for any $e$ prove that the difference between $f(x)$ and $f(x+e)$ is the same as the directional derivative of $f(x+te)$ in the direction of $e$

Question

Let $f : \Bbb R^n → \Bbb R$ and $x ∈ \Bbb R^n$. Suppose that for any unit vector $e ∈ \Bbb R^n$, the directional derivative $\nabla_e f(x + te)$ exists for all $t ∈ [0, 1]$. Prove that $f(x+e)−f(x)=\nabla_e f(x+te)$ for some $t ∈ (0, 1)$.

So I think that this is basically the mean value theorem but with directional derivatives? But my knowledge of directional derivatives is limited and I am unsure of where to even begin on this one.

I know that $(x+te)∈(x,x+e)$ but that doesn't imply anything about $f(x+te)$.

And just writing stuff down I have: $\nabla_e f(x+te)=\frac{df}{dx_1}(x_1+te_1)\, e_1+...+\frac{df}{dx_n}(x_n+te_n)\, e_n$

Answer 1

Note that $x$ and $e$ are fixed here. Consider the auxiliary function $$\phi(t):=f(x+t e)\qquad(0\leq t\leq 1)\ .$$ Then $$\phi'(t)=\lim_{h\to0}{\phi(t+h)-\phi(t)\over h}=\lim_{h\to0}{f(x+te +he)-f(x+te)\over h}=\nabla_ef(x+te)\ .$$ Now "put it all together".