This is an microeconomics question, and I'm having trouble with the derivative of an integral function. The question is as follows:
From the following indirect-utility function, obtain the marshallian demand for x.
$V(p,b) = G\left(A(p) + \frac{\bar{b}^\eta b^{1 - \eta}}{1-\eta} \right)$
Where $ A(p) = \int_{p}^{p^G} x(\epsilon,\bar{b})d\epsilon$
So I went through Roy's Identity: $- x(p,b) = \frac{\partial v /\partial p}{\partial v / \partial b}$
$\partial v(p,b)/ \partial b = G'(.) \bar{b}^\eta b^{-\eta}$
$\partial v(p,b)/ \partial p = G'(.) \frac{\partial A(p)}{\partial p} $
The $G'()s$ will cancel out , but I'm having trouble with $ \frac{\partial A(p)}{\partial p} $
I tried applying the first theorem of calculus as this: $ \int_{p}^{p^G} x(\epsilon,\bar{b})d\epsilon = F(x(p^G,\bar{b})) - F(x(p,\bar{b}) $
From this I got $ \frac{\partial A(p)}{\partial p} = G p^{G-1} x(p^G,\bar{b}) - x(p,\bar{b}) $.
But ending up with two different price vectors ($ x(p,\bar{b})$ and $ x(p^g,\bar{b} $)) on x don't look like it's the way to go.
Any input would be greatly appreciated on how to handle $ \frac{\partial A(p)}{\partial p} $
Thanks!
Comment on mlc's hint: seems i'm misapplying Leibniz' rule: I'm getting: $ \frac{x(p^g,\bar{b})(p^G)' - x(p,\bar{b})*p'}{\partial p}$
Under appropriate conditions, (a simplified version of) the Leibniz's rule states $$\frac{d}{dx} \int_{a(x)}^{b(x)} f(t) dt = f(x,b(x)) b^\prime(x) - f(x,a(x)) a^\prime(x)$$
Apply this rule to $$A(p) = \int_{p}^{p^G} x(\epsilon,\bar{b})d\epsilon$$ and obtain $$\frac{d}{dp} A(p) = - x(p,\bar{b})$$