<p>Well going through the derivatives, simply put. View $U$ as a function. <strong>A.K.A $U=U(q_1,q_2)$</strong> and let $U_1=U(q_1)$ and $U_2=U(q_2)$. So to find $U_1$ and $U_2$ respectively, just take the partial derivatives in respect to the  given variables</p>

<p>$$U_1=\frac{\partial U(q_1,q_2)}{\partial q_1 }=(1-\beta)q_1^{-\beta}+ \frac{\partial}{\partial q_1} \gamma q_2=(1-\beta)q_1^{-\beta}$$</p>

<p>Since $\frac{\partial}{\partial q_1}\gamma q_2=0$</p>

<p>Similarly,
$$U_2=\frac{\partial U(q_1,q_2)}{\partial q_2}= \frac{\partial}{\partial q_2}q_1^{1-\beta}+\gamma=\gamma$$</p>

<p>since $\frac{\partial}{\partial q_2} q_1^{1-\beta}=0$</p>

<p>Hope this helps </p>

CC BY-SA 3.0
2017-11-10T17:50:34.610Z
This makes sense. I was just a little confused because in the cobb-douglas derivation of U_1 q1 and q2 remained - but now I know its b/c we applied the product rule for derivation. Hey thanks!
Romaion
CC BY-SA 3.0
2017-11-10T05:11:21.033Z
2017-11-10T05:11:21.033Z
Bensstats
<p>I'm having difficultly understanding what my professor taught in class.</p>

<p><a href="https://i.sstatic.net/F0QeH.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/F0QeH.png" alt="enter image description here"></a>
I thought, like cobb-douglas, when finding $U_1$ we take partial derivative with respect to $q_1$ and hold everything else constant. Why does gamma($q_2$) disappear? Same with $U_2$. and what is gamma? a variable?</p>

CC BY-SA 3.0
2017-11-10T04:23:11.530Z
q2 disappears because it's being held constant. The derivative of q2 is trivial.
serakfalcon
CC BY-SA 3.0
2017-11-10T04:25:56.203Z
Hey, welcome to Economics.SE! you might want to review how to take partial derivative. MRS is just $\frac{\partial U/\partial q_1}{ \partial U/ \partial q_2}$ that is all your professor is doing.
EconJohn
<utility><preferences>
MRS for quasi-linear preferences
Romaion
# MRS for quasi-linear preferences

(Asked by: username_0 on 2017-11-10)

I'm having difficultly understanding what my professor taught in class.

[![enter image description here](https://i.sstatic.net/F0QeH.png)](https://i.sstatic.net/F0QeH.png)
I thought, like cobb-douglas, when finding $U\_1$ we take partial derivative with respect to $q\_1$ and hold everything else constant. Why does gamma($q\_2$) disappear? Same with $U\_2$. and what is gamma? a variable?

**Tags:** utility, preferences

> **Comments:**
> **username_1**: q2 disappears because it's being held constant. The derivative of q2 is trivial.  
> **username_2**: Hey, welcome to Economics.SE! you might want to review how to take partial derivative. MRS is just $\frac{\partial U/\partial q_1}{ \partial U/ \partial q_2}$ that is all your professor is doing.  

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## 1 answers

### Answer by username_3 on 2017-11-10. Score: 2 | Accepted answer

Well going through the derivatives, simply put. View $U$ as a function. **A.K.A $U=U(q\_1,q\_2)$** and let $U\_1=U(q\_1)$ and $U\_2=U(q\_2)$. So to find $U\_1$ and $U\_2$ respectively, just take the partial derivatives in respect to the given variables

$$U\_1=\frac{\partial U(q\_1,q\_2)}{\partial q\_1 }=(1-\beta)q\_1^{-\beta}+ \frac{\partial}{\partial q\_1} \gamma q\_2=(1-\beta)q\_1^{-\beta}$$

Since $\frac{\partial}{\partial q\_1}\gamma q\_2=0$

Similarly,
$$U\_2=\frac{\partial U(q\_1,q\_2)}{\partial q\_2}= \frac{\partial}{\partial q\_2}q\_1^{1-\beta}+\gamma=\gamma$$

since $\frac{\partial}{\partial q\_2} q\_1^{1-\beta}=0$

Hope this helps

> **Comments:**
> **username_0**: This makes sense. I was just a little confused because in the cobb-douglas derivation of U_1 q1 and q2 remained - but now I know its b/c we applied the product rule for derivation. Hey thanks!  

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