[proofplan] Let $L = Df_a$ be the linear derivative map supplied by differentiability of $f$ at $a$. First we show directly from the definition of differentiability that, for every direction $u \in \mathbb{R}^n$, the [directional derivative](/page/Directional%20Derivative) $D_u f(a)$ exists and equals $L(u)$. Applying this identity to $u = \alpha v+\beta w$, to $u=v$, and to $u=w$, the desired formula follows from the linearity of $L$. [/proofplan]

[step:Identify each directional derivative with the derivative map applied to the direction]Since $f$ is differentiable at $a$, there exists a [linear map](/page/Linear%20Map) \begin{align*} L := Df_a: \mathbb{R}^n \to \mathbb{R} \end{align*} such that \begin{align*} \lim_{h \to 0} \frac{f(a+h)-f(a)-L(h)}{|h|} = 0. \end{align*} Fix $u \in \mathbb{R}^n$. Because $U$ is open and $a \in U$, there exists $\varepsilon > 0$ such that $a+tu \in U$ whenever $|t|<\varepsilon$. If $u=0$, then \begin{align*} \frac{f(a+t0)-f(a)}{t} = 0 \end{align*} for all $0<|t|<\varepsilon$, so $D_0 f(a)=0=L(0)$. Assume now that $u \neq 0$. For $0<|t|<\varepsilon$, apply the differentiability expansion with $h=tu$: \begin{align*} \frac{f(a+tu)-f(a)}{t} = L(u) + \frac{f(a+tu)-f(a)-L(tu)}{t}. \end{align*} Taking absolute values of the error term and using $|tu|=|t|\,|u|$, we get \begin{align*} \left|\frac{f(a+tu)-f(a)-L(tu)}{t}\right| = |u|\,\left|\frac{f(a+tu)-f(a)-L(tu)}{|tu|}\right|. \end{align*} As $t \to 0$, we have $tu \to 0$, so the last expression tends to $0$ by differentiability at $a$. Hence \begin{align*} D_u f(a) = \lim_{t \to 0} \frac{f(a+tu)-f(a)}{t} = L(u). \end{align*}[/step]

[guided]The goal of this step is to justify the bridge between two notions: the Fréchet derivative $Df_a$, which is a linear map on $\mathbb{R}^n$, and the directional derivative $D_u f(a)$, which is a one-variable derivative along the line through $a$ in direction $u$. Since $f$ is differentiable at $a$, the definition gives a linear map \begin{align*} L := Df_a: \mathbb{R}^n \to \mathbb{R} \end{align*} with the approximation property \begin{align*} \lim_{h \to 0} \frac{f(a+h)-f(a)-L(h)}{|h|} = 0. \end{align*} This means that $L(h)$ is the first-order part of the increment $f(a+h)-f(a)$, and the remaining error is negligible compared with $|h|$. Fix a direction $u \in \mathbb{R}^n$. Because $U$ is open and $a \in U$, there is some radius $r>0$ such that $B(a,r) \subset U$. If $u \neq 0$, choose $\varepsilon = r/|u|$; then $|t|<\varepsilon$ implies $|tu|<r$, hence $a+tu \in U$. If $u=0$, any positive $\varepsilon$ works. Thus the one-variable difference quotient defining $D_u f(a)$ is defined for all sufficiently small nonzero $t$. First consider the zero direction. If $u=0$, then for every nonzero $t$ sufficiently close to $0$, \begin{align*} \frac{f(a+t0)-f(a)}{t} = \frac{f(a)-f(a)}{t} = 0. \end{align*} Therefore $D_0 f(a)=0$. Since $L$ is linear, $L(0)=0$, so $D_0 f(a)=L(0)$. Now assume $u \neq 0$. For $0<|t|<\varepsilon$, set $h=tu$. The differentiability expansion at $a$ gives \begin{align*} f(a+tu)-f(a) = L(tu) + \bigl(f(a+tu)-f(a)-L(tu)\bigr). \end{align*} Divide by $t$. Since $L$ is linear, $L(tu)=tL(u)$, and therefore \begin{align*} \frac{f(a+tu)-f(a)}{t} = L(u) + \frac{f(a+tu)-f(a)-L(tu)}{t}. \end{align*} It remains to prove that the second term tends to $0$ as $t \to 0$. Taking absolute values and using $|tu|=|t|\,|u|$, we obtain \begin{align*} \left|\frac{f(a+tu)-f(a)-L(tu)}{t}\right| = |u|\,\left|\frac{f(a+tu)-f(a)-L(tu)}{|tu|}\right|. \end{align*} As $t \to 0$, the vector $tu$ tends to $0$ in $\mathbb{R}^n$. The defining limit for differentiability therefore implies \begin{align*} \left|\frac{f(a+tu)-f(a)-L(tu)}{|tu|}\right| \to 0. \end{align*} Multiplication by the fixed number $|u|$ preserves convergence to $0$, so the error term in the directional difference quotient vanishes. Hence \begin{align*} D_u f(a) = \lim_{t \to 0} \frac{f(a+tu)-f(a)}{t} = L(u). \end{align*} This proves, for every $u \in \mathbb{R}^n$, that the directional derivative exists and is obtained by applying the derivative map $Df_a$ to the direction $u$.[/guided]

[step:Apply linearity of the derivative map to the chosen directions] Apply the identity from the previous step with $u=\alpha v+\beta w$, $u=v$, and $u=w$. We obtain \begin{align*} D_{\alpha v+\beta w} f(a) = L(\alpha v+\beta w), \end{align*} \begin{align*} D_v f(a) = L(v), \end{align*} and \begin{align*} D_w f(a) = L(w). \end{align*} Since $L: \mathbb{R}^n \to \mathbb{R}$ is linear, \begin{align*} L(\alpha v+\beta w) = \alpha L(v)+\beta L(w). \end{align*} Substituting the three directional-derivative identities gives \begin{align*} D_{\alpha v+\beta w} f(a) = \alpha D_v f(a)+\beta D_w f(a). \end{align*} This is the desired linearity in the direction. [/step]