Just an observation to avoid a possible confusion.</p>
<p>In your text <strong>we have a partial derivative</strong>.
But the argument above is completely correct, even if we have a partial derivative.</p>
<p><span class="math-container">$\frac{\partial h}{\partial p_{x}}∆P_x$</span> is not the differential of the function of several variables <span class="math-container">$ h(P_x, P_y, U)$</span>, of course. But <span class="math-container">$\frac{\partial h}{\partial p_{x}}∆P_x$</span> <strong>is nevertheless a differential</strong>, the differential of the function of <strong>one</strong> variable <span class="math-container">$h(P_x, \bar P_y, \bar U)$</span>, where <span class="math-container">$\bar P_y$</span> and <span class="math-container">$\bar U$</span> are fixed value.</p>
<p>A partial derivative is, actually, <strong>a derivative of a function of one variable,</strong> by definition, because it is the derivative of the function which is obtained keeping fixed all the variables, except the variable with respect to we are taking the derivative.</p>
<p>In our case, the function is <span class="math-container">$h(P_x, \bar P_y, \bar U)$</span>.
As <span class="math-container">$\frac{\partial h}{\partial p_{x}}∆P_x$</span>  is the differential of the function  of <strong>one</strong> variable <span class="math-container">$h(P_x, \bar P_y, \bar U)$</span>, what we said above  about the differential of a function of one variable applies.</p>

CC BY-SA 4.0
2023-02-10T13:23:59.690Z
This seems like an exceptional answer, I really appreciate you taking the time to clarify the matter for me. I just had one question about your note at the end which I will add in a separate comment:
CormJack
CC BY-SA 4.0
2023-02-10T13:29:39.343Z
1) You are correct, that if I was calculating/graphing the Hicksian demand curve for $x$, then I should technically have $h(P_x, \bar P_y, \bar U)$, i.e. clearly show $p_x$ and $U$ are fixed. Should I do that, is my notation technically wrong?
CormJack
CC BY-SA 4.0
2023-02-10T13:32:35.857Z
2) In the observation, were you trying to say that given these other variables are technically fixed, that I should use the notation $\frac{dh}{dp_x}$ because technically the function  $h(P_x, \bar P_y, \bar U)$ is just of one variable. Or were you saying that whenever we take a partial derivative, e.g. $\frac{\partial h}{\partial p_{x}}∆P_x$ built into the mechanics is the assumption that the over variables of $h$ are temporarily fixed (in order for us to just take the rate of change in one direction). I was slightly confused so I'd really appreciate the additional clarification!
CormJack
CC BY-SA 4.0
2023-02-10T13:47:00.447Z
Edit: I didn't mean to include $∆P_x$ above with the partial derivative. I meant just to give an example of the partial derivative not the differential....But actually now I'm here what would that be, a "partial differential?" or given the other variables are fixed for the hicksian demand, would that just be the "differential" for $h$.
CormJack
CC BY-SA 4.0
2023-02-10T13:49:16.867Z
Question 1). I hope I understand you exactly, you want to know if you should indicate that a variable is fixed when you take a partial derivative. No, it is implicit, for the very definition of partial derivative. And, in general, we mustn't indicate in notation a fixed variable every time we use it, if it has been said and it is clear from the context. To indicate it in notation every time will be heavy unnecessary notation. So, I think that your notation is correct. As for question 2) I'll read it and I answer you later, in few time. Thank you for your kindness.
BakerStreet
CC BY-SA 4.0
2023-02-10T14:03:38.040Z
Amazing, I appreciate it!
CormJack
CC BY-SA 4.0
2023-02-10T17:03:04.497Z
Question 2). When you have a partial derivative it is implicit that the variables are fixed, except the variable with respect to you are differentiating. It is correct to use the symbol $\partial$ not $df/dx$. A partial derivative is defined for a function of **several variables**, say $f(x,y,z)$.  In this case the partial derivative of $f$ with respect, say, $x$ is defined as the derivative of $f$  when $x$ changes and the other variables are fixed. Whenever we have a function of several variables we must use the symbol $\partial$.
Your notations are correct.
BakerStreet
CC BY-SA 4.0
2023-02-10T17:13:18.667Z
As for the *differential* for functions of several variable, this is a more complicated matter, it is not as in the case of function of one single variable. If you haven't already studied  functions of several variable, at the moment don't worry, don't mind the differential. When you will study functions of several variables  the matter  will be clear, but it is not a simple matter, all becomes more complicated.  Think, for example, that for function of several variables there is an important difference between differentiation  (taking the derivatives) and the differential.
BakerStreet
CC BY-SA 4.0
2023-02-10T17:15:32.820Z
In short, at the moment don't worry, wait to study functions of several variables. These should be studied on a good book of mathematical analysis, or of more advanced calculus.
BakerStreet
CC BY-SA 4.0
2023-02-27T15:50:41.200Z
Okay got it! Can you recommend me a book you know would have a good explanation of the differential in it? I'm pretty sure I should understand this, because I have studied advanced calculus and linear algebra and some analysis...But my mathematical education has been quite non standard, so I have some random wholes...as perhaps you might have gathered from my questions.
CormJack
CC BY-SA 4.0
2023-02-27T16:16:16.877Z
What do you refer to as 'diffrential'? Diffrential for functions of several variables or differentiation of functions one single variable? They are quite different matters.
BakerStreet
CC BY-SA 4.0
2023-02-27T17:05:23.883Z
This video (linked below) covers both single variable and multivariable differential as I understand it. And regarding the multivariable differential if we had something like $f(x(t), y(t))$ then I always thought the differential would be $\frac{\partial{f}}{\partial{x}}\frac{\partial{x}}{\partial{t}}∆t +\frac{\partial{f}}{\partial{y}}\frac{\partial{y}}{\partial{t}}∆t$  or perhaps it should be written as $\frac{\partial{f}}{\partial{x}}\frac{dx}{dt}∆t +\frac{\partial{f}}{\partial{y}}\frac{dx}{dt}∆t$ given that $x$ and $y$ are functions of single variables.
CormJack
CC BY-SA 4.0
2023-02-27T17:07:00.420Z
So yes that covers my understanding of the differential, same as outlined in this video. However the concepts don't seem wildly different to me in the multivariable case compared to the single variable case, so perhaps I'm missing something, or a lot? https://youtu.be/Z0BOldpndXA?t=207
CormJack
CC BY-SA 4.0
2023-02-27T17:29:52.683Z
The function you have mentioned is actually a function of one  single variable, $t$. Yes, the differential for function of several variables is different, even if the it takes a form similar to the differential of one variables, because it is written using derivatives. Aniway, it seemed to me that you wrote that $\Delta y$ is the diffrential of a function of several variables, it is not. One needs to study separately functions of one variables, and then functions of several variables, where the differential is defined.
BakerStreet
CC BY-SA 4.0
2023-02-27T17:43:04.773Z
I don't know many books of calculus in English, as for first/second year analysis there are many very good books in my language, and we use mainly these. Well-known books in English are Rudin, *Principles of Analysis*,  and Apostol, I know them, but I don't recommend them as first calculus books. A well- known book is Spivak, *Calculus*, for function of one variable, you can see it. Or you can see Courant, which covers both one variable and several variables, it is well-known but I haven't seen it https://www.amazon.it/gp/product/354065058X/ref=ox_sc_act_image_11?smid=A11IL2PNWYJU7H&psc=1
BakerStreet
CC BY-SA 4.0
2023-02-27T17:58:30.287Z
A good idea would be also, not for theory (without proofs) but for exercises, a lot of applications and solved problems, and a concise exposition of concepts, Schaum's Advanced Calculus, (Wrede and Spiegel authors), both for one and for several variables, very useful. –These are books of mathematical snalysis, not Mathematics for economist, which is a different matter. But I think that in these books you can find what are you asking, it books of mathematics for economist  you may not find a comprehensive treatment no, often,  the diffential for function of several variables.
BakerStreet
CC BY-SA 4.0
2023-02-27T22:27:11.723Z
Aha amazing thank you, I attached an image of the book you recommend, and responded to some other stuff in our chat thread!
CormJack
CC BY-SA 4.0
2023-02-09T14:26:28.823Z
2023-02-09T15:11:22.107Z
2023-02-09T15:11:22.107Z
BakerStreet
<p>Substitution Effect (SE) for a price increase of <span class="math-container">$P_x$</span> to <span class="math-container">$P_x'$</span> can be written as:</p>
<p><span class="math-container">$h(P_x', P_y, U) - h(P_x, P_y, U) = ∆h$</span>, where <span class="math-container">$h$</span> is Hicksian demand. Correct?</p>
<p>The Slutsky equations decomposes a change in Marshallian Demand into the SE and Income Effect (IE), where the SE is written as: <span class="math-container">$\frac{\partial h}{\partial p_{x}}$</span></p>
<p><strong>Question:</strong></p>
<p>For small <span class="math-container">$∆P_x$</span> is this correct: <span class="math-container">$∆h = h(P_x', P_y, U) - h(P_x, P_y, U) \approx \frac{\partial h}{\partial p_{x}}∆P_x$</span></p>
<p>I.e. Unless our Hicksian demand was linear, <strong>it would be inappropriate to actually calculate the SE for large changes in P using:</strong></p>
<p><span class="math-container">$\frac{\partial h}{\partial p_{x}}∆P_x$</span></p>
<p>This would only give us a decent linear approximation of the value of the SE for small changes in <span class="math-container">$P_x$</span>. Am i correct, or am i getting my calculus confused? Thanks!</p>

<utility><demand><welfare-economics>
Calculating the substitution effect with the derivative
CormJack
# Calculating the substitution effect with the derivative

(Asked by: username_0 on 2023-02-08)

Substitution Effect (SE) for a price increase of $P\_x$ to $P\_x'$ can be written as:

$h(P\_x', P\_y, U) - h(P\_x, P\_y, U) = ∆h$, where $h$ is Hicksian demand. Correct?

The Slutsky equations decomposes a change in Marshallian Demand into the SE and Income Effect (IE), where the SE is written as: $\frac{\partial h}{\partial p\_{x}}$

**Question:**

For small $∆P\_x$ is this correct: $∆h = h(P\_x', P\_y, U) - h(P\_x, P\_y, U) \approx \frac{\partial h}{\partial p\_{x}}∆P\_x$

I.e. Unless our Hicksian demand was linear, **it would be inappropriate to actually calculate the SE for large changes in P using:**

$\frac{\partial h}{\partial p\_{x}}∆P\_x$

This would only give us a decent linear approximation of the value of the SE for small changes in $P\_x$. Am i correct, or am i getting my calculus confused? Thanks!

**Tags:** utility, demand, welfare-economics

---

## 1 answers

### Answer by username_1 on 2023-02-09. Score: 4 | Accepted answer

You wrote:

> Unless our Hicksian demand was linear, **it would be inappropriate to
> actually calculate the SE for large changes in P using:**
>
> $\frac{\partial h}{\partial p\_{x}}∆P\_x$

You are right, a differential of a function of one variable $f(x)$ is a good approximation of the increase of the function **only for small changes of $x$.**

To summarize the reason why this approximation is not good for large changes of $x$, I synthesize below, in the picture, the geometric meaning of the differential of a function $y=f(x)$ of **one** single variable.

[![enter image description here](https://i.sstatic.net/o0aeJ.png)](https://i.sstatic.net/o0aeJ.png)

As I think you already know, the derivative of a function $f: \mathbb {R} \rightarrow \mathbb {R}$ in a point $x\_0$ represents geometrically **the slope of the tangent line** to the function at $x\_0$ (the trigonometric tangent of the angle $\alpha$ in the picture above).

The **differential $dy$ of $f(x)$** in $x\_0$ is, by definition:

$$dy= f'(x\_0) \Delta x \;\;\;\;\;\;\;(1)$$

where $f'(x\_0)$ is the derivative of $f$ at $x\_0$.

When $x$ increase from $x\_0$ to $x\_0+\Delta x$, the function increases from $f(x\_0)$ to $f(x\_0+\Delta x)$, and its increase is $\Delta Y$, which in the picture is equal to the segment $\bar {BS}$.

The differential $dy$ in the picture is equal (according to $(1)$) to the segment $TS$, that is the increase of the **ordinate of the tangent line** following the increase $\Delta x$.

The segment $\bar {BT}= \bar {BS} -\bar {TS}$ represents the **'error'** we make when approximating the increase of the function $\Delta Y$ with the differential $dy$.

As can be seen in the picture, this error becomes larger and larger as $\Delta x$ becomes larger, and smaller as $\Delta x$ becomes smaller, and tends to $0$ as $\Delta x \rightarrow 0$.

Of course this is a geometric, informal argument, but what we said can be showed formally and rigorously.

$$\;$$

**OBSERVATION.** Just an observation to avoid a possible confusion.

In your text **we have a partial derivative**.
But the argument above is completely correct, even if we have a partial derivative.

$\frac{\partial h}{\partial p\_{x}}∆P\_x$ is not the differential of the function of several variables $ h(P\_x, P\_y, U)$, of course. But $\frac{\partial h}{\partial p\_{x}}∆P\_x$ **is nevertheless a differential**, the differential of the function of **one** variable $h(P\_x, \bar P\_y, \bar U)$, where $\bar P\_y$ and $\bar U$ are fixed value.

A partial derivative is, actually, **a derivative of a function of one variable,** by definition, because it is the derivative of the function which is obtained keeping fixed all the variables, except the variable with respect to we are taking the derivative.

In our case, the function is $h(P\_x, \bar P\_y, \bar U)$.
As $\frac{\partial h}{\partial p\_{x}}∆P\_x$ is the differential of the function of **one** variable $h(P\_x, \bar P\_y, \bar U)$, what we said above about the differential of a function of one variable applies.

> **Comments:**
> **username_0**: This seems like an exceptional answer, I really appreciate you taking the time to clarify the matter for me. I just had one question about your note at the end which I will add in a separate comment:  
> **username_0**: 1) You are correct, that if I was calculating/graphing the Hicksian demand curve for $x$, then I should technically have $h(P_x, \bar P_y, \bar U)$, i.e. clearly show $p_x$ and $U$ are fixed. Should I do that, is my notation technically wrong?  
> **username_0**: 2) In the observation, were you trying to say that given these other variables are technically fixed, that I should use the notation $\frac{dh}{dp_x}$ because technically the function  $h(P_x, \bar P_y, \bar U)$ is just of one variable. Or were you saying that whenever we take a partial derivative, e.g. $\frac{\partial h}{\partial p_{x}}∆P_x$ built into the mechanics is the assumption that the over variables of $h$ are temporarily fixed (in order for us to just take the rate of change in one direction). I was slightly confused so I'd really appreciate the additional clarification!  
> **username_0**: Edit: I didn't mean to include $∆P_x$ above with the partial derivative. I meant just to give an example of the partial derivative not the differential....But actually now I'm here what would that be, a "partial differential?" or given the other variables are fixed for the hicksian demand, would that just be the "differential" for $h$.  
> **username_1**: Question 1). I hope I understand you exactly, you want to know if you should indicate that a variable is fixed when you take a partial derivative. No, it is implicit, for the very definition of partial derivative. And, in general, we mustn't indicate in notation a fixed variable every time we use it, if it has been said and it is clear from the context. To indicate it in notation every time will be heavy unnecessary notation. So, I think that your notation is correct. As for question 2) I'll read it and I answer you later, in few time. Thank you for your kindness.  
> **username_0**: Amazing, I appreciate it!  
> **username_1**: Question 2). When you have a partial derivative it is implicit that the variables are fixed, except the variable with respect to you are differentiating. It is correct to use the symbol $\partial$ not $df/dx$. A partial derivative is defined for a function of **several variables**, say $f(x,y,z)$.  In this case the partial derivative of $f$ with respect, say, $x$ is defined as the derivative of $f$  when $x$ changes and the other variables are fixed. Whenever we have a function of several variables we must use the symbol $\partial$. Your notations are correct.  
> **username_1**: As for the *differential* for functions of several variable, this is a more complicated matter, it is not as in the case of function of one single variable. If you haven't already studied  functions of several variable, at the moment don't worry, don't mind the differential. When you will study functions of several variables  the matter  will be clear, but it is not a simple matter, all becomes more complicated.  Think, for example, that for function of several variables there is an important difference between differentiation  (taking the derivatives) and the differential.  
> **username_1**: In short, at the moment don't worry, wait to study functions of several variables. These should be studied on a good book of mathematical analysis, or of more advanced calculus.  
> **username_0**: Okay got it! Can you recommend me a book you know would have a good explanation of the differential in it? I'm pretty sure I should understand this, because I have studied advanced calculus and linear algebra and some analysis...But my mathematical education has been quite non standard, so I have some random wholes...as perhaps you might have gathered from my questions.  
> **username_1**: What do you refer to as 'diffrential'? Diffrential for functions of several variables or differentiation of functions one single variable? They are quite different matters.  
> **username_0**: This video (linked below) covers both single variable and multivariable differential as I understand it. And regarding the multivariable differential if we had something like $f(x(t), y(t))$ then I always thought the differential would be $\frac{\partial{f}}{\partial{x}}\frac{\partial{x}}{\partial{t}}∆t +\frac{\partial{f}}{\partial{y}}\frac{\partial{y}}{\partial{t}}∆t$  or perhaps it should be written as $\frac{\partial{f}}{\partial{x}}\frac{dx}{dt}∆t +\frac{\partial{f}}{\partial{y}}\frac{dx}{dt}∆t$ given that $x$ and $y$ are functions of single variables.  
> **username_0**: So yes that covers my understanding of the differential, same as outlined in this video. However the concepts don't seem wildly different to me in the multivariable case compared to the single variable case, so perhaps I'm missing something, or a lot? https://youtu.be/Z0BOldpndXA?t=207  
> **username_1**: The function you have mentioned is actually a function of one  single variable, $t$. Yes, the differential for function of several variables is different, even if the it takes a form similar to the differential of one variables, because it is written using derivatives. Aniway, it seemed to me that you wrote that $\Delta y$ is the diffrential of a function of several variables, it is not. One needs to study separately functions of one variables, and then functions of several variables, where the differential is defined.  
> **username_1**: I don't know many books of calculus in English, as for first/second year analysis there are many very good books in my language, and we use mainly these. Well-known books in English are Rudin, *Principles of Analysis*,  and Apostol, I know them, but I don't recommend them as first calculus books. A well- known book is Spivak, *Calculus*, for function of one variable, you can see it. Or you can see Courant, which covers both one variable and several variables, it is well-known but I haven't seen it https://www.amazon.it/gp/product/354065058X/ref=ox_sc_act_image_11?smid=A11IL2PNWYJU7H&psc=1  
> **username_1**: A good idea would be also, not for theory (without proofs) but for exercises, a lot of applications and solved problems, and a concise exposition of concepts, Schaum's Advanced Calculus, (Wrede and Spiegel authors), both for one and for several variables, very useful. –These are books of mathematical snalysis, not Mathematics for economist, which is a different matter. But I think that in these books you can find what are you asking, it books of mathematics for economist  you may not find a comprehensive treatment no, often,  the diffential for function of several variables.  
> **username_0**: Aha amazing thank you, I attached an image of the book you recommend, and responded to some other stuff in our chat thread!  

---

