Find the directional derivative of the function at the given point in the direction of the vector v. f(x, y, z) = xe^y + ye^z + ze^x, (0, 0, 0), v = 4, 3, −2

Answer:[tex]D_{\vec{u}}f(0,0,0)=\frac{5}{\sqrt{29}}[/tex]Step-by-step explanation:We need to find the directional derivative of the function at the given point in the direction of the vector v. [tex]f(x, y, z)=xe^{y} + ye^{z} + ze^{x}[/tex] ,point (0, 0, 0) and [tex]v=<4, 3, -2>[/tex]  By Theorem: If f is a differentiable function of x , y and z , then f has a directional derivative for  any unit vector [tex]\overrightarrow{v} =<v_1,v_2,v_3>[/tex] and [tex]D_{\overrightarrow{u}}f(x,y,z)=f_{x}(x,y,z)u_1+f_{y}(x,y,z)u_2+f_{z}(x,y,z)u_3[/tex] where [tex]\overrightarrow{u}=\frac{\overrightarrow{v}}{||v||}[/tex] since,  [tex]v=<4, 3, -2>[/tex] then [tex]\overrightarrow{u}=\frac{\overrightarrow{v}}{||v||}[/tex] [tex]\overrightarrow{u}=< \frac{4}{\sqrt{4^{2}+3^{2}+(-2)^{2}}},\frac{3}{\sqrt{4^{2}+3^{2}+(-2)^{2}}},\frac{-2}{\sqrt{4^{2}+3^{2}+(-2)^{2}}} >[/tex] [tex]\overrightarrow{u}=< \frac{4}{\sqrt{29}},\frac{3}{\sqrt{29}},\frac{-2}{\sqrt{29} }>[/tex] The partial derivatives are[tex]f_{x}(x,y,z)=e^{y}+ze^{x}[/tex]   [tex]f_{y}(x,y,z)=xe^{y}+e^{z}[/tex][tex]f_{z}(x,y,z)=ye^{z}+e^{x}[/tex]Then the directional derivative is [tex]D_{\vec{u}}f(x,y,z)=(e^{y}+ze^{x})(\frac{4}{\sqrt{29}})+(xe^{y}+e^{z})(\frac{3}{\sqrt{29}})+(ye^{z}+e^{x})(\frac{-2}{\sqrt{29}})[/tex]so, directional derivative at point (0,0,0)[tex]D_{\vec{u}}f(0,0,0)=(e^{0}+0e^{0})(\frac{4}{\sqrt{29}})+(0e^{0}+e^{0})(\frac{3}{\sqrt{29}})+(0e^{0}+e^{0})(\frac{-2}{\sqrt{29}})[/tex][tex]D_{\vec{u}}f(0,0,0)=\frac{4}{\sqrt{29}}+\frac{3}{\sqrt{29}}+\frac{-2}{\sqrt{29}}[/tex][tex]D_{\vec{u}}f(0,0,0)=\frac{4+3-2}{\sqrt{29}}[/tex][tex]D_{\vec{u}}f(0,0,0)=\frac{5}{\sqrt{29}}[/tex]