<p><a href="https://i.sstatic.net/RVLcl.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/RVLcl.png" alt="Lecture slides"></a>Just a small question about the differentiation technique used in the consumer based asset pricing model. </p>

<p>I need to maximize the following equation: 
$U(c_t) + E_t[\beta \cdot u(c_{t+1})]$</p>

<p>$$\begin{align} \text{such that} \qquad c_t &amp; = e_t - p_t \xi \\
c_{t+1} &amp; = e_{t + 1} + x_{t + 1} \xi \end{align}$$</p>

<p><s> where $t$ is a time subscript, and $S$ stand for Xi, the greek letter. I'm new to this site, so haven't figured out all the neat tools for symbols. </s></p>

<p>I can use substitution or a Lagrange formula, I know how both work. What puzzles me though is part of the solution. From what I know about partial differentiation, the derivative of the first part $u(c_t)$ with respect to $c_t$ (when using the lagrange method), or with respect to $\xi$ if you use substitution, becomes $u$, but my book (and several online sources) give $u'(c_t)$, and so the solution to the optimization problem becomes $u'(c_t)p_t = E_t[\beta \cdot u'(c_{t+1})]$</p>

<p>I'm sure I'm missing a very simple differentiation rule, but I can't find it anywhere. </p>

<p>Thanks for your attention:)</p>

<p>Best regards,</p>

<p>Anne</p>

CC BY-SA 3.0
2016-09-13T02:36:36.677Z
What do you mean by "becomes u"
VCG
CC BY-SA 3.0
2016-09-13T03:25:29.773Z
It's a long sentence, bit unclear maybe. I mean the derivative of u(ct) with respect to (ct) is u.
Jansen
CC BY-SA 3.0
2016-09-13T05:59:53.810Z
edited for clarity; let me know if something ended up iffy
Kitsune Cavalry
CC BY-SA 3.0
2016-09-13T06:02:02.243Z
To write your equations, you can use Latex' maths. That being said, isn't your problem simply related to the notation? $u^\prime$ usually indicates the first derivative of $u$. $u^\prime(c_t)=\frac{du}{dc_t}(c_t)$, if you want.
clem steredenn
CC BY-SA 3.0
2016-09-13T06:06:04.563Z
@KitsuneCavalry interesting, it seems SE then completely discarded my edit. Or maybe it's in the review queue... but it would somehow conflict with yours...
clem steredenn
CC BY-SA 3.0
2016-09-13T07:07:13.387Z
@bilbo_pingouin guess that is it. Because later on we want to use other values for (ct), it stays that way instead of performing the actual differentiation. In this notation it's easier to spot that than in the notation used in the lecture. I'll upload a photo
Jansen
CC BY-SA 3.0
2016-09-13T11:10:01.000Z
"it stays that way instead of performing the actual differentiation" again, what do you mean by that? the derivative of $u(c_t)$ is not u. They aren't multiplied.
VCG
CC BY-SA 3.0
2016-09-13T17:18:49.690Z
@bilbo_pingouin I wasn't aware you had an edit of this question.
Kitsune Cavalry
<asset-pricing>
Consumer based asset pricing model, differentiation problem
Jansen
# Consumer based asset pricing model, differentiation problem

(Asked by: username_0 on 2016-09-12)

[![Lecture slides](https://i.sstatic.net/RVLcl.png)](https://i.sstatic.net/RVLcl.png)Just a small question about the differentiation technique used in the consumer based asset pricing model.

I need to maximize the following equation:
$U(c\_t) + E\_t[\beta \cdot u(c\_{t+1})]$

$$\begin{align} \text{such that} \qquad c\_t & = e\_t - p\_t \xi \\
c\_{t+1} & = e\_{t + 1} + x\_{t + 1} \xi \end{align}$$

~~where $t$ is a time subscript, and $S$ stand for Xi, the greek letter. I'm new to this site, so haven't figured out all the neat tools for symbols.~~

I can use substitution or a Lagrange formula, I know how both work. What puzzles me though is part of the solution. From what I know about partial differentiation, the derivative of the first part $u(c\_t)$ with respect to $c\_t$ (when using the lagrange method), or with respect to $\xi$ if you use substitution, becomes $u$, but my book (and several online sources) give $u'(c\_t)$, and so the solution to the optimization problem becomes $u'(c\_t)p\_t = E\_t[\beta \cdot u'(c\_{t+1})]$

I'm sure I'm missing a very simple differentiation rule, but I can't find it anywhere.

Thanks for your attention:)

Best regards,

Anne

**Tags:** asset-pricing

> **Comments:**
> **username_1**: What do you mean by "becomes u"  
> **username_0**: It's a long sentence, bit unclear maybe. I mean the derivative of u(ct) with respect to (ct) is u.  
> **username_2**: edited for clarity; let me know if something ended up iffy  
> **username_3**: To write your equations, you can use Latex' maths. That being said, isn't your problem simply related to the notation? $u^\prime$ usually indicates the first derivative of $u$. $u^\prime(c_t)=\frac{du}{dc_t}(c_t)$, if you want.  
> **username_3**: @KitsuneCavalry interesting, it seems SE then completely discarded my edit. Or maybe it's in the review queue... but it would somehow conflict with yours...  
> **username_0**: @bilbo_pingouin guess that is it. Because later on we want to use other values for (ct), it stays that way instead of performing the actual differentiation. In this notation it's easier to spot that than in the notation used in the lecture. I'll upload a photo  
> **username_1**: "it stays that way instead of performing the actual differentiation" again, what do you mean by that? the derivative of $u(c_t)$ is not u. They aren't multiplied.  
> **username_2**: @bilbo_pingouin I wasn't aware you had an edit of this question.  

---

## 0 answers

