$\Omega^1_{K|k}\otimes_KL\rightarrow\Omega^1_{L|k}$ is an isomorphism when $K\subset L$ is finite

Let $k\subset K\subset L$ be field extensions. Then we have the following exact sequence of $L$-vector spaces: $$\Omega^1_{K|k}\otimes_KL\rightarrow\Omega^1_{L|k}\rightarrow \Omega^1_{L|K}\rightarrow 0.$$ If $K\subset L$ is algebraic then $\Omega^1_{L|K}=0$ and so we have a surjective $L$-linear map $\Omega^1_{K|k}\otimes_KL\rightarrow\Omega^1_{L|k}.$ I want to prove that if $K\subset L$ is finite and separated then this map is an isomorphism.

If $K\subset L$ is finite and separable then $L\cong K[x]/(P)$ for some polynomial $P.$ I thought of using the conormal exact sequence $(P)/(P^2)\rightarrow \Omega^1_{K[x]|k}\otimes_{K[x]}L\rightarrow \Omega^1_{L|k}\rightarrow 0$ but that doesn't seem to yield anything interesting.
Any help would be appreciated!

Instead of the conormal exact sequence, you can use the cotangent exact sequence

$$L \otimes_K \Omega_{K/k}^1 \to \Omega^1_{L/k} \to \Omega^1_{L/K} \to 0$$

Since everything is just a field, we have smoothness (every vector space is free), so this sequence is actually exact on the left as well:

$$0 \to L \otimes_K \Omega_{K/k}^1 \to \Omega^1_{L/k} \to \Omega^1_{L/K} \to 0.$$

Since $L/K$ is separable, $\Omega_{L/K}^1 = 0$ (easy exercise using explicit realization of differentials and separable polynomial), from which you obtain your desired isomorphism.