Assume $\displaystyle{(x, y)}$ represents the position of an object and $\displaystyle{f(x,y)}$ is a scalar field on the $\displaystyle{x}$ $\displaystyle{y}$ plane. Then $\displaystyle{\frac{\partial f}{\partial x}}$ represents the change of $\displaystyle{f}$ per unit length along the positive $\displaystyle{x}$ direction. In other words, it is the spatial rate of change of $\displaystyle{f}$ along the $\displaystyle{x}$ direction.

Similarly, derivative $\displaystyle{\frac{\partial f}{\partial y}}$ represents the spatial rate of change of $\displaystyle{f}$ along the $\displaystyle{y}$ direction.

For an arbitrary direction, due to the nature of displacement, the change of $\displaystyle{f}$ is $\displaystyle{\delta f = \frac{\partial f}{\partial x} \delta x + \frac{\partial f}{\partial x} \delta y}$ when the object has finished moving $\displaystyle{\delta x}$ in $\displaystyle{x}$ direction and then $\displaystyle{\delta y}$ in $\displaystyle{y}$ direction.

Then, the spatial rate of change of $\displaystyle{f}$ is \displaystyle{ \begin{aligned} &\frac{\delta f}{\sqrt{(\delta x)^2 + (\delta y)^2}} \\ &= \frac{\partial f}{\partial x} \frac{\delta x}{\sqrt{(\delta x)^2 + (\delta y)^2}} + \frac{\partial f}{\partial x} \frac{\delta y}{\sqrt{(\delta x)^2 + (\delta y)^2}} \\ \end{aligned} }

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For simplicity, denote the resultant displacement as $\displaystyle{\mathbf{v}}$: $\displaystyle{\mathbf{v} = (\delta x, \delta y)}$

and define $\displaystyle{\nabla f(x)}$ as $\displaystyle{\left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)}$

Then, the change of the $\displaystyle{f}$ due to the displacement $\displaystyle{\mathbf{v}}$ is \displaystyle{\begin{aligned} \left(\delta f\right)_{\mathbf{v}} &= \frac{\partial f}{\partial x} \delta x + \frac{\partial f}{\partial x} \delta y \\ &= \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial x}\right) \cdot (\delta x, \delta y) \\ &= \left(\nabla f\right) \cdot \mathbf{v} \\ \end{aligned}}

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So the spatial rate of change $\displaystyle{f}$ along the direction of the vector $\displaystyle{\mathbf{v}}$ is \displaystyle{\begin{aligned} D_{\mathbf{v}}(f) &= \frac{\left(\delta f\right)_{\mathbf{v}}}{|\mathbf{v}|} \\ &= \frac{\partial f}{\partial x} \frac{\delta x}{\sqrt{(\delta x)^2 + (\delta y)^2}} + \frac{\partial f}{\partial x} \frac{\delta y}{\sqrt{(\delta x)^2 + (\delta y)^2}} \\ &= \left(\nabla f\right) \cdot \frac{\mathbf{v}}{|\mathbf{v}|} \\ &= \left(\nabla f\right) \cdot \hat{\mathbf{v}} \\ \end{aligned}} $\displaystyle{D_{\mathbf{v}}(f)}$ is called directional derivative.

— Me@2016-02-06 09:49:22 PM

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This is the reason that $\displaystyle{\nabla f}$ is in the steepest direction.

If $\displaystyle{\hat{\mathbf{v}}}$ is chosen to be parallel to $\displaystyle{\nabla f}$, the directional derivative $\displaystyle{\left(\nabla f\right) \cdot \hat{\mathbf{v}}}$ would be maximized.

— Me@2021-08-20 05:20:02 PM

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