TITLE: Derivative of an integral with change of variable.
QUESTION [5 upvotes]: I have this problem.
I want to compute the following integral in terms of $z$, where $x = az$
$$
\frac{d}{da}\int a^2x^2dx
$$
Computing the integral first, the derivative as the last passage, I obtain what I believe to be the correct solution, i.e.,
$$
\frac{d}{da}\int a^2x^2dx = \frac{d}{da}\left(a^2 \frac{x^3}{3}\right) = \frac{d}{da}\left(a^2 \frac{a^3z^3}{3} + c\right) = 5a^4\frac{z^3}{3}
$$
and the same by substituting first in the integral, keeping in mind that $dx = a dz$,
$$
\frac{d}{da}\int a^2x^2dx = \frac{d}{da}\int a^2(az)^2 adz \frac{d}{da}\int a^5z^2 dz = \frac{d}{da}\left(a^5 \frac{z^3}{3}\right) = 5a^4\frac{z^3}{3}.
$$
Things start to get messy when I take the derivative inside the integral, i.e.,
$$
\frac{d}{da}\int a^2x^2dx = \int 2ax^2dx = \int 2a a^2z^2 adz = 2a^4\frac{z^3}{3}.
$$
One might argue that I need to keep in consideration also of $d/da x$, since $x$ depends on $a$, after all. However, I still obtain a wrong result
$$
\frac{d}{da}\int a^2x^2dx = \int a^2 \frac{d}{da}\left(x^2\right) + 2a x^2 dx = \int \left( a^2 \frac{d}{da}\left(a^2z^2\right) + 2a a^2z^2  \right)adz \\
= \int 2a^4z^2 + 2a^4z^2dz = 4a^4\frac{z^3}{3}.
$$
Since I am obtaining three different results, I must be making at least two different mistakes; I wonder where these mistakes are. (sorry for being lazy and not including the constant $c$ from the integrals)

REPLY [2 votes]: We substitute $x=az$ right from the beginning and obtain
\begin{align*}
\color{blue}{\frac{d}{da}\int a^2x^2dx}&=\frac{d}{da}\int a^2(az)^2d(az)\\
&=\frac{d}{da}\left(a^5\int z^2 dz\right)\tag{1}\\
&=\left(\int z^2 dz\right)\left(\frac{d}{da} a^5\right)\tag{2}\\
&\,\,\color{blue}{=\left(\frac{1}{3}z^3+C\right)\left(5a^4\right)}\tag{3}\\
\end{align*}
Comment:

In (1) we use $d(az)=adz$ and factor out $a$.

In (2) we factor out the constant part.

In (3) we integrate and differentiate.



Next we look at OPs final approach somewhat more detailed. Since OP wants to compute the integral with respect to $z$ we consider $x=x(z)=az$ as function of $z$. We obtain
\begin{align*}
\frac{d}{da}\int a^2x^2\,dx&=\frac{d}{da}\int a^2x^2(z)\,d\left(x(z)\right)\\
&=\frac{d}{da}\int a^2 x^2(z) x^{\prime}(z)\,dz\tag{4}
\end{align*}

We are now ready to do the differentiation with respect to $a$ before the integration. To better see what's going on we write
$x=x(z,a)$ and $x_z(z,a)$ instead of $x^{\prime}(z)$ to indicate that we differentiate with respect to $z$.

We obtain from (4) by successively applying the multiplication rule $(u\,v)^{\prime}=u^\prime v+uv^\prime$
\begin{align*}
\color{blue}{\frac{d}{da}}&\color{blue}{\int a^2 x^2(z,a) x_z(z,a)dz}\\
&=\int \frac{d}{da} \left(a^2 \cdot \left(x^2(z,a) x_z(z,a)\right)\right)dz\\
&=\int 2a \left(x^2(z,a) x_z(z,a)\right)dz\\
&\qquad +\int a^2\frac{d}{da}\left(x^2(z,a) x_z(z,a)\right)dz\\
&=\int 2a \left(x^2(z,a) x_z(z,a)\right)dz\\
&\qquad +\int a^2 \left(2x(z,a) \left(\frac{d}{da}x(z,a)\right) x_z(z,a) +x^2(z,a)\frac{d}{da} x_z(z,a)\right)dz\\
&=\int 2a \left((az)^2 a\right)dz\\
&\qquad +\int a^2 \left(2(az)\left(z\right) a + (az)^2\cdot 1\right)dz\\
&=2a^4\int z^2\,dz+2a^4\int z^2\,dz + a^4 \int z^2dz\\
&=5a^4\int z^2dz\\
&\,\,\color{blue}{=5a^4\left(\frac{1}{3}z^3+C\right)}
\end{align*}
in accordance with (3).