Sum of partial derivatives Suppose that
$$\mu_i(x)=x_i \int_0^1 t^{n-1} \rho(tx) dt$$
where $\rho$ is a function on $\mathbb R^n$ and $tx=(tx_1,\dots,tx_n)\in \mathbb R^n$. Show that
$$\sum_{i=1}^n \frac{\partial\mu_i}{\partial x_i}=\rho .$$
This problem looks simple, but I am having difficulty in showing the result.
I guess that the first step is to find $\frac{\partial\mu_i}{\partial x_i}$ using the product rule.
 A: let $\rho_k$ represent the derivative of $\rho$ wrt the $k^{\text{th}}$ argument
$$
\frac{\partial \mu_i}{\partial x_i} = \int_0^1t^{n-1}\rho(tx) dt + x_i\int_0^1t^n \rho_i(tx)dt \tag{1}
$$
and for the first integral, by parts we have
$$
 \int_0^1t^{n-1}\rho(tx) dt = \frac{\rho(x)}n -\frac1{n}\int_0^1 t^n\frac{d\rho(tx)}{dt}dt
$$
where
$$
\frac{d\rho(tx)}{dt} = \sum_{i=1}^n x_i\rho_i(tx)
$$
thus
$$
\sum_{i=1}^n  \int_0^1t^{n-1}\rho(tx) dt = \rho(x) - \int_0^1 t^n\frac{d\rho(tx)}{dt}dt \\
= \rho(x) -\int_0^1 t^n  \sum_{i=1}^n x_i\rho_i(tx) dt \\
= \rho(x)- \sum_{i=1}^n x_i\int_0^1 t^n \rho_i(tx) dt
$$ 
giving, from summation of (1)
$$
 \sum_{i=1}^n\frac{\partial \mu_i}{\partial x_i} =\rho(x)
$$
A: $$\dfrac{\partial\mu_i}{\partial x_i}=\int_0^1 t^{n-1} \rho(tx) dt + x_i\int_0^1 t^{n} \rho'(tx) dt,$$
$$\sum_{i=1}^n\dfrac{\partial\mu_i}{\partial x_i} = n\int_0^1 t^{n-1} \rho(tx) dt +\sum_{i=1}^n x_i\int_0^1 t^{n} \rho'(tx) dt = \int_0^1 \rho(tx) dt^n + \int_0^1 t^{n} \dfrac{d\rho(tx)}{dt} dt = t^{n} \rho(tx)\biggr|_0^1 - \int_0^1 t^{n} \dfrac{d\rho(tx)}{dt} dt + \int_0^1 t^{n} \dfrac{d\rho(tx)}{dt} dt = \rho(x)$$
