In Exercises 21-28, find the directional derivative of the...

In Exercises $21-28$, find the directional derivative of the given function at the specified point $P$ and in the direction of the given unit vector $\mathbf{u}$. $$ f(x, y)=\sqrt{\frac{y}{x}} \text { at } P=(4,9), \mathbf{u}=\left\langle-\frac{\sqrt{17}}{17},-\frac{4 \sqrt{17}}{17}\right\rangle $$

In Exercises $21-28$, find the directional derivative of the given function at the specified point $P$ and in the direction of the given unit vector $\mathbf{u}$.
$$
f(x, y)=\sqrt{\frac{y}{x}} \text { at } P=(4,9), \mathbf{u}=\left\langle-\frac{\sqrt{17}}{17},-\frac{4 \sqrt{17}}{17}\right\rangle
$$

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Key Concepts

Directional Derivative

The directional derivative measures the rate at which a function changes at a given point when moving in a specific direction. It is calculated by taking the dot product of the gradient of the function and a unit vector in the direction of interest.

Gradient

The gradient is a vector composed of the partial derivatives of a function with respect to its variables. It indicates the direction of the steepest ascent and its magnitude represents the rate of increase of the function in that direction.

Partial Derivative

Partial derivatives are the derivatives of a multivariable function with respect to one variable at a time, holding the other variables constant. They form the components of the gradient vector.

Unit Vector

A unit vector has a magnitude of one and is used to specify direction without scaling. In the context of directional derivatives, employing a unit vector ensures that the computed rate of change is solely influenced by the function's behavior and not by the length of the vector.

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