Show that $\|Du_{\lambda}\|_{L^2(\mathbb{R}^n)} = \|Du\|_{L^2(\mathbb{R}^n)}$ 
Let $x \in \mathbb{R}^n$. Given $$u(x) := \left(\frac 1{1+|x|^2} \right)^{\frac{n-2}2}, \quad u_\lambda(x):=\left(\frac \lambda{\lambda^2+|x|^2} \right)^{\frac{n-2}2},$$ I need to show that  $\|Du_{\lambda}\|_{L^2(\mathbb{R}^n)} = \|Du\|_{L^2(\mathbb{R}^n)}$. (Note that $Du=\nabla u = \sum_{i=1}^n u_{x_i}$.)

Equivalently, I can show $\|Du_{\lambda}\|_{L^2(\mathbb{R}^n)}^2 = \|Du\|_{L^2(\mathbb{R}^n)}^2$. My following derivatives are $$\small {u_{x_i}=(2-n)\left(\frac 1{1+|x|^2} \right)^{\frac{n-2}2}\left(\frac 1{1+|x|^2} \right) |x| x_i  \text{ and } (u_\lambda)_{x_i}=(2-n)\left(\frac {\lambda}{\lambda^2+|x|^2} \right)^{\frac{n-2}2}\left(\frac 1{\lambda^2+|x|^2} \right) |x| x_i,}$$ for $i \in \{1,\ldots,n\}$. Thus, $$\small{Du =(2-n)\left(\frac 1{1+|x|^2} \right)^{\frac{n-2}2}\left(\frac {|x|^2}{1+|x|^2} \right) \text{ and }Du_{\lambda}=(2-n)\left(\frac {\lambda}{\lambda^2+|x|^2} \right)^{\frac{n-2}2}\left(\frac {|x|^2}{\lambda^2+|x|^2} \right).}$$
Therefore, $$\small{\|Du\|_{L^2(\mathbb{R}^n)}^2=\int_{\mathbb{R}^n} (2-n)^2\left( \frac 1{1+|x|^2}\right)^{n-2} \frac {|x|^4}{1+|x|^2} \, dx, \\\|Du_{\lambda}\|_{L^2(\mathbb{R}^n)}^2=\int_{\mathbb{R}^n} (2-n)^2\left( \frac {\lambda}{\lambda^2+|x|^2}\right)^{n-2} \frac {|x|^4}{\lambda^2+|x|^2} \, dx}.$$ For $\|Du_{\lambda}\|_{L^2(\mathbb{R}^n)}$, I perform a substitution of $u_i = \frac{x_i}{\lambda}$ and $du_i=\frac{dx_i}{\lambda}$. Then $|u|=\frac{|x|}{\lambda}$ and $du=\frac{dx}{\lambda^n}$. When I apply these substitutions, and simplify algebraically, I obtained $$\|Du\|_{L^2(\mathbb{R}^n)}^2=\int_{\mathbb{R}^n} (2-n)^2 \color{red}{\lambda^2} \frac {|u|^4}{(1+|u|^2)^n} \, du.$$ The problem is that I get an extra $\lambda^2$ that is not supposed to be there...
 A: Notice that
$$
u\left(\frac{x}{\lambda}\right)=\lambda^{\frac{n-2}{2}}u_\lambda(x),
$$
and therefore
$$
u_\lambda(x)=\lambda^{-\frac{n-2}{2}}u\left(\frac{x}{\lambda}\right).
$$
It follows that
$$
D_iu_\lambda(x)=\lambda^{-\frac{n}{2}}D_iu\left(\frac{x}{\lambda}\right)\quad \forall i=1,2,\ldots,n,
$$
and
$$
\|Du_\lambda\|^2_{L^2(\mathbb{R}^n)}=\sum_{i=1}^n\int_{\mathbb{R}^n}|D_iu_\lambda(x)|^2\,dx=\sum_{i=1}^n\int_{\mathbb{R}^n}\lambda^{-n}\left|D_iu\left(\frac{x}{\lambda}\right)\right|^2\,dx=\sum_{i=1}^n\lambda^{-n}\int_{\mathbb{R}^n}\left|D_iu\left(\frac{x}{\lambda}\right)\right|^2\,dx.
$$
Setting $x=\lambda y$, we have:
$$
\int_{\mathbb{R}^n}\left|D_iu\left(\frac{x}{\lambda}\right)\right|^2\,dx=\lambda^n\int_{\mathbb{R}^n}|D_iu(y)|^2\,dy \quad \forall i=1,2,\ldots,n.
$$
Thus
\begin{eqnarray}
\|Du_\lambda\|^2_{L^2(\mathbb{R}^n)}&=&\sum_{i=1}^n\lambda^{-n}\int_{\mathbb{R}^n}\left|D_iu\left(\frac{x}{\lambda}\right)\right|^2\,dx=\sum_{i=1}^n\lambda^{-n}\lambda^n\int_{\mathbb{R}^n}\left|D_iu(y)\right|^2\,dy\\
&=&\sum_{i=1}^n\int_{\mathbb{R}^n}\left|D_iu(y)\right|^2\,dy=\|Du\|_{L^2(\mathbb{R}^n)}.
\end{eqnarray}
