Short Answer

Find the directional derivative of the given function at the specified point P in the direction of the given vector. Note: The given vectors may not be unit vectors.
$f (x, y, z) = \sqrt{\frac{x y}{z}}, P = (2, 3, 1), v = ⟨ 2, 1, - 2 ⟩$

$D_{u} f (2, 3, 1) = \frac{10}{\sqrt{6}}$

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Step 1. Given Information

$The directional derivative of a function f (x, y, z) at a point P = (x_{0}, y_{0}, z_{0}) in the direction of a unit vector u = ⟨ a, b, c ⟩ is given by D_{u} f (x_{0}, y_{0}, z_{0}) = lim_{h \to 0} \frac{f (x_{0} + a h, y_{0} + b h, z_{0} + c h) - f (x_{0}, y_{0}, z_{0})}{h} Here f (x, y, z) = \sqrt{\frac{x y}{z}}, and P = (2, 3, 1) . A l s o t h e v e c t o r i s v = (2, 1, - 2)$

Step 2. Solution

$The unit vector " u " in the direction of " v " is given by \begin{array}{l} u = \frac{1}{∥ v ∥} v \\ = \frac{1}{\sqrt{2^{2} + 1^{2} + (- 2)^{2}}} ⟨ 2, 1, - 2 ⟩ \\ = \frac{1}{3} ⟨ 2, 1, - 2 ⟩ \\ N o w b y d e f i n i t i o n : \\ \begin{array}{l} D_{u} f (2, 3, 1) = lim_{h \to 0} \frac{f (2 + \frac{2 h}{3}, 3 + \frac{h}{3}, 1 - \frac{2 h}{3}) - f (2, 3, 1)}{h} \\ = lim_{h \to 0} \frac{\sqrt{(2 + \frac{2 h}{3}) (3 + \frac{h}{3}) \div (1 - \frac{2 h}{3})} - \sqrt{\frac{6}{1}}}{h} \\ = lim_{h \to 0} \frac{\sqrt{6 + \frac{8 h}{3} + \frac{2 h}{9} - \sqrt{6 - 4 h}}}{h \sqrt{1 - \frac{2 h}{3}}} \\ R a t i o n a l i z e t h e n u m e r a t o r g i v e s : \\ \begin{array}{l} D_{u} f (2, 3, 1) = lim_{h \to 0} \frac{\sqrt{6 + \frac{8 h}{3} + \frac{2 h^{2}}{9}} - \sqrt{6 - 4 h}}{h \sqrt{1 - \frac{2 h}{3}}} \times \frac{\sqrt{6 + \frac{8 h}{3} + \frac{2 h^{2}}{9}} + \sqrt{6 - 4 h}}{\sqrt{6 + \frac{8 h}{3} + \frac{2 h^{2}}{9}} + \sqrt{6 - 4 h}} \\ = lim_{h \to 0} \frac{6 + \frac{8 h}{3} - \frac{2 h^{2}}{9} - 6 + 4 h}{h \sqrt{1 - \frac{2 h}{3}} (\sqrt{6 + \frac{8 h}{3} + \frac{2 h^{2}}{9}} + \sqrt{6 - 4 h})} \\ = lim_{h \to 0} \frac{\sqrt{\frac{20}{3} - \frac{2}{9} h}}{\sqrt{1 - \frac{2 h}{3}} (\sqrt{6 + \frac{8 h}{3} + \frac{2 h^{2}}{9}} + \sqrt{6 - 4 h})} \\ T a k i n g l i m i t a s h \to 0 g i v e s : D_{u} f (2, 3, 1) = \frac{10}{\sqrt{6}} \end{array} \end{array} \end{array}$

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Find the directional derivative of the given function at the specified point P in the direction of the given vector. Note: The given vectors may not be unit vectors.
$f (x, y, z) = \sqrt{\frac{x y}{z}}, P = (2, 3, 1), v = ⟨ 2, 1, - 2 ⟩$

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Find the directional derivative of the given function at the specified point P in the direction of the given vector. Note: The given vectors may not be unit vectors. f(x,y,z)=xyz,P=(2,3,1),v=⟨2,1,−2⟩

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Find the directional derivative of the given function at the specified point P in the direction of the given vector. Note: The given vectors may not be unit vectors.
$f (x, y, z) = \sqrt{\frac{x y}{z}}, P = (2, 3, 1), v = ⟨ 2, 1, - 2 ⟩$