# Differentiation of an integral with implicitly defined variable

• Nov 14th 2012, 02:25 AM
PeterPain
Differentiation of an integral with implicitly defined variable
Hey folks,

I am currently writing on a paper in economics and came up with the following problem I couldn't solve so far.

I would like to differentiate the integral from the following equation with regards to q_i

Attachment 25708

while ^z_i is implicitly defined by the following equation

Attachment 25709

The result is supposed to be the following equation, though I do not understand the way to arrive at it.

Attachment 25710

I would be glad if someone could help me to understand the way to the derivative. Please take into consideration that I have no background in mathematics :)

Cheers,
Peter
• Nov 17th 2012, 01:38 AM
hollywood
Re: Differentiation of an integral with implicitly defined variable
There are a lot of subscripts and functions that make this look harder than it really is.

You have: $V^i=\int_{\hat{z}_i}^{z} (R^i(q_i,q_j,z_i)-D_i) f(z_i) dz_i$

Now you want to take the partial derivative with respect to $q_i$. In other words, you want to see how $V_i$ changes when $q_i$ changes and all the other variables are held constant.

$\frac{\partial{V^i}}{\partial{q_i}} = \frac{\partial}{\partial{q_i}} \int_{\hat{z}_i}^{z} (R^i(q_i,q_j,z_i)-D_i) f(z_i) dz_i$

Assuming you can exchange the derivative with the integral,

$\frac{\partial{V^i}}{\partial{q_i}} = \int_{\hat{z}_i}^{z} \frac{\partial}{\partial{q_i}} $(R^i(q_i,q_j,z_i)-D_i) f(z_i)$ dz_i$

Since $f(z_i)$ doesn't vary with respect to $q_i$, we can treat it like a constant. The derivative of a constant times a function is the constant times the derivative:

$\frac{\partial{V^i}}{\partial{q_i}} = \int_{\hat{z}_i}^{z} \frac{\partial}{\partial{q_i}} $(R^i(q_i,q_j,z_i)-D_i)$ f(z_i) dz_i$

And lastly, $D_i$ is a constant, so the derivative of a function minus a constant is just the derivative of that function:

$\frac{\partial{V^i}}{\partial{q_i}} = \int_{\hat{z}_i}^{z} \frac{\partial{R^i(q_i,q_j,z_i)}}{\partial{q_i}} f(z_i) dz_i$

And I assume that in your notation, $V_i^i=\frac{\partial{V^i}}{\partial{q_i}}$ and $R_i^i(q_i,q_j,z_i)=\frac{\partial{R^i(q_i,q_j,z_i) }}{\partial{q_i}}$, so the result is:

$V_i^i = \int_{\hat{z}_i}^{z} R_i^i(q_i,q_j,z_i) f(z_i) dz_i$

- Hollywood
• Nov 17th 2012, 03:02 AM
PeterPain
Re: Differentiation of an integral with implicitly defined variable
Thanks a lot, Hollywood! You have been a great help to me.

Cheers,
Peter