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Despite Will Sawin's answer in comment, it was not so obvious for me at first glance, so I give a detailed (at least for $X$ affine) proof here.

Lemma. Let $A$ be a $K$-algebra and let $A\to B$ be fpqc. Then the following sequence is exact $$ A((t))\to B((t))\rightrightarrows B((t))\otimes_{A((t))}B((t)). $$

Proof. Since $A\to B$ is fpqc, Grothendieck's fpqc-descent shows that the sequence $$ A\to B\rightrightarrows B\otimes_A B $$ is exact. Consider the sequence $$ A[[t]]\to B[[t]]\rightrightarrows B[[t]]\otimes_{A[[t]]}B[[t]]. $$ We claim that this sequence is also exact.

The first arrow $A[[t]]\to B[[t]]$ is injective since $A\to B$ is injective and since $A[[t]] = \prod_{n\in \mathbf{N}}A$ and $B[[t]] = \prod_{n\in \mathbf{N}}B$ as $A$-modules. It remains to show that for $f(t)\in B[[t]]$, $$f(t)\otimes 1 = 1\otimes f(t)\in B[[t]]\otimes_{A[[t]]}B[[t]] \quad\Longrightarrow \quad f(t)\in A[[t]]. $$ Suppose that $f(t) = \sum_{n\in \mathbf{N}} b_n t^n\in B[[t]]$ satisfies the condition $f(t)\otimes 1 = 1\otimes f(t)$. It is enough to show that $f(t) \mod{t^N} \in A[t]/t^N$ for each $N\ge 1$. Indeed, if we tensorise the first exact sequence by $\otimes_{A}A[t]/t^N$, we get $$ A[t]/t^N\to B[t]/t^N\rightrightarrows B[t]/t^N\otimes_{A[t]/t^N} B[t]/t^N, $$ which implies that $b_n\in A$ for $n < N$. Hence $f(t)\in A[[t]]$.

Inverting $t$, we deduce the exactness of $$ A((t))\to B((t))\rightrightarrows B((t))\otimes_{A((t))}B((t)). $$ End of Proof

Now if $X = \mathrm{Spec}\, C$ is an affine $K$-scheme and A,B are as above, then the exactness of the following sequence follows from the lemma $$ \mathrm{Hom}(C, A((t)))\to \mathrm{Hom}(C, B((t)))\rightrightarrows \mathrm{Hom}(C, B((t))\otimes_{A((t))} B((t))). $$ Hence, if $X$ is affine, $LX$ is automatically an fpqc sheaf. The general case follows from the standard procedure of glueing.

Despite Will Sawin's answer in comment, it was not so obvious for me at first glance, so I give a detailed (at least for $X$ affine) proof here.

Lemma. Let $A$ be a $K$-algebra and let $A\to B$ be fpqc. Then the following sequence is exact $$ A((t))\to B((t))\rightrightarrows (B\otimes_{A}B)((t)). $$

Proof. Since $A\to B$ is fpqc, Grothendieck's fpqc-descent shows that the sequence $$ A\to B\rightrightarrows B\otimes_A B $$ is exact. Consider the sequence $$ A[[t]]\to B[[t]]\rightrightarrows (B\otimes_{A}B)[[t]]. $$ We claim that this sequence is also exact.

The first arrow $A[[t]]\to B[[t]]$ is injective since $A\to B$ is injective and since $A[[t]] = \prod_{n\in \mathbf{N}}A$ and $B[[t]] = \prod_{n\in \mathbf{N}}B$ as $A$-modules. It remains to show that for $f(t)\in B[[t]]$, $$f(t)\otimes 1 = 1\otimes f(t)\in (B\otimes_{A}B)[[t]] \quad\Longrightarrow \quad f(t)\in A[[t]]. $$ Suppose that $f(t) = \sum_{n\in \mathbf{N}} b_n t^n\in B[[t]]$ satisfies the condition $f(t)\otimes 1 = 1\otimes f(t)$. It is enough to show that $f(t) \mod{t^N} \in A[t]/t^N$ for each $N\ge 1$. Indeed, if we tensorise the first exact sequence by $\otimes_{A}A[t]/t^N$, we get $$ A[t]/t^N\to B[t]/t^N\rightrightarrows (B\otimes_{A} B)[t]/t^N, $$ which implies that $b_n\in A$ for $n < N$. Hence $f(t)\in A[[t]]$.

Inverting $t$, we deduce the exactness of $$ A((t))\to B((t))\rightrightarrows (B\otimes_{A}B)((t)). $$ End of Proof

Now if $X = \mathrm{Spec}\, C$ is an affine $K$-scheme and A,B are as above, then the exactness of the following sequence follows from the lemma $$ \mathrm{Hom}(C, A((t)))\to \mathrm{Hom}(C, B((t)))\rightrightarrows \mathrm{Hom}(C, (B\otimes_{A} B)((t))). $$ Hence, if $X$ is affine, $LX$ is automatically an fpqc sheaf. The general case follows from the standard procedure of glueing.

Despite Will Sawin's answer in comment, it was not so obvious for me at first glance, so I give a detailed (at least for $X$ affine) proof here.

Lemma. Let $A$ be a $K$-algebra and let $A\to B$ be fpqc. Then the following sequence is exact $$ A((t))\to B((t))\rightrightarrows B((t))\otimes_{A((t))}B((t)). $$

Proof. Since $A\to B$ is fpqc, Grothendieck's fpqc-descent shows that the sequence $$ A\to B\rightrightarrows B\otimes_A B $$ is exact. Consider the sequence $$ A[[t]]\to B[[t]]\rightrightarrows B[[t]]\otimes_{A[[t]]}B[[t]]. $$ We claim that this sequence is also exact.

The first arrow $A[[t]]\to B[[t]]$ is injective since $A\to B$ is injective and since $A[[t]] = \prod_{n\in \mathbf{N}}A$ and $B[[t]] = \prod_{n\in \mathbf{N}}B$ as $A$-modules. It remains to show that for $f(t)\in B[[t]]$, $$f(t)\otimes 1 = 1\otimes f(t)\in B[[t]]\otimes_{A[[t]]}B[[t]] \quad\Longrightarrow \quad f(t)\in A[[t]]. $$ Suppose that $f(t) = \sum_{n\in \mathbf{N}} b_n t^n\in B[[t]]$ satisfies the condition $f(t)\otimes 1 = 1\otimes f(t)$. It is enough to show that $f(t) \mod{t^N} \in A[t]/t^N$ for each $N\ge 1$. Indeed, if we tensorise the first exact sequence by $\otimes_{A}A[t]/t^N$, we get $$ A[t]/t^N\to B[t]/t^N\rightrightarrows B[t]/t^N\otimes_{A[t]/t^N} B[t]/t^N, $$ which implies that $b_n\in A$ for $n < N$. Hence $f(t)\in A[[t]]$.

Inverting $t$, we deduce the exactness of $$ A((t))\to B((t))\rightrightarrows B((t))\otimes_{A((t))}B((t)). $$ End of Proof

Now if $X = \mathrm{Spec}\, C$ is an affine $K$-scheme and A,B are as above, then the exactness of the following sequence is obviously exact $$ \mathrm{Hom}(C, A((t)))\to \mathrm{Hom}(C, B((t)))\rightrightarrows \mathrm{Hom}(C, B((t))\otimes_{A((t))} B((t))). $$ Hence, if $X$ is affine, $LX$ is automatically an fpqc sheaf. The general case follows from the standard procedure of glueing.

Despite Will Sawin's answer in comment, it was not so obvious for me at first glance, so I give a detailed (at least for $X$ affine) proof here.

Lemma. Let $A$ be a $K$-algebra and let $A\to B$ be fpqc. Then the following sequence is exact $$ A((t))\to B((t))\rightrightarrows B((t))\otimes_{A((t))}B((t)). $$

Proof. Since $A\to B$ is fpqc, Grothendieck's fpqc-descent shows that the sequence $$ A\to B\rightrightarrows B\otimes_A B $$ is exact. Consider the sequence $$ A[[t]]\to B[[t]]\rightrightarrows B[[t]]\otimes_{A[[t]]}B[[t]]. $$ We claim that this sequence is also exact.

The first arrow $A[[t]]\to B[[t]]$ is injective since $A\to B$ is injective and since $A[[t]] = \prod_{n\in \mathbf{N}}A$ and $B[[t]] = \prod_{n\in \mathbf{N}}B$ as $A$-modules. It remains to show that for $f(t)\in B[[t]]$, $$f(t)\otimes 1 = 1\otimes f(t)\in B[[t]]\otimes_{A[[t]]}B[[t]] \quad\Longrightarrow \quad f(t)\in A[[t]]. $$ Suppose that $f(t) = \sum_{n\in \mathbf{N}} b_n t^n\in B[[t]]$ satisfies the condition $f(t)\otimes 1 = 1\otimes f(t)$. It is enough to show that $f(t) \mod{t^N} \in A[t]/t^N$ for each $N\ge 1$. Indeed, if we tensorise the first exact sequence by $\otimes_{A}A[t]/t^N$, we get $$ A[t]/t^N\to B[t]/t^N\rightrightarrows B[t]/t^N\otimes_{A[t]/t^N} B[t]/t^N, $$ which implies that $b_n\in A$ for $n < N$. Hence $f(t)\in A[[t]]$.

Inverting $t$, we deduce the exactness of $$ A((t))\to B((t))\rightrightarrows B((t))\otimes_{A((t))}B((t)). $$ End of Proof

Now if $X = \mathrm{Spec}\, C$ is an affine $K$-scheme and A,B are as above, then the exactness of the following sequence follows from the lemma $$ \mathrm{Hom}(C, A((t)))\to \mathrm{Hom}(C, B((t)))\rightrightarrows \mathrm{Hom}(C, B((t))\otimes_{A((t))} B((t))). $$ Hence, if $X$ is affine, $LX$ is automatically an fpqc sheaf. The general case follows from the standard procedure of glueing.

It was not so obvious for me at first glance, so I give a detailed (at least for $X$ affine) proof here.

Lemma. Let $A$ be a $K$-algebra and let $A\to B$ be fpqc. Then the following sequence is exact $$ A((t))\to B((t))\rightrightarrows B((t))\otimes_{A((t))}B((t)). $$

Proof. Since $A\to B$ is fpqc, Grothendieck's fpqc-descent shows that the sequence $$ A\to B\rightrightarrows B\otimes_A B $$ is exact. Consider the sequence $$ A[[t]]\to B[[t]]\rightrightarrows B[[t]]\otimes_{A[[t]]}B[[t]]. $$ We claim that this sequence is also exact.

The first arrow $A[[t]]\to B[[t]]$ is injective since $A\to B$ is injective and since $A[[t]] = \prod_{n\in \mathbf{N}}A$ and $B[[t]] = \prod_{n\in \mathbf{N}}B$ as $A$-modules. It remains to show that for $f(t)\in B[[t]]$, $$f(t)\otimes 1 = 1\otimes f(t)\in B[[t]]\otimes_{A[[t]]}B[[t]] \quad\Longrightarrow \quad f(t)\in A[[t]]. $$ Suppose that $f(t) = \sum_{n\in \mathbf{N}} b_n t^n\in B[[t]]$ satisfies the condition $f(t)\otimes 1 = 1\otimes f(t)$. It is enough to show that $f(t) \mod{t^N} \in A[t]/t^N$ for each $N\ge 1$. Indeed, if we tensorise the first exact sequence by $\otimes_{A}A[t]/t^N$, we get $$ A[t]/t^N\to B[t]/t^N\rightrightarrows B[t]/t^N\otimes_{A[t]/t^N} B[t]/t^N, $$ which implies that $b_n\in A$ for $n < N$. Hence $f(t)\in A[[t]]$.

Inverting $t$, we deduce the exactness of $$ A((t))\to B((t))\rightrightarrows B((t))\otimes_{A((t))}B((t)). $$ End of Proof

Now if $X = \mathrm{Spec}\, C$ is an affine $K$-scheme and A,B are as above, then the exactness of the following sequence is obviously exact $$ \mathrm{Hom}(C, A((t)))\to \mathrm{Hom}(C, B((t)))\rightrightarrows \mathrm{Hom}(C, B((t))\otimes_{A((t))} B((t))). $$ Hence, if $X$ is affine, $LX$ is automatically an fpqc sheaf. The general case follows from the standard procedure of glueing.

Despite Will Sawin's answer in comment, it was not so obvious for me at first glance, so I give a detailed (at least for $X$ affine) proof here.

Lemma. Let $A$ be a $K$-algebra and let $A\to B$ be fpqc. Then the following sequence is exact $$ A((t))\to B((t))\rightrightarrows B((t))\otimes_{A((t))}B((t)). $$

Proof. Since $A\to B$ is fpqc, Grothendieck's fpqc-descent shows that the sequence $$ A\to B\rightrightarrows B\otimes_A B $$ is exact. Consider the sequence $$ A[[t]]\to B[[t]]\rightrightarrows B[[t]]\otimes_{A[[t]]}B[[t]]. $$ We claim that this sequence is also exact.

The first arrow $A[[t]]\to B[[t]]$ is injective since $A\to B$ is injective and since $A[[t]] = \prod_{n\in \mathbf{N}}A$ and $B[[t]] = \prod_{n\in \mathbf{N}}B$ as $A$-modules. It remains to show that for $f(t)\in B[[t]]$, $$f(t)\otimes 1 = 1\otimes f(t)\in B[[t]]\otimes_{A[[t]]}B[[t]] \quad\Longrightarrow \quad f(t)\in A[[t]]. $$ Suppose that $f(t) = \sum_{n\in \mathbf{N}} b_n t^n\in B[[t]]$ satisfies the condition $f(t)\otimes 1 = 1\otimes f(t)$. It is enough to show that $f(t) \mod{t^N} \in A[t]/t^N$ for each $N\ge 1$. Indeed, if we tensorise the first exact sequence by $\otimes_{A}A[t]/t^N$, we get $$ A[t]/t^N\to B[t]/t^N\rightrightarrows B[t]/t^N\otimes_{A[t]/t^N} B[t]/t^N, $$ which implies that $b_n\in A$ for $n < N$. Hence $f(t)\in A[[t]]$.

Inverting $t$, we deduce the exactness of $$ A((t))\to B((t))\rightrightarrows B((t))\otimes_{A((t))}B((t)). $$ End of Proof

Now if $X = \mathrm{Spec}\, C$ is an affine $K$-scheme and A,B are as above, then the exactness of the following sequence is obviously exact $$ \mathrm{Hom}(C, A((t)))\to \mathrm{Hom}(C, B((t)))\rightrightarrows \mathrm{Hom}(C, B((t))\otimes_{A((t))} B((t))). $$ Hence, if $X$ is affine, $LX$ is automatically an fpqc sheaf. The general case follows from the standard procedure of glueing.