I need to solve the partial derivative with respect to $C_i$ of the integral provided below. I'd be more than happy for some help. [1]: https://i.stack.imgur.com/pKSNr.png
$$\frac{\partial}{\partial C_i} \left[\left(\int_{i=0}^1 C_i^{\tfrac{\eta-1}\eta} \, di\right)^{\tfrac\eta{\eta-1}}\right]$$
Based on the book's answer, we get: [2]: https://i.stack.imgur.com/ncN8w.png
$$\frac\eta{\eta-1} \left[\int_{j=0}^1 C_j^{\tfrac{\eta-1}\eta}\,dj\right]^{\tfrac1{\eta-1}} \frac{\eta-1}\eta C_i^{-\tfrac1\eta}$$
Can someone explain how do we get there? And why does the subscript change from $i$ to $j$?
The indexing notation is a little bit strange since you're integrating over it so i substitute it with common argument of function. We define functional notation for $$ f = f(x) \quad x\in[0,1] $$ like $$ J[f] := \bigg( \int_0^1 f(x)^{\frac{\eta-1}{\eta}}dx \bigg) ^{\frac{\eta}{\eta-1}}. $$ Technically what you're looking for is not like usual derivative, it is $$ \frac{\partial}{\partial C} J[C] (x). $$ in one sense this is functional derivative $\frac{\delta J}{\delta C}$. also in this case you can treat it like a variable because of simple operations. but to be more precise if $$ J[f] = \big( J_1[f] \big)^{\frac{\eta}{\eta-1}} \quad \text{where} \quad J_1[f] = \int_0^1 f(x)^{\frac{\eta-1}{\eta}}dx $$ then by chain rule (functional derivative chain rule) $$ \frac{\partial}{\partial C} J[C] \stackrel{*}{=} {\frac{\eta}{\eta-1}} \big( J_1[C] \big)^{\frac{\eta}{\eta-1}-1} \frac{\partial}{\partial C} J_1[C] = {\frac{\eta}{\eta-1}} \big( J_1[C] \big)^{\frac{1}{\eta-1}} \frac{\partial}{\partial C} J_1[C] $$ and for functional $J_1$ we have: $$ \frac{\partial}{\partial C} J_1[C] = \frac{\delta}{\delta C} \int_0^1 C(x)^{\frac{\eta-1}{\eta}}dx \stackrel{**}{=} \frac{\eta-1}{\eta} C(x)^{\frac{\eta-1}{\eta}-1}=\frac{\eta-1}{\eta} C(x)^{\frac{-1}{\eta}} \tag{1}\label{1} $$ after combining these you get the answer: $$ \frac{\partial}{\partial C} J[C] (x)= {\frac{\eta}{\eta-1}} \bigg( \int_0^1 C(y)^{\frac{\eta-1}{\eta}}dy \bigg)^{\frac{1}{\eta-1}} \frac{\eta-1}{\eta} C(x)^{\frac{-1}{\eta}}. $$
Proof of $**$ in $\eqref{1}$ with definition of functional derivative $$ J_1[C+\delta C] = \int_0^1 [C+\delta C]^{\frac{\eta-1}{\eta}}dx \\ = \int_0^1 [C^{\frac{\eta-1}{\eta}}+\frac{\eta-1}{\eta} C^{\frac{\eta-1}{\eta}-1} \delta C + \dots ]dx =\int_0^1 [C^{\frac{\eta-1}{\eta}}+\frac{\eta-1}{\eta} C^{\frac{-1}{\eta}} \delta C + \dots ]dx \\ = J_1[C] + \int_0^1 \underbrace{\bigg( \frac{\eta-1}{\eta} C^{\frac{-1}{\eta}} \bigg)}_{\frac{\delta J_1}{\delta C}} \delta C dx + O(\delta^2 C) $$
$*$ If you look at functional derivative chain rule, you'll see that there is integration over argument of internal functional, which we don't have here. the point here is $J$ is not a functional of $J_1$. what we did there was just usual derivative of a function. (this could be proved similar to above)