方向导数 directional derivative

最近在看花书，其中 4.3 中提到了方向导数

当 α = 0 \alpha = 0 α=0 时， ∂ ∂ α f ( x + α u ) = u T ∇ x f ( x ) \frac{\partial}{\partial \alpha}f(\textbf{x} + \alpha \textbf{u}) = \textbf{u}^T \nabla_xf(\textbf{x}) ∂α∂f(x+αu)=uT∇xf(x) 是怎么得出的？根据全微分
∂ f ( x + α u ) = ∂ f ( x 1 + α u 1 ) ∂ x 1 ⋅ ∂ ( x 1 + α u 1 ) + ⋯ + ∂ f ( x n + α u n ) ∂ x n ⋅ ∂ ( x n + α u n ) ∂ f ( x + α u ) ∂ α = ∂ f ( x 1 + α u 1 ) ∂ x 1 ⋅ ∂ ( x 1 + α u 1 ) ∂ α + ⋯ + ∂ f ( x n + α u n ) ∂ x n ⋅ ∂ ( x n + α u n ) ∂ α = ∂ f ( x 1 + α u 1 ) ∂ x 1 ⋅ u 1 + ⋯ + ∂ f ( x n + α u n ) ∂ x n ⋅ u n lim ⁡ α → 0 ∂ f ( x + α u ) ∂ α = ∂ f ( x 1 ) ∂ x 1 ⋅ u 1 + ⋯ + ∂ f ( x n ) ∂ x n ⋅ u n = u T ∇ x f ( x ) \partial f(\textbf{x} + \alpha \textbf{u}) = \frac{\partial f(x_1 + \alpha u_1)}{\partial x_1}·\partial(x_1 + \alpha u_1) + \dots + \frac{\partial f(x_n + \alpha u_n)}{\partial x_n}·\partial(x_n + \alpha u_n) \\ \begin{aligned} \frac{\partial f(\textbf{x} + \alpha \textbf{u})}{\partial \alpha} &= \frac{\partial f(x_1 + \alpha u_1)}{\partial x_1}·\frac{\partial(x_1 + \alpha u_1)}{\partial \alpha} + \dots + \frac{\partial f(x_n + \alpha u_n)}{\partial x_n}·\frac{\partial(x_n + \alpha u_n)}{\partial \alpha} \\ &= \frac{\partial f(x_1 + \alpha u_1)}{\partial x_1}·u_1 + \dots + \frac{\partial f(x_n + \alpha u_n)}{\partial x_n}·u_n \end{aligned} \\ \lim_{\alpha \rightarrow 0} \frac{\partial f(\textbf{x} + \alpha \textbf{u})}{\partial \alpha} = \frac{\partial f(x_1)}{\partial x_1}·u_1 + \dots + \frac{\partial f(x_n)}{\partial x_n}·u_n = \textbf{u}^T \nabla_xf(\textbf{x}) ∂f(x+αu)=∂x1∂f(x1+αu1)⋅∂(x1+αu1)+⋯+∂xn∂f(xn+αun)⋅∂(xn+αun)∂α∂f(x+αu)=∂x1∂f(x1+αu1)⋅∂α∂(x1+αu1)+⋯+∂xn∂f(xn+αun)⋅∂α∂(xn+αun)=∂x1∂f(x1+αu1)⋅u1+⋯+∂xn∂f(xn+αun)⋅unα→0lim∂α∂f(x+αu)=∂x1∂f(x1)⋅u1+⋯+∂xn∂f(xn)⋅un=uT∇xf(x)
Reference: https://www.youtube.com/watch?v=-DumtBiW4HE

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