Commutator Group of $\operatorname{GL}_2(\mathbb{R})$ is $\operatorname{SL}_2(\mathbb{R})$ 
Let $\operatorname{GL}_2(\mathbb{R})$ be the general linear group of $2\times2$ matrices and $\operatorname{SL}_2(\mathbb{R})$ the special linear group of $2 \times 2$ matrices. Show that the commutator subgroup of $\operatorname{GL}_2(\mathbb{R})$ is $\operatorname{SL}_2(\mathbb{R})$.

I can show that the commutator subgroup is contained in $\operatorname{SL}_2(\mathbb{R})$ as if $A,B \in \operatorname{GL}_2(\mathbb{R})$ then
$$
 \det(ABA^{-1}B^{-1}) = \det(A)\det(B)\det(A^{-1})\det(B^{-1}) = 1.
$$
But how can I show the reverse inclusion? That is, that $\operatorname{SL}_2(\mathbb{R})$ is contained in the commutator subgroup of $\operatorname{GL}_2(\mathbb{R})$.
Any help will be appreciated.
 A: If $x\in\mathbb R$, then$$\begin{pmatrix}1&x\\0&1\end{pmatrix}=\begin{pmatrix}2&0\\0&1\end{pmatrix}\begin{pmatrix}1&x\\0&1\end{pmatrix}\begin{pmatrix}2&0\\0&1\end{pmatrix}^{-1}\begin{pmatrix}1&x\\0&1\end{pmatrix}^{-1}$$and therefore $\left(\begin{smallmatrix}1&x\\0&1\end{smallmatrix}\right)$ is a product of commutators. For the same reason $\left(\begin{smallmatrix}1&0\\x&1\end{smallmatrix}\right)$ is a product of commutators. On the other hand, if $x\in\mathbb{R}\setminus\{0\}$,$$\begin{pmatrix}x&0\\0&x^{-1}\end{pmatrix}=\begin{pmatrix}x&0\\0&1\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}x&0\\0&1\end{pmatrix}^{-1}\begin{pmatrix}0&1\\1&0\end{pmatrix}^{-1}.$$
Now, let $\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)\in SL_2(\mathbb{R})$. If $a\neq0$, then$$\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}1&0\\\frac ca&1\end{pmatrix}\begin{pmatrix}1&ab\\0&1\end{pmatrix}\begin{pmatrix}a&0\\0&a^{-1}\end{pmatrix},$$which is a product of commutators. Otherwise, $b\neq0$ and$$\begin{pmatrix}0&b\\c&d\end{pmatrix}=\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}1&-\frac db\\0&1\end{pmatrix}\begin{pmatrix}b^{-1}&0\\0&b\end{pmatrix}.$$Finally, $\left(\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}\right)$ is a commutator, since it is equal to$$\begin{pmatrix}1&2\\0&1\end{pmatrix}\begin{pmatrix}-1&0\\1&2\end{pmatrix}\begin{pmatrix}1&2\\0&1\end{pmatrix}^{-1}\begin{pmatrix}-1&0\\1&2\end{pmatrix}^{-1}.$$
A: Let SL $=$ https://people.brandeis.edu/~igusa/Math131b/SL.pdf,
GL $=$ https://ysharifi.wordpress.com/2011/01/29/commutator-subgroup-of-the-general-linear-group/, and
Pontryagin $=$ Pontryagin, L.: Topological Groups, Princeton University Press, 1946.
The following proof is based on the proof in GL.
Let $F$ be a field and $G'$ be the commutator subgroup of a group $G$.
$(A, B \in GL(n,F))\Rightarrow \det (ABA^{-1}B^{-1})=1$, so $GL(n,F)' \subset SL(n,F)$.
$SL(n,F)=SL(n,F)'$  [SL, p.3, Theorem 14.8]
$\subset GL(n,F)'$ [Pontryagin, p.14, Definition 8].
Remark. For other related topics, read Example 6.165 in https://sites.google.com/view/lcwangpress/%E9%A6%96%E9%A0%81/papers/mathematical-methods
