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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #1381 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=x^{4}ln(y^{2})e^{(z)} - cos(y^{2})cos(z)ln(x)\\&\text{In the direction of }\overrightarrow{v}=(-20,14,0).\end{align*}$

Possible Answers:

${\frac{(0.82cos(y^{2})cos(z))}{x}- 3.28x^{3}ln(y^{2})e^{(z)} +\frac{ (1.14x^{4}e^{(z)})}{y}+ 1.14ysin(y^{2})cos(z)ln(x)}$

${97.6x^{3}ln(y^{2})e^{(z)} -\frac{ (24.4cos(y^{2})cos(z))}{x}+ 24.4x^{4}ln(y^{2})e^{(z)} + 24.4cos(y^{2})ln(x)sin(z) +\frac{ (48.8x^{4}e^{(z)})}{y}+ 48.8ysin(y^{2})cos(z)ln(x)}$

${\frac{(20cos(y^{2})cos(z))}{x}- 80x^{3}ln(y^{2})e^{(z)} +\frac{ (28x^{4}e^{(z)})}{y}+ 28ysin(y^{2})cos(z)ln(x)}$

${\frac{(195x^{3}e^{(z)})}{y}-\frac{ (48.8ysin(y^{2})sin(z))}{x}}$

Correct answer:

${\frac{(0.82cos(y^{2})cos(z))}{x}- 3.28x^{3}ln(y^{2})e^{(z)} +\frac{ (1.14x^{4}e^{(z)})}{y}+ 1.14ysin(y^{2})cos(z)ln(x)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-20)^2+(14)^2+(0)^2}=24.41\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-20}{24.41}=-0.82\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{14}{24.41}=0.57\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{24.41}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.82)(4x^{3}ln(y^{2})e^{(z)} -\frac{ (cos(y^{2})cos(z))}{x})+(0.57)(\frac{(2x^{4}e^{(z)})}{y}+ 2ysin(y^{2})cos(z)ln(x))+(0)(x^{4}ln(y^{2})e^{(z)} + cos(y^{2})ln(x)sin(z))\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(0.82cos(y^{2})cos(z))}{x}- 3.28x^{3}ln(y^{2})e^{(z)} +\frac{ (1.14x^{4}e^{(z)})}{y}+ 1.14ysin(y^{2})cos(z)ln(x)\end{align*}$

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Example Question #1382 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2x^{2}ln(z^{2})e^{(y)} - 3^{(z^{4})}x^{4}sin(y)\\&\text{In the direction of }\overrightarrow{v}=(-16,0,-16).\end{align*}$

Possible Answers:

${2.84\cdot 3^{(z^{4})}x^{3}sin(y) -\frac{ (2.84x^{2}e^{(y)})}{z}- 2.84xln(z^{2})e^{(y)} + 3.12\cdot 3^{(z^{4})}x^{4}z^{3}sin(y)}$

${64\cdot 3^{(z^{4})}x^{3}sin(y) -\frac{ (64x^{2}e^{(y)})}{z}- 64xln(z^{2})e^{(y)} + 70.3\cdot 3^{(z^{4})}x^{4}z^{3}sin(y)}$

${\frac{(181xe^{(y)})}{z}- 398\cdot 3^{(z^{4})}x^{3}z^{3}cos(y)}$

${45.3x^{2}ln(z^{2})e^{(y)} - 22.6\cdot 3^{(z^{4})}x^{4}cos(y) - 90.5\cdot 3^{(z^{4})}x^{3}sin(y) +\frac{ (90.5x^{2}e^{(y)})}{z}+ 90.5xln(z^{2})e^{(y)} - 99.4\cdot 3^{(z^{4})}x^{4}z^{3}sin(y)}$

Correct answer:

${2.84\cdot 3^{(z^{4})}x^{3}sin(y) -\frac{ (2.84x^{2}e^{(y)})}{z}- 2.84xln(z^{2})e^{(y)} + 3.12\cdot 3^{(z^{4})}x^{4}z^{3}sin(y)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-16)^2+(0)^2+(-16)^2}=22.63\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-16}{22.63}=-0.71\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{22.63}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-16}{22.63}=-0.71\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.71)(4xln(z^{2})e^{(y)} - 4\cdot 3^{(z^{4})}x^{3}sin(y))+(0)(2x^{2}ln(z^{2})e^{(y)} - 1\cdot 3^{(z^{4})}x^{4}cos(y))+(-0.71)(\frac{(4x^{2}e^{(y)})}{z}- 4.39\cdot 3^{(z^{4})}x^{4}z^{3}sin(y))\\&D_{\overrightarrow{u}}(x,y,z)=2.84\cdot 3^{(z^{4})}x^{3}sin(y) -\frac{ (2.84x^{2}e^{(y)})}{z}- 2.84xln(z^{2})e^{(y)} + 3.12\cdot 3^{(z^{4})}x^{4}z^{3}sin(y)\end{align*}$

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Example Question #1383 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-4^xz^{3}sin(y)\\&\text{In the direction of }\overrightarrow{v}=(0,20,-2).\end{align*}$

Possible Answers:

${-83.6\cdot 4^xz^{2}cos(y)}$

${6\cdot 4^xz^{2}sin(y) - 20\cdot 4^xz^{3}cos(y)}$

${- 20.1\cdot 4^xz^{3}cos(y) - 60.3\cdot 4^xz^{2}sin(y) - 27.9\cdot 4^xz^{3}sin(y)}$

${0.3\cdot 4^xz^{2}sin(y) - 1\cdot 4^xz^{3}cos(y)}$

Correct answer:

${0.3\cdot 4^xz^{2}sin(y) - 1\cdot 4^xz^{3}cos(y)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(20)^2+(-2)^2}=20.1\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{20.1}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{20}{20.1}=1\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-2}{20.1}=-0.1\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-1.39\cdot 4^xz^{3}sin(y))+(1)(-1\cdot 4^xz^{3}cos(y))+(-0.1)(-3\cdot 4^xz^{2}sin(y))\\&D_{\overrightarrow{u}}(x,y,z)=0.3\cdot 4^xz^{2}sin(y) - 1\cdot 4^xz^{3}cos(y)\end{align*}$

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Example Question #317 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- 4y^{4}cos(z) - ln(x^{3})ln(y^{3})e^{(z)}\\&\text{In the direction of }\overrightarrow{v}=(-10,0,-9).\end{align*}$

Possible Answers:

${\frac{(30ln(y^{3})e^{(z)})}{x}- 36y^{4}sin(z) + 9ln(x^{3})ln(y^{3})e^{(z)}}$

${-\frac{(121e^{(z)})}{(xy)}}$

${\frac{(2.22ln(y^{3})e^{(z)})}{x}- 2.68y^{4}sin(z) + 0.67ln(x^{3})ln(y^{3})e^{(z)}}$

${53.8y^{4}sin(z) - 215y^{3}cos(z) -\frac{ (40.3ln(y^{3})e^{(z)})}{x}-\frac{ (40.3ln(x^{3})e^{(z)})}{y}- 13.4ln(x^{3})ln(y^{3})e^{(z)}}$

Correct answer:

${\frac{(2.22ln(y^{3})e^{(z)})}{x}- 2.68y^{4}sin(z) + 0.67ln(x^{3})ln(y^{3})e^{(z)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-10)^2+(0)^2+(-9)^2}=13.45\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-10}{13.45}=-0.74\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{13.45}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-9}{13.45}=-0.67\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.74)(-\frac{(3ln(y^{3})e^{(z)})}{x})+(0)(- 16y^{3}cos(z) -\frac{ (3ln(x^{3})e^{(z)})}{y})+(-0.67)(4y^{4}sin(z) - ln(x^{3})ln(y^{3})e^{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(2.22ln(y^{3})e^{(z)})}{x}- 2.68y^{4}sin(z) + 0.67ln(x^{3})ln(y^{3})e^{(z)}\end{align*}$

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Example Question #1384 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=e^{(x^{3})}e^{(y^{4})}e^{(z^{4})}\\&\text{In the direction of }\overrightarrow{v}=(-6,-3,0).\end{align*}$

Possible Answers:

${- 2.67x^{2}e^{(x^{3})}e^{(y^{4})}e^{(z^{4})} - 1.8y^{3}e^{(x^{3})}e^{(y^{4})}e^{(z^{4})}}$

${322x^{2}y^{3}z^{3}e^{(x^{3})}e^{(y^{4})}e^{(z^{4})}}$

${- 18x^{2}e^{(x^{3})}e^{(y^{4})}e^{(z^{4})} - 12y^{3}e^{(x^{3})}e^{(y^{4})}e^{(z^{4})}}$

${20.1x^{2}e^{(x^{3})}e^{(y^{4})}e^{(z^{4})} + 26.8y^{3}e^{(x^{3})}e^{(y^{4})}e^{(z^{4})} + 26.8z^{3}e^{(x^{3})}e^{(y^{4})}e^{(z^{4})}}$

Correct answer:

${- 2.67x^{2}e^{(x^{3})}e^{(y^{4})}e^{(z^{4})} - 1.8y^{3}e^{(x^{3})}e^{(y^{4})}e^{(z^{4})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-6)^2+(-3)^2+(0)^2}=6.71\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-6}{6.71}=-0.89\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-3}{6.71}=-0.45\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{6.71}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.89)(3x^{2}e^{(x^{3})}e^{(y^{4})}e^{(z^{4})})+(-0.45)(4y^{3}e^{(x^{3})}e^{(y^{4})}e^{(z^{4})})+(0)(4z^{3}e^{(x^{3})}e^{(y^{4})}e^{(z^{4})})\\&D_{\overrightarrow{u}}(x,y,z)=- 2.67x^{2}e^{(x^{3})}e^{(y^{4})}e^{(z^{4})} - 1.8y^{3}e^{(x^{3})}e^{(y^{4})}e^{(z^{4})}\end{align*}$

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Example Question #1385 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- 5y^{2}cos(z) - y^{3}z^{2}cos(x)\\&\text{In the direction of }\overrightarrow{v}=(0,13,-17).\end{align*}$

Possible Answers:

${34y^{3}zcos(x) - 130ycos(z) - 85y^{2}sin(z) - 39y^{2}z^{2}cos(x)}$

${107y^{2}sin(z) - 214ycos(z) - 42.8y^{3}zcos(x) - 64.2y^{2}z^{2}cos(x) + 21.4y^{3}z^{2}sin(x)}$

${128y^{2}zsin(x)}$

${1.58y^{3}zcos(x) - 6.1ycos(z) - 3.95y^{2}sin(z) - 1.83y^{2}z^{2}cos(x)}$

Correct answer:

${1.58y^{3}zcos(x) - 6.1ycos(z) - 3.95y^{2}sin(z) - 1.83y^{2}z^{2}cos(x)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(13)^2+(-17)^2}=21.4\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{21.4}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{13}{21.4}=0.61\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-17}{21.4}=-0.79\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(y^{3}z^{2}sin(x))+(0.61)(- 10ycos(z) - 3y^{2}z^{2}cos(x))+(-0.79)(5y^{2}sin(z) - 2y^{3}zcos(x))\\&D_{\overrightarrow{u}}(x,y,z)=1.58y^{3}zcos(x) - 6.1ycos(z) - 3.95y^{2}sin(z) - 1.83y^{2}z^{2}cos(x)\end{align*}$

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Example Question #1386 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3^zln(x^{4})ln(y^{2}) - 2xsin(y^{3})sin(z^{2})\\&\text{In the direction of }\overrightarrow{v}=(0,-12,-16).\end{align*}$

Possible Answers:

${\frac{(176\cdot 3^z)}{(xy)}- 240y^{2}zcos(y^{3})cos(z^{2})}$

${22\cdot 3^zln(x^{4})ln(y^{2}) - 40sin(y^{3})sin(z^{2}) +\frac{ (80\cdot 3^zln(y^{2}))}{x}+\frac{ (40\cdot 3^zln(x^{4}))}{y}- 120xy^{2}cos(y^{3})sin(z^{2}) - 80xzcos(z^{2})sin(y^{3})}$

${3.6xy^{2}cos(y^{3})sin(z^{2}) -\frac{ (1.2\cdot 3^zln(x^{4}))}{y}- 0.879\cdot 3^zln(x^{4})ln(y^{2}) + 3.2xzcos(z^{2})sin(y^{3})}$

${0.703\cdot 3^zln(x^{4})ln(y^{2}) +\frac{ (0.72\cdot 3^zln(x^{4}))}{y}- 2.16xy^{2}cos(y^{3})sin(z^{2}) - 2.56xzcos(z^{2})sin(y^{3})}$

Correct answer:

${3.6xy^{2}cos(y^{3})sin(z^{2}) -\frac{ (1.2\cdot 3^zln(x^{4}))}{y}- 0.879\cdot 3^zln(x^{4})ln(y^{2}) + 3.2xzcos(z^{2})sin(y^{3})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-12)^2+(-16)^2}=20\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{20}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-12}{20}=-0.6\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-16}{20}=-0.8\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{(4\cdot 3^zln(y^{2}))}{x}- 2sin(y^{3})sin(z^{2}))+(-0.6)(\frac{(2\cdot 3^zln(x^{4}))}{y}- 6xy^{2}cos(y^{3})sin(z^{2}))+(-0.8)(1.1\cdot 3^zln(x^{4})ln(y^{2}) - 4xzcos(z^{2})sin(y^{3}))\\&D_{\overrightarrow{u}}(x,y,z)=3.6xy^{2}cos(y^{3})sin(z^{2}) -\frac{ (1.2\cdot 3^zln(x^{4}))}{y}- 0.879\cdot 3^zln(x^{4})ln(y^{2}) + 3.2xzcos(z^{2})sin(y^{3})\end{align*}$

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Example Question #1387 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{(3cos(z^{2})sin(x))}{y}+ e^{(y^{3})}e^{(x)}sin(z)\\&\text{In the direction of }\overrightarrow{v}=(0,-7,12).\end{align*}$

Possible Answers:

${41.7y^{2}e^{(y^{3})}e^{(x)}cos(z) +\frac{ (83.3zsin(z^{2})cos(x))}{y^{2}}}$

${\frac{(41.7cos(z^{2})cos(x))}{y}+ 13.9e^{(y^{3})}e^{(x)}cos(z) -\frac{ (41.7cos(z^{2})sin(x))}{y^{2}}+ 13.9e^{(y^{3})}e^{(x)}sin(z) + 41.7y^{2}e^{(y^{3})}e^{(x)}sin(z) -\frac{ (83.3zsin(z^{2})sin(x))}{y}}$

${0.86e^{(y^{3})}e^{(x)}cos(z) +\frac{ (1.5cos(z^{2})sin(x))}{y^{2}}- 1.5y^{2}e^{(y^{3})}e^{(x)}sin(z) -\frac{ (5.16zsin(z^{2})sin(x))}{y}}$

${12e^{(y^{3})}e^{(x)}cos(z) +\frac{ (21cos(z^{2})sin(x))}{y^{2}}- 21y^{2}e^{(y^{3})}e^{(x)}sin(z) -\frac{ (72zsin(z^{2})sin(x))}{y}}$

Correct answer:

${0.86e^{(y^{3})}e^{(x)}cos(z) +\frac{ (1.5cos(z^{2})sin(x))}{y^{2}}- 1.5y^{2}e^{(y^{3})}e^{(x)}sin(z) -\frac{ (5.16zsin(z^{2})sin(x))}{y}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-7)^2+(12)^2}=13.89\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{13.89}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-7}{13.89}=-0.5\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{12}{13.89}=0.86\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{(3cos(z^{2})cos(x))}{y}+ e^{(y^{3})}e^{(x)}sin(z))+(-0.5)(3y^{2}e^{(y^{3})}e^{(x)}sin(z) -\frac{ (3cos(z^{2})sin(x))}{y^{2}})+(0.86)(e^{(y^{3})}e^{(x)}cos(z) -\frac{ (6zsin(z^{2})sin(x))}{y})\\&D_{\overrightarrow{u}}(x,y,z)=0.86e^{(y^{3})}e^{(x)}cos(z) +\frac{ (1.5cos(z^{2})sin(x))}{y^{2}}- 1.5y^{2}e^{(y^{3})}e^{(x)}sin(z) -\frac{ (5.16zsin(z^{2})sin(x))}{y}\end{align*}$

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Example Question #1388 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3z^{3}sin(y^{3})tan(x) - cos(z^{2})ln(y^{3})e^{(x)}\\&\text{In the direction of }\overrightarrow{v}=(0,4,-10).\end{align*}$

Possible Answers:

${7.78z^{2}sin(y^{3})tan(x) -\frac{ (0.411cos(z^{2})e^{(x)})}{y}+ 1.23y^{2}z^{3}cos(y^{3})tan(x) + 1.73zln(y^{3})sin(z^{2})e^{(x)}}$

${291y^{2}z^{2}cos(y^{3})\cdot (tan(x)^{2} + 1) +\frac{ (64.6zsin(z^{2})e^{(x)})}{y}}$

${3.33y^{2}z^{3}cos(y^{3})tan(x) - 8.37z^{2}sin(y^{3})tan(x) -\frac{ (1.11cos(z^{2})e^{(x)})}{y}- 1.86zln(y^{3})sin(z^{2})e^{(x)}}$

${32.3z^{3}sin(y^{3})\cdot (tan(x)^{2} + 1) -\frac{ (32.3cos(z^{2})e^{(x)})}{y}+ 96.9z^{2}sin(y^{3})tan(x) - 10.8cos(z^{2})ln(y^{3})e^{(x)} + 96.9y^{2}z^{3}cos(y^{3})tan(x) + 21.5zln(y^{3})sin(z^{2})e^{(x)}}$

Correct answer:

${3.33y^{2}z^{3}cos(y^{3})tan(x) - 8.37z^{2}sin(y^{3})tan(x) -\frac{ (1.11cos(z^{2})e^{(x)})}{y}- 1.86zln(y^{3})sin(z^{2})e^{(x)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(4)^2+(-10)^2}=10.77\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{10.77}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{4}{10.77}=0.37\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-10}{10.77}=-0.93\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(3z^{3}sin(y^{3})\cdot (tan(x)^{2} + 1) - cos(z^{2})ln(y^{3})e^{(x)})+(0.37)(9y^{2}z^{3}cos(y^{3})tan(x) -\frac{ (3cos(z^{2})e^{(x)})}{y})+(-0.93)(9z^{2}sin(y^{3})tan(x) + 2zln(y^{3})sin(z^{2})e^{(x)})\\&D_{\overrightarrow{u}}(x,y,z)=3.33y^{2}z^{3}cos(y^{3})tan(x) - 8.37z^{2}sin(y^{3})tan(x) -\frac{ (1.11cos(z^{2})e^{(x)})}{y}- 1.86zln(y^{3})sin(z^{2})e^{(x)}\end{align*}$

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Example Question #1389 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-ln(z^{2})ln(x)sin(y)\\&\text{In the direction of }\overrightarrow{v}=(10,5,0).\end{align*}$

Possible Answers:

${- 5ln(z^{2})cos(y)ln(x) -\frac{ (10ln(z^{2})sin(y))}{x}}$

${-\frac{(22.4cos(y))}{(xz)}}$

${-\frac{ (22.4ln(x)sin(y))}{z}- 11.2ln(z^{2})cos(y)ln(x) -\frac{ (11.2ln(z^{2})sin(y))}{x}}$

${- 0.45ln(z^{2})cos(y)ln(x) -\frac{ (0.89ln(z^{2})sin(y))}{x}}$

Correct answer:

${- 0.45ln(z^{2})cos(y)ln(x) -\frac{ (0.89ln(z^{2})sin(y))}{x}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(10)^2+(5)^2+(0)^2}=11.18\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{10}{11.18}=0.89\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{5}{11.18}=0.45\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{11.18}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.89)(-\frac{(ln(z^{2})sin(y))}{x})+(0.45)(-1ln(z^{2})cos(y)ln(x))+(0)(-\frac{(2ln(x)sin(y))}{z})\\&D_{\overrightarrow{u}}(x,y,z)=- 0.45ln(z^{2})cos(y)ln(x) -\frac{ (0.89ln(z^{2})sin(y))}{x}\end{align*}$

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Calculus 3 : Partial Derivatives

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