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Calculus 3 : Directional Derivatives

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Example Questions

Example Question #1301 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=yln(x^{2})sin(z^{4}) + 5cos(y^{3})ln(x)sin(z)\\&\text{In the direction of }\overrightarrow{v}=(-15,9,0).\end{align*}$

Possible Answers:

${9ln(x^{2})sin(z^{4}) -\frac{ (75cos(y^{3})sin(z))}{x}-\frac{ (30ysin(z^{4}))}{x}- 135y^{2}sin(y^{3})ln(x)sin(z)}$

${\frac{(140z^{3}cos(z^{4}))}{x}-\frac{ (262y^{2}sin(y^{3})cos(z))}{x}}$

${17.5ln(x^{2})sin(z^{4}) + 87.4cos(y^{3})cos(z)ln(x) +\frac{ (87.4cos(y^{3})sin(z))}{x}+\frac{ (35ysin(z^{4}))}{x}+ 70yz^{3}cos(z^{4})ln(x^{2}) - 262y^{2}sin(y^{3})ln(x)sin(z)}$

${0.51ln(x^{2})sin(z^{4}) -\frac{ (4.3cos(y^{3})sin(z))}{x}-\frac{ (1.72ysin(z^{4}))}{x}- 7.65y^{2}sin(y^{3})ln(x)sin(z)}$

Correct answer:

${0.51ln(x^{2})sin(z^{4}) -\frac{ (4.3cos(y^{3})sin(z))}{x}-\frac{ (1.72ysin(z^{4}))}{x}- 7.65y^{2}sin(y^{3})ln(x)sin(z)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-15)^2+(9)^2+(0)^2}=17.49\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-15}{17.49}=-0.86\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{9}{17.49}=0.51\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{17.49}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.86)(\frac{(5cos(y^{3})sin(z))}{x}+\frac{ (2ysin(z^{4}))}{x})+(0.51)(ln(x^{2})sin(z^{4}) - 15y^{2}sin(y^{3})ln(x)sin(z))+(0)(5cos(y^{3})cos(z)ln(x) + 4yz^{3}cos(z^{4})ln(x^{2}))\\&D_{\overrightarrow{u}}(x,y,z)=0.51ln(x^{2})sin(z^{4}) -\frac{ (4.3cos(y^{3})sin(z))}{x}-\frac{ (1.72ysin(z^{4}))}{x}- 7.65y^{2}sin(y^{3})ln(x)sin(z)\end{align*}$

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Example Question #1301 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-cos(x^{4})cos(z^{4})ln(y^{4})\\&\text{In the direction of }\overrightarrow{v}=(8,19,0).\end{align*}$

Possible Answers:

${0.608x^{3}cos(z^{4})ln(y^{4})sin(x^{4}) -\frac{ (3.39cos(x^{4})cos(z^{4}))}{y}}$

${82.5x^{3}cos(z^{4})ln(y^{4})sin(x^{4}) -\frac{ (82.5cos(x^{4})cos(z^{4}))}{y}+ 82.5z^{3}cos(x^{4})ln(y^{4})sin(z^{4})}$

${1.56x^{3}cos(z^{4})ln(y^{4})sin(x^{4}) -\frac{ (3.68cos(x^{4})cos(z^{4}))}{y}}$

${-\frac{(1322x^{3}z^{3}sin(x^{4})sin(z^{4}))}{y}}$

Correct answer:

${1.56x^{3}cos(z^{4})ln(y^{4})sin(x^{4}) -\frac{ (3.68cos(x^{4})cos(z^{4}))}{y}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(8)^2+(19)^2+(0)^2}=20.62\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{8}{20.62}=0.39\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{19}{20.62}=0.92\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{20.62}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.39)(4x^{3}cos(z^{4})ln(y^{4})sin(x^{4}))+(0.92)(-\frac{(4cos(x^{4})cos(z^{4}))}{y})+(0)(4z^{3}cos(x^{4})ln(y^{4})sin(z^{4}))\\&D_{\overrightarrow{u}}(x,y,z)=1.56x^{3}cos(z^{4})ln(y^{4})sin(x^{4}) -\frac{ (3.68cos(x^{4})cos(z^{4}))}{y}\end{align*}$

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Example Question #3671 : Calculus 3

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- 2cos(z^{3})tan(y) - 3^xsin(z^{2})e^{(y^{2})}\\&\text{In the direction of }\overrightarrow{v}=(-4,-18,0).\end{align*}$

Possible Answers:

${36cos(z^{3})\cdot (tan(y)^{2} + 1) + 4.39\cdot 3^xsin(z^{2})e^{(y^{2})} + 36\cdot 3^xysin(z^{2})e^{(y^{2})}}$

${111z^{2}sin(z^{3})tan(y) - 20.3\cdot 3^xsin(z^{2})e^{(y^{2})} - 36.9cos(z^{3})\cdot (tan(y)^{2} + 1) - 36.9\cdot 3^xzcos(z^{2})e^{(y^{2})} - 36.9\cdot 3^xysin(z^{2})e^{(y^{2})}}$

${-81\cdot 3^xyzcos(z^{2})e^{(y^{2})}}$

${1.96cos(z^{3})\cdot (tan(y)^{2} + 1) + 0.242\cdot 3^xsin(z^{2})e^{(y^{2})} + 1.96\cdot 3^xysin(z^{2})e^{(y^{2})}}$

Correct answer:

${1.96cos(z^{3})\cdot (tan(y)^{2} + 1) + 0.242\cdot 3^xsin(z^{2})e^{(y^{2})} + 1.96\cdot 3^xysin(z^{2})e^{(y^{2})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-4)^2+(-18)^2+(0)^2}=18.44\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-4}{18.44}=-0.22\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-18}{18.44}=-0.98\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{18.44}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.22)(-1.1\cdot 3^xsin(z^{2})e^{(y^{2})})+(-0.98)(- 2cos(z^{3})\cdot (tan(y)^{2} + 1) - 2\cdot 3^xysin(z^{2})e^{(y^{2})})+(0)(6z^{2}sin(z^{3})tan(y) - 2\cdot 3^xzcos(z^{2})e^{(y^{2})})\\&D_{\overrightarrow{u}}(x,y,z)=1.96cos(z^{3})\cdot (tan(y)^{2} + 1) + 0.242\cdot 3^xsin(z^{2})e^{(y^{2})} + 1.96\cdot 3^xysin(z^{2})e^{(y^{2})}\end{align*}$

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Example Question #1311 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=yln(x^{2})sin(z^{4}) + 5cos(y^{3})ln(x)sin(z)\\&\text{In the direction of }\overrightarrow{v}=(-15,9,0).\end{align*}$

Possible Answers:

${17.5ln(x^{2})sin(z^{4}) + 87.4cos(y^{3})cos(z)ln(x) +\frac{ (87.4cos(y^{3})sin(z))}{x}+\frac{ (35ysin(z^{4}))}{x}+ 70yz^{3}cos(z^{4})ln(x^{2}) - 262y^{2}sin(y^{3})ln(x)sin(z)}$

${\frac{(140z^{3}cos(z^{4}))}{x}-\frac{ (262y^{2}sin(y^{3})cos(z))}{x}}$

${0.51ln(x^{2})sin(z^{4}) -\frac{ (4.3cos(y^{3})sin(z))}{x}-\frac{ (1.72ysin(z^{4}))}{x}- 7.65y^{2}sin(y^{3})ln(x)sin(z)}$

${9ln(x^{2})sin(z^{4}) -\frac{ (75cos(y^{3})sin(z))}{x}-\frac{ (30ysin(z^{4}))}{x}- 135y^{2}sin(y^{3})ln(x)sin(z)}$

Correct answer:

${0.51ln(x^{2})sin(z^{4}) -\frac{ (4.3cos(y^{3})sin(z))}{x}-\frac{ (1.72ysin(z^{4}))}{x}- 7.65y^{2}sin(y^{3})ln(x)sin(z)}$

Explanation:

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Example Question #1312 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-cos(x^{4})cos(z^{4})ln(y^{4})\\&\text{In the direction of }\overrightarrow{v}=(8,19,0).\end{align*}$

Possible Answers:

${-\frac{(1322x^{3}z^{3}sin(x^{4})sin(z^{4}))}{y}}$

${0.608x^{3}cos(z^{4})ln(y^{4})sin(x^{4}) -\frac{ (3.39cos(x^{4})cos(z^{4}))}{y}}$

${82.5x^{3}cos(z^{4})ln(y^{4})sin(x^{4}) -\frac{ (82.5cos(x^{4})cos(z^{4}))}{y}+ 82.5z^{3}cos(x^{4})ln(y^{4})sin(z^{4})}$

${1.56x^{3}cos(z^{4})ln(y^{4})sin(x^{4}) -\frac{ (3.68cos(x^{4})cos(z^{4}))}{y}}$

Correct answer:

${1.56x^{3}cos(z^{4})ln(y^{4})sin(x^{4}) -\frac{ (3.68cos(x^{4})cos(z^{4}))}{y}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(8)^2+(19)^2+(0)^2}=20.62\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{8}{20.62}=0.39\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{19}{20.62}=0.92\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{20.62}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.39)(4x^{3}cos(z^{4})ln(y^{4})sin(x^{4}))+(0.92)(-\frac{(4cos(x^{4})cos(z^{4}))}{y})+(0)(4z^{3}cos(x^{4})ln(y^{4})sin(z^{4}))\\&D_{\overrightarrow{u}}(x,y,z)=1.56x^{3}cos(z^{4})ln(y^{4})sin(x^{4}) -\frac{ (3.68cos(x^{4})cos(z^{4}))}{y}\end{align*}$

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Example Question #1313 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3^{(x^{3})}z^{4}sin(y)\\&\text{In the direction of }\overrightarrow{v}=(0,9,15).\end{align*}$

Possible Answers:

${231\cdot 3^{(x^{3})}x^{2}z^{3}cos(y)}$

${0.51\cdot 3^{(x^{3})}z^{4}cos(y) + 3.44\cdot 3^{(x^{3})}z^{3}sin(y)}$

${0.26\cdot 3^{(x^{3})}z^{4}cos(y) + 2.96\cdot 3^{(x^{3})}z^{3}sin(y)}$

${17.5\cdot 3^{(x^{3})}z^{4}cos(y) + 70\cdot 3^{(x^{3})}z^{3}sin(y) + 57.6\cdot 3^{(x^{3})}x^{2}z^{4}sin(y)}$

Correct answer:

${0.51\cdot 3^{(x^{3})}z^{4}cos(y) + 3.44\cdot 3^{(x^{3})}z^{3}sin(y)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(9)^2+(15)^2}=17.49\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{17.49}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{9}{17.49}=0.51\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{15}{17.49}=0.86\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(3.3\cdot 3^{(x^{3})}x^{2}z^{4}sin(y))+(0.51)(3^{(x^{3})}z^{4}cos(y))+(0.86)(4\cdot 3^{(x^{3})}z^{3}sin(y))\\&D_{\overrightarrow{u}}(x,y,z)=0.51\cdot 3^{(x^{3})}z^{4}cos(y) + 3.44\cdot 3^{(x^{3})}z^{3}sin(y)\end{align*}$

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Example Question #241 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{ln(z^{4})}{x^{2}}- 2yz^{3}sin(x^{2})\\&\text{In the direction of }\overrightarrow{v}=(0,-14,-6).\end{align*}$

Possible Answers:

${\frac{60.9}{(x^{2}z)}-\frac{ (30.5ln(z^{4}))}{x^{3}}- 30.5z^{3}sin(x^{2}) - 91.4yz^{2}sin(x^{2}) - 60.9xyz^{3}cos(x^{2})}$

${-183xz^{2}cos(x^{2})}$

${\frac{0.608}{(x^{2}z)}- 1.69z^{3}sin(x^{2}) - 0.913yz^{2}sin(x^{2})}$

${1.84z^{3}sin(x^{2}) -\frac{ 1.56}{(x^{2}z)}+ 2.34yz^{2}sin(x^{2})}$

Correct answer:

${1.84z^{3}sin(x^{2}) -\frac{ 1.56}{(x^{2}z)}+ 2.34yz^{2}sin(x^{2})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-14)^2+(-6)^2}=15.23\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{15.23}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-14}{15.23}=-0.92\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-6}{15.23}=-0.39\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-\frac{ (2ln(z^{4}))}{x^{3}}- 4xyz^{3}cos(x^{2}))+(-0.92)(-2z^{3}sin(x^{2}))+(-0.39)(\frac{4.0}{(x^{2}z)}- 6yz^{2}sin(x^{2}))\\&D_{\overrightarrow{u}}(x,y,z)=1.84z^{3}sin(x^{2}) -\frac{ 1.56}{(x^{2}z)}+ 2.34yz^{2}sin(x^{2})\end{align*}$

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Example Question #1315 : Partial Derivatives

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

Possible Answers:

${0.548cos(z^{4})e^{(x^{3})}cos(y) + 1.35x^{2}cos(z^{4})e^{(x^{3})}sin(y)}$

${-161x^{2}z^{3}sin(z^{4})e^{(x^{3})}cos(y)}$

${0.74cos(z^{4})e^{(x^{3})}cos(y) - 2.01x^{2}cos(z^{4})e^{(x^{3})}sin(y)}$

${13.4cos(z^{4})e^{(x^{3})}cos(y) + 40.3x^{2}cos(z^{4})e^{(x^{3})}sin(y) - 53.8z^{3}sin(z^{4})e^{(x^{3})}sin(y)}$

Correct answer:

${0.74cos(z^{4})e^{(x^{3})}cos(y) - 2.01x^{2}cos(z^{4})e^{(x^{3})}sin(y)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-9)^2+(10)^2+(0)^2}=13.45\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-9}{13.45}=-0.67\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{10}{13.45}=0.74\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{13.45}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.67)(3x^{2}cos(z^{4})e^{(x^{3})}sin(y))+(0.74)(cos(z^{4})e^{(x^{3})}cos(y))+(0)(-4z^{3}sin(z^{4})e^{(x^{3})}sin(y))\\&D_{\overrightarrow{u}}(x,y,z)=0.74cos(z^{4})e^{(x^{3})}cos(y) - 2.01x^{2}cos(z^{4})e^{(x^{3})}sin(y)\end{align*}$

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Example Question #1316 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3e^{(z)}sin(x)tan(y) -\frac{ (3^ye^{(x)})}{z^{2}}\\&\text{In the direction of }\overrightarrow{v}=(-5,4,0).\end{align*}$

Possible Answers:

${1.83e^{(z)}cos(x)tan(y) + 1.15e^{(z)}sin(x)\cdot (tan(y)^{2} + 1) -\frac{ (1.03\cdot 3^ye^{(x)})}{z^{2}}}$

${19.2e^{(z)}cos(x)\cdot (tan(y)^{2} + 1) +\frac{ (14.1\cdot 3^ye^{(x)})}{z^{3}}}$

${1.86e^{(z)}sin(x)\cdot (tan(y)^{2} + 1) - 2.34e^{(z)}cos(x)tan(y) +\frac{ (0.0989\cdot 3^ye^{(x)})}{z^{2}}}$

${19.2e^{(z)}cos(x)tan(y) + 19.2e^{(z)}sin(x)\cdot (tan(y)^{2} + 1) + 19.2e^{(z)}sin(x)tan(y) -\frac{ (13.4\cdot 3^ye^{(x)})}{z^{2}}+\frac{ (12.8\cdot 3^ye^{(x)})}{z^{3}}}$

Correct answer:

${1.86e^{(z)}sin(x)\cdot (tan(y)^{2} + 1) - 2.34e^{(z)}cos(x)tan(y) +\frac{ (0.0989\cdot 3^ye^{(x)})}{z^{2}}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-5)^2+(4)^2+(0)^2}=6.4\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-5}{6.4}=-0.78\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{4}{6.4}=0.62\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{6.4}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.78)(3e^{(z)}cos(x)tan(y) -\frac{ (1\cdot 3^ye^{(x)})}{z^{2}})+(0.62)(3e^{(z)}sin(x)\cdot (tan(y)^{2} + 1) -\frac{ (1.1\cdot 3^ye^{(x)})}{z^{2}})+(0)(3e^{(z)}sin(x)tan(y) +\frac{ (2\cdot 3^ye^{(x)})}{z^{3}})\\&D_{\overrightarrow{u}}(x,y,z)=1.86e^{(z)}sin(x)\cdot (tan(y)^{2} + 1) - 2.34e^{(z)}cos(x)tan(y) +\frac{ (0.0989\cdot 3^ye^{(x)})}{z^{2}}\end{align*}$

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Example Question #1317 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3^{(y^{4})}ln(z^{3})sin(x^{4}) + 4sin(x^{2})tan(y^{2})tan(z^{2})\\&\text{In the direction of }\overrightarrow{v}=(-5,-2,0).\end{align*}$

Possible Answers:

${\frac{(284\cdot 3^{(y^{4})}x^{3}y^{3}cos(x^{4}))}{z}+ 172xyzcos(x^{2})\cdot (tan(y^{2})^{2} + 1)\cdot (tan(z^{2})^{2} + 1)}$

${- 3.72\cdot 3^{(y^{4})}x^{3}cos(x^{4})ln(z^{3}) - 1.63\cdot 3^{(y^{4})}y^{3}ln(z^{3})sin(x^{4}) - 7.44xcos(x^{2})tan(y^{2})tan(z^{2}) - 2.96ysin(x^{2})tan(z^{2})\cdot (tan(y^{2})^{2} + 1)}$

${\frac{(16.2\cdot 3^{(y^{4})}sin(x^{4}))}{z}+ 21.6\cdot 3^{(y^{4})}x^{3}cos(x^{4})ln(z^{3}) + 23.7\cdot 3^{(y^{4})}y^{3}ln(z^{3})sin(x^{4}) + 43.1xcos(x^{2})tan(y^{2})tan(z^{2}) + 43.1ysin(x^{2})tan(z^{2})\cdot (tan(y^{2})^{2} + 1) + 43.1zsin(x^{2})tan(y^{2})\cdot (tan(z^{2})^{2} + 1)}$

${3.46\cdot 3^{(y^{4})}x^{3}cos(x^{4})ln(z^{3}) + 0.602\cdot 3^{(y^{4})}y^{3}ln(z^{3})sin(x^{4}) + 6.92xcos(x^{2})tan(y^{2})tan(z^{2}) + 1.1ysin(x^{2})tan(z^{2})\cdot (tan(y^{2})^{2} + 1)}$

Correct answer:

${- 3.72\cdot 3^{(y^{4})}x^{3}cos(x^{4})ln(z^{3}) - 1.63\cdot 3^{(y^{4})}y^{3}ln(z^{3})sin(x^{4}) - 7.44xcos(x^{2})tan(y^{2})tan(z^{2}) - 2.96ysin(x^{2})tan(z^{2})\cdot (tan(y^{2})^{2} + 1)}$

Explanation:

$\begin{align*}\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-2}{5.39}=-0.37\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{5.39}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.93)(4\cdot 3^{(y^{4})}x^{3}cos(x^{4})ln(z^{3}) + 8xcos(x^{2})tan(y^{2})tan(z^{2}))+(-0.37)(4.39\cdot 3^{(y^{4})}y^{3}ln(z^{3})sin(x^{4}) + 8ysin(x^{2})tan(z^{2})\cdot (tan(y^{2})^{2} + 1))+(0)(\frac{(3\cdot 3^{(y^{4})}sin(x^{4}))}{z}+ 8zsin(x^{2})tan(y^{2})\cdot (tan(z^{2})^{2} + 1))\\&D_{\overrightarrow{u}}(x,y,z)=- 3.72\cdot 3^{(y^{4})}x^{3}cos(x^{4})ln(z^{3}) - 1.63\cdot 3^{(y^{4})}y^{3}ln(z^{3})sin(x^{4}) - 7.44xcos(x^{2})tan(y^{2})tan(z^{2}) - 2.96ysin(x^{2})tan(z^{2})\cdot (tan(y^{2})^{2} + 1)\end{align*}$

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Calculus 3 : Directional Derivatives

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Example Question #1301 : Partial Derivatives

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