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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #1251 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-y^{2}z^{4}cos(x)\\&\text{In the direction of }\overrightarrow{v}=(17,0,3).\end{align*}$

Possible Answers:

${17.3y^{2}z^{4}sin(x) - 69y^{2}z^{3}cos(x) - 34.5yz^{4}cos(x)}$

${0.98y^{2}z^{4}sin(x) - 0.68y^{2}z^{3}cos(x)}$

${17y^{2}z^{4}sin(x) - 12y^{2}z^{3}cos(x)}$

${138yz^{3}sin(x)}$

Correct answer:

${0.98y^{2}z^{4}sin(x) - 0.68y^{2}z^{3}cos(x)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(17)^2+(0)^2+(3)^2}=17.26\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{17}{17.26}=0.98\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{17.26}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{3}{17.26}=0.17\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.98)(y^{2}z^{4}sin(x))+(0)(-2yz^{4}cos(x))+(0.17)(-4y^{2}z^{3}cos(x))\\&D_{\overrightarrow{u}}(x,y,z)=0.98y^{2}z^{4}sin(x) - 0.68y^{2}z^{3}cos(x)\end{align*}$

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Example Question #1252 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3^{(y^{4})}e^{(x^{2})}e^{(z)}\\&\text{In the direction of }\overrightarrow{v}=(12,14,0).\end{align*}$

Possible Answers:

${162\cdot3^{(y^{4})}xy^{3}e^{(x^{2})}e^{(z)}}$

${1.3\cdot3^{(y^{4})}xe^{(x^{2})}e^{(z)} + 3.34\cdot3^{(y^{4})}y^{3}e^{(x^{2})}e^{(z)}}$

${0.845\cdot3^{(y^{4})}xe^{(x^{2})}e^{(z)} + 2.54\cdot3^{(y^{4})}y^{3}e^{(x^{2})}e^{(z)}}$

${18.4\cdot3^{(y^{4})}e^{(x^{2})}e^{(z)} + 36.9\cdot3^{(y^{4})}xe^{(x^{2})}e^{(z)} + 81\cdot3^{(y^{4})}y^{3}e^{(x^{2})}e^{(z)}}$

Correct answer:

${1.3\cdot3^{(y^{4})}xe^{(x^{2})}e^{(z)} + 3.34\cdot3^{(y^{4})}y^{3}e^{(x^{2})}e^{(z)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(12)^2+(14)^2+(0)^2}=18.44\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{12}{18.44}=0.65\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{14}{18.44}=0.76\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{18.44}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.65)(2\cdot3^{(y^{4})}xe^{(x^{2})}e^{(z)})+(0.76)(4.39\cdot3^{(y^{4})}y^{3}e^{(x^{2})}e^{(z)})+(0)(3^{(y^{4})}e^{(x^{2})}e^{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=1.3\cdot3^{(y^{4})}xe^{(x^{2})}e^{(z)} + 3.34\cdot3^{(y^{4})}y^{3}e^{(x^{2})}e^{(z)}\end{align*}$

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Example Question #1253 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2^{(y^{3})}sin(z^{3})ln(x)\\&\text{In the direction of }\overrightarrow{v}=(-19,-3,0).\end{align*}$

Possible Answers:

${-\frac{ (0.99\cdot2^{(y^{3})}sin(z^{3}))}{x}- 0.333\cdot2^{(y^{3})}y^{2}sin(z^{3})ln(x)}$

${\frac{(19.2\cdot2^{(y^{3})}sin(z^{3}))}{x}+ 57.7\cdot2^{(y^{3})}z^{2}cos(z^{3})ln(x) + 40\cdot2^{(y^{3})}y^{2}sin(z^{3})ln(x)}$

${-\frac{ (19\cdot2^{(y^{3})}sin(z^{3}))}{x}- 6.24\cdot2^{(y^{3})}y^{2}sin(z^{3})ln(x)}$

${\frac{(120\cdot2^{(y^{3})}y^{2}z^{2}cos(z^{3}))}{x}}$

Correct answer:

${-\frac{ (0.99\cdot2^{(y^{3})}sin(z^{3}))}{x}- 0.333\cdot2^{(y^{3})}y^{2}sin(z^{3})ln(x)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-19)^2+(-3)^2+(0)^2}=19.24\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-19}{19.24}=-0.99\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-3}{19.24}=-0.16\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{19.24}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.99)(\frac{(2^{(y^{3})}sin(z^{3}))}{x})+(-0.16)(2.08\cdot2^{(y^{3})}y^{2}sin(z^{3})ln(x))+(0)(3\cdot2^{(y^{3})}z^{2}cos(z^{3})ln(x))\\&D_{\overrightarrow{u}}(x,y,z)=-\frac{ (0.99\cdot2^{(y^{3})}sin(z^{3}))}{x}- 0.333\cdot2^{(y^{3})}y^{2}sin(z^{3})ln(x)\end{align*}$

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Example Question #1254 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=x^{2}ln(y^{3})sin(z)\\&\text{In the direction of }\overrightarrow{v}=(-13,0,9).\end{align*}$

Possible Answers:

${0.57x^{2}ln(y^{3})cos(z) - 1.64xln(y^{3})sin(z)}$

${31.6xln(y^{3})sin(z) + 15.8x^{2}ln(y^{3})cos(z) +\frac{ (47.4x^{2}sin(z))}{y}}$

${\frac{(94.9xcos(z))}{y}}$

${1.34xln(y^{3})sin(z) + 0.325x^{2}ln(y^{3})cos(z)}$

Correct answer:

${0.57x^{2}ln(y^{3})cos(z) - 1.64xln(y^{3})sin(z)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-13)^2+(0)^2+(9)^2}=15.81\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-13}{15.81}=-0.82\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{15.81}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{9}{15.81}=0.57\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.82)(2xln(y^{3})sin(z))+(0)(\frac{(3x^{2}sin(z))}{y})+(0.57)(x^{2}ln(y^{3})cos(z))\\&D_{\overrightarrow{u}}(x,y,z)=0.57x^{2}ln(y^{3})cos(z) - 1.64xln(y^{3})sin(z)\end{align*}$

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Example Question #1255 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=ln(x^{2})e^{(y^{3})}e^{(z^{2})}\\&\text{In the direction of }\overrightarrow{v}=(0,-10,-2).\end{align*}$

Possible Answers:

${- 0.4zln(x^{2})e^{(y^{3})}e^{(z^{2})} - 2.94y^{2}ln(x^{2})e^{(y^{3})}e^{(z^{2})}}$

${0.08zln(x^{2})e^{(y^{3})}e^{(z^{2})} + 2.88y^{2}ln(x^{2})e^{(y^{3})}e^{(z^{2})}}$

${\frac{(20.4e^{(y^{3})}e^{(z^{2})})}{x}+ 20.4zln(x^{2})e^{(y^{3})}e^{(z^{2})} + 30.6y^{2}ln(x^{2})e^{(y^{3})}e^{(z^{2})}}$

${\frac{(122y^{2}ze^{(y^{3})}e^{(z^{2})})}{x}}$

Correct answer:

${- 0.4zln(x^{2})e^{(y^{3})}e^{(z^{2})} - 2.94y^{2}ln(x^{2})e^{(y^{3})}e^{(z^{2})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-10)^2+(-2)^2}=10.2\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{10.2}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-10}{10.2}=-0.98\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-2}{10.2}=-0.2\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{(2e^{(y^{3})}e^{(z^{2})})}{x})+(-0.98)(3y^{2}ln(x^{2})e^{(y^{3})}e^{(z^{2})})+(-0.2)(2zln(x^{2})e^{(y^{3})}e^{(z^{2})})\\&D_{\overrightarrow{u}}(x,y,z)=- 0.4zln(x^{2})e^{(y^{3})}e^{(z^{2})} - 2.94y^{2}ln(x^{2})e^{(y^{3})}e^{(z^{2})}\end{align*}$

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Example Question #1256 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-ln(z^{4})e^{(y^{3})}\\&\text{In the direction of }\overrightarrow{v}=(1,0,16).\end{align*}$

Possible Answers:

${-\frac{(64e^{(y^{3})})}{z}}$

${-\frac{ (64.1e^{(y^{3})})}{z}- 48.1y^{2}ln(z^{4})e^{(y^{3})}}$

${-\frac{(4e^{(y^{3})})}{z}}$

{0}

Correct answer:

${-\frac{(4e^{(y^{3})})}{z}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(1)^2+(0)^2+(16)^2}=16.03\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{1}{16.03}=0.06\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{16.03}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{16}{16.03}=1\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.06)(0.0)+(0)(-3y^{2}ln(z^{4})e^{(y^{3})})+(1)(-\frac{(4e^{(y^{3})})}{z})\\&D_{\overrightarrow{u}}(x,y,z)=-\frac{(4e^{(y^{3})})}{z}\end{align*}$

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Example Question #1257 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-3^{(y^{2})}ln(z^{3})\\&\text{In the direction of }\overrightarrow{v}=(0,-7,7).\end{align*}$

Possible Answers:

{0}

${1.56\cdot3^{(y^{2})}yln(z^{3}) -\frac{ (2.13\cdot3^{(y^{2})})}{z}}$

${-\frac{ (29.7\cdot3^{(y^{2})})}{z}- 21.8\cdot3^{(y^{2})}yln(z^{3})}$

${15.4\cdot3^{(y^{2})}yln(z^{3}) -\frac{ (21\cdot3^{(y^{2})})}{z}}$

Correct answer:

${1.56\cdot3^{(y^{2})}yln(z^{3}) -\frac{ (2.13\cdot3^{(y^{2})})}{z}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-7)^2+(7)^2}=9.9\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{9.9}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-7}{9.9}=-0.71\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{7}{9.9}=0.71\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(0.0)+(-0.71)(-2.2\cdot3^{(y^{2})}yln(z^{3}))+(0.71)(-\frac{(3\cdot3^{(y^{2})})}{z})\\&D_{\overrightarrow{u}}(x,y,z)=1.56\cdot3^{(y^{2})}yln(z^{3}) -\frac{ (2.13\cdot3^{(y^{2})})}{z}\end{align*}$

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Example Question #3621 : Calculus 3

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3^{(x^{2})}\cdot3^ze^{(y^{2})}\\&\text{In the direction of }\overrightarrow{v}=(19,0,16).\end{align*}$

Possible Answers:

${120\cdot3^z\cdot3^{(x^{2})}xye^{(y^{2})}}$

${17.6\cdot3^z\cdot3^{(x^{2})}e^{(y^{2})} + 41.7\cdot3^z\cdot3^{(x^{2})}xe^{(y^{2})}}$

${27.3\cdot3^z\cdot3^{(x^{2})}e^{(y^{2})} + 54.6\cdot3^z\cdot3^{(x^{2})}xe^{(y^{2})} + 49.7\cdot3^z\cdot3^{(x^{2})}ye^{(y^{2})}}$

${0.703\cdot3^z\cdot3^{(x^{2})}e^{(y^{2})} + 1.67\cdot3^z\cdot3^{(x^{2})}xe^{(y^{2})}}$

Correct answer:

${0.703\cdot3^z\cdot3^{(x^{2})}e^{(y^{2})} + 1.67\cdot3^z\cdot3^{(x^{2})}xe^{(y^{2})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(19)^2+(0)^2+(16)^2}=24.84\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{19}{24.84}=0.76\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{24.84}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{16}{24.84}=0.64\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.76)(2.2\cdot3^z\cdot3^{(x^{2})}xe^{(y^{2})})+(0)(2\cdot3^z\cdot3^{(x^{2})}ye^{(y^{2})})+(0.64)(1.1\cdot3^z\cdot3^{(x^{2})}e^{(y^{2})})\\&D_{\overrightarrow{u}}(x,y,z)=0.703\cdot3^z\cdot3^{(x^{2})}e^{(y^{2})} + 1.67\cdot3^z\cdot3^{(x^{2})}xe^{(y^{2})}\end{align*}$

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Example Question #1258 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-4^{(z^{4})}e^{(y^{3})}e^{(x)}\\&\text{In the direction of }\overrightarrow{v}=(0,9,-11).\end{align*}$

Possible Answers:

${- 1.19\cdot4^{(z^{4})}y^{2}e^{(y^{3})}e^{(x)} - 3.29\cdot4^{(z^{4})}z^{3}e^{(y^{3})}e^{(x)}}$

${- 14.2\cdot4^{(z^{4})}e^{(y^{3})}e^{(x)} - 42.6\cdot4^{(z^{4})}y^{2}e^{(y^{3})}e^{(x)} - 78.8\cdot4^{(z^{4})}z^{3}e^{(y^{3})}e^{(x)}}$

${4.27\cdot4^{(z^{4})}z^{3}e^{(y^{3})}e^{(x)} - 1.89\cdot4^{(z^{4})}y^{2}e^{(y^{3})}e^{(x)}}$

${-236\cdot4^{(z^{4})}y^{2}z^{3}e^{(y^{3})}e^{(x)}}$

Correct answer:

${4.27\cdot4^{(z^{4})}z^{3}e^{(y^{3})}e^{(x)} - 1.89\cdot4^{(z^{4})}y^{2}e^{(y^{3})}e^{(x)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(9)^2+(-11)^2}=14.21\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{14.21}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{9}{14.21}=0.63\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-11}{14.21}=-0.77\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-1\cdot4^{(z^{4})}e^{(y^{3})}e^{(x)})+(0.63)(-3\cdot4^{(z^{4})}y^{2}e^{(y^{3})}e^{(x)})+(-0.77)(-5.55\cdot4^{(z^{4})}z^{3}e^{(y^{3})}e^{(x)})\\&D_{\overrightarrow{u}}(x,y,z)=4.27\cdot4^{(z^{4})}z^{3}e^{(y^{3})}e^{(x)} - 1.89\cdot4^{(z^{4})}y^{2}e^{(y^{3})}e^{(x)}\end{align*}$

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Example Question #193 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=x^{2}e^{(y^{4})}e^{(z^{2})}\\&\text{In the direction of }\overrightarrow{v}=(0,16,-16).\end{align*}$

Possible Answers:

${2.84x^{2}y^{3}e^{(y^{4})}e^{(z^{2})} - 1.42x^{2}ze^{(y^{4})}e^{(z^{2})}}$

${45.3xe^{(y^{4})}e^{(z^{2})} + 45.3x^{2}ze^{(y^{4})}e^{(z^{2})} + 90.5x^{2}y^{3}e^{(y^{4})}e^{(z^{2})}}$

${1.01x^{2}ze^{(y^{4})}e^{(z^{2})} + 2.02x^{2}y^{3}e^{(y^{4})}e^{(z^{2})}}$

${362xy^{3}ze^{(y^{4})}e^{(z^{2})}}$

Correct answer:

${2.84x^{2}y^{3}e^{(y^{4})}e^{(z^{2})} - 1.42x^{2}ze^{(y^{4})}e^{(z^{2})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(16)^2+(-16)^2}=22.63\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{22.63}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{16}{22.63}=0.71\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-16}{22.63}=-0.71\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(2xe^{(y^{4})}e^{(z^{2})})+(0.71)(4x^{2}y^{3}e^{(y^{4})}e^{(z^{2})})+(-0.71)(2x^{2}ze^{(y^{4})}e^{(z^{2})})\\&D_{\overrightarrow{u}}(x,y,z)=2.84x^{2}y^{3}e^{(y^{4})}e^{(z^{2})} - 1.42x^{2}ze^{(y^{4})}e^{(z^{2})}\end{align*}$

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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #1251 : Partial Derivatives

Example Question #1252 : Partial Derivatives

Example Question #1253 : Partial Derivatives

Example Question #1254 : Partial Derivatives

Example Question #1255 : Partial Derivatives

Example Question #1256 : Partial Derivatives

Example Question #1257 : Partial Derivatives

Example Question #3621 : Calculus 3

Example Question #1258 : Partial Derivatives

Example Question #193 : Directional Derivatives

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