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

enter image description here 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?

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    $\begingroup$ q2 disappears because it's being held constant. The derivative of q2 is trivial. $\endgroup$ Nov 10, 2017 at 4:23
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    $\begingroup$ 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. $\endgroup$
    – EconJohn
    Nov 10, 2017 at 4:25

1 Answer 1


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

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    $\begingroup$ 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! $\endgroup$
    – Romaion
    Nov 10, 2017 at 17:50

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