Commutator Group of $GL_2(R)$ is $SL_2(R)$

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.

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$$\begin{pmatrix}1&2\\0&1\end{pmatrix}\begin{pmatrix}-1&0\\1&2\end{pmatrix}\begin{pmatrix}1&2\\0&1\end{pmatrix}^{-1}\begin{pmatrix}-1&0\\1&2\end{pmatrix}^{-1}.$$