Let $G$ be a linear algebraic group over an algebraically closed field $k$, and let $e$ be the identity of $G$. Then $k$ is a $k[G]$-module via the action $f \cdot a = f(e)a$. The tangent space of $G$ at $e$ can be identified with the $k[G]$-module $\textrm{Der}_k(k[G],k)$, where $k$ in the second argument is regarded as a $k[G]$-module. So every $f \in k[G]$ is associated with an element $\delta(f) \in k$.
There is supposed to be (Humphreys, Linear Algebraic Groups) an injective homomorphism of $k$-vector spaces $$\eta: \textrm{Der}_k(k[G],k) \rightarrow \textrm{Der}_k(k[G],k[G])$$ given by the formula $$\eta(\delta)(f)(x) = \delta(\lambda_{x^{-1}}(f))$$ where $\lambda_{x^{-1}}$ is the $k$-algebra automorphism of $k[G]$ given by $\lambda_{x^{-1}}(h)(y) = h(xy)$. What I don't understand is why the assignment $x \mapsto \delta(\lambda_{x^{-1}}f)$ actually defines an element of $k[G]$, i.e. a regular function from $G$ to $k$.
Here is a very simple example. Let $G = k$, so $k[G] = k[X]$. Let $\delta: k[X] \rightarrow k$ be the $k$-derivation given by $\delta(f(X)) = f'(0)$. Then $\eta(\delta): k[G] \rightarrow k[G]$ is the function $$\eta(\delta)(f(X))(x) = \delta(\lambda_{-x}f(X)) = \delta(f(X-x)) = f'(-x)$$ and the mapping $x \mapsto f'(-x)$ is certainly regular.