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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #1351 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2^{(y^{2})}xsin(z^{4}) - 4y^{2}ln(z)\\&\text{In the direction of }\overrightarrow{v}=(-13,-19,0).\end{align*}$

Possible Answers:

${128\cdot 2^{(y^{2})}yz^{3}cos(z^{4})}$

${23\cdot 2^{(y^{2})}sin(z^{4}) -\frac{ (92.1y^{2})}{z}- 184yln(z) + 31.9\cdot 2^{(y^{2})}xysin(z^{4}) + 92.1\cdot 2^{(y^{2})}xz^{3}cos(z^{4})}$

${6.64yln(z) - 0.56\cdot 2^{(y^{2})}sin(z^{4}) - 1.15\cdot 2^{(y^{2})}xysin(z^{4})}$

${152yln(z) - 13\cdot 2^{(y^{2})}sin(z^{4}) - 26.3\cdot 2^{(y^{2})}xysin(z^{4})}$

Correct answer:

${6.64yln(z) - 0.56\cdot 2^{(y^{2})}sin(z^{4}) - 1.15\cdot 2^{(y^{2})}xysin(z^{4})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-13)^2+(-19)^2+(0)^2}=23.02\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-13}{23.02}=-0.56\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-19}{23.02}=-0.83\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{23.02}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.56)(2^{(y^{2})}sin(z^{4}))+(-0.83)(1.39\cdot 2^{(y^{2})}xysin(z^{4}) - 8yln(z))+(0)(4\cdot 2^{(y^{2})}xz^{3}cos(z^{4}) -\frac{ (4y^{2})}{z})\\&D_{\overrightarrow{u}}(x,y,z)=6.64yln(z) - 0.56\cdot 2^{(y^{2})}sin(z^{4}) - 1.15\cdot 2^{(y^{2})}xysin(z^{4})\end{align*}$

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Example Question #1352 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- 2cos(z^{3})tan(y^{2}) - y^{4}ln(x^{3})ln(z^{4})\\&\text{In the direction of }\overrightarrow{v}=(0,-14,-11).\end{align*}$

Possible Answers:

${\frac{(2.48y^{4}ln(x^{3}))}{z}+ 3.16ycos(z^{3})\cdot (tan(y^{2})^{2} + 1) + 3.16y^{3}ln(x^{3})ln(z^{4}) - 3.72z^{2}tan(y^{2})sin(z^{3})}$

${2.31z^{2}tan(y^{2})sin(z^{3}) - 2.5ycos(z^{3})\cdot (tan(y^{2})^{2} + 1) - 2.5y^{3}ln(x^{3})ln(z^{4}) -\frac{ (1.54y^{4}ln(x^{3}))}{z}}$

${-\frac{(854y^{3})}{(xz)}}$

${107z^{2}tan(y^{2})sin(z^{3}) -\frac{ (53.4y^{4}ln(z^{4}))}{x}- 71.2ycos(z^{3})\cdot (tan(y^{2})^{2} + 1) - 71.2y^{3}ln(x^{3})ln(z^{4}) -\frac{ (71.2y^{4}ln(x^{3}))}{z}}$

Correct answer:

${\frac{(2.48y^{4}ln(x^{3}))}{z}+ 3.16ycos(z^{3})\cdot (tan(y^{2})^{2} + 1) + 3.16y^{3}ln(x^{3})ln(z^{4}) - 3.72z^{2}tan(y^{2})sin(z^{3})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-14)^2+(-11)^2}=17.8\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{17.8}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-14}{17.8}=-0.79\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-11}{17.8}=-0.62\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-\frac{(3y^{4}ln(z^{4}))}{x})+(-0.79)(- 4ycos(z^{3})\cdot (tan(y^{2})^{2} + 1) - 4y^{3}ln(x^{3})ln(z^{4}))+(-0.62)(6z^{2}tan(y^{2})sin(z^{3}) -\frac{ (4y^{4}ln(x^{3}))}{z})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(2.48y^{4}ln(x^{3}))}{z}+ 3.16ycos(z^{3})\cdot (tan(y^{2})^{2} + 1) + 3.16y^{3}ln(x^{3})ln(z^{4}) - 3.72z^{2}tan(y^{2})sin(z^{3})\end{align*}$

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Example Question #1353 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-2^{(y^{2})}sin(z^{4})e^{(x^{2})}\\&\text{In the direction of }\overrightarrow{v}=(13,0,-1).\end{align*}$

Possible Answers:

${0.32\cdot 2^{(y^{2})}z^{3}cos(z^{4})e^{(x^{2})} - 2\cdot 2^{(y^{2})}xsin(z^{4})e^{(x^{2})}}$

${-145\cdot 2^{(y^{2})}xyz^{3}cos(z^{4})e^{(x^{2})}}$

${- 52.2\cdot 2^{(y^{2})}z^{3}cos(z^{4})e^{(x^{2})} - 26.1\cdot 2^{(y^{2})}xsin(z^{4})e^{(x^{2})} - 18.1\cdot 2^{(y^{2})}ysin(z^{4})e^{(x^{2})}}$

${4\cdot 2^{(y^{2})}z^{3}cos(z^{4})e^{(x^{2})} - 26\cdot 2^{(y^{2})}xsin(z^{4})e^{(x^{2})}}$

Correct answer:

${0.32\cdot 2^{(y^{2})}z^{3}cos(z^{4})e^{(x^{2})} - 2\cdot 2^{(y^{2})}xsin(z^{4})e^{(x^{2})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(13)^2+(0)^2+(-1)^2}=13.04\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{13}{13.04}=1\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{13.04}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-1}{13.04}=-0.08\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(1)(-2\cdot 2^{(y^{2})}xsin(z^{4})e^{(x^{2})})+(0)(-1.39\cdot 2^{(y^{2})}ysin(z^{4})e^{(x^{2})})+(-0.08)(-4\cdot 2^{(y^{2})}z^{3}cos(z^{4})e^{(x^{2})})\\&D_{\overrightarrow{u}}(x,y,z)=0.32\cdot 2^{(y^{2})}z^{3}cos(z^{4})e^{(x^{2})} - 2\cdot 2^{(y^{2})}xsin(z^{4})e^{(x^{2})}\end{align*}$

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Example Question #1354 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- z^{4}ln(x^{2})e^{(y)} - cos(y^{3})e^{(x^{4})}cos(z)\\&\text{In the direction of }\overrightarrow{v}=(-5,0,14).\end{align*}$

Possible Answers:

${-\frac{ (119z^{3}e^{(y)})}{x}- 178x^{3}y^{2}sin(y^{3})e^{(x^{4})}sin(z)}$

${0.884cos(y^{3})e^{(x^{4})}sin(z) - 3.53z^{3}ln(x^{2})e^{(y)} -\frac{ (0.231z^{4}e^{(y)})}{x}- 0.462x^{3}cos(y^{3})e^{(x^{4})}cos(z)}$

${14.9cos(y^{3})e^{(x^{4})}sin(z) - 14.9z^{4}ln(x^{2})e^{(y)} - 59.5z^{3}ln(x^{2})e^{(y)} -\frac{ (29.7z^{4}e^{(y)})}{x}- 59.5x^{3}cos(y^{3})e^{(x^{4})}cos(z) + 44.6y^{2}sin(y^{3})e^{(x^{4})}cos(z)}$

${0.94cos(y^{3})e^{(x^{4})}sin(z) - 3.76z^{3}ln(x^{2})e^{(y)} +\frac{ (0.68z^{4}e^{(y)})}{x}+ 1.36x^{3}cos(y^{3})e^{(x^{4})}cos(z)}$

Correct answer:

${0.94cos(y^{3})e^{(x^{4})}sin(z) - 3.76z^{3}ln(x^{2})e^{(y)} +\frac{ (0.68z^{4}e^{(y)})}{x}+ 1.36x^{3}cos(y^{3})e^{(x^{4})}cos(z)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-5)^2+(0)^2+(14)^2}=14.87\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-5}{14.87}=-0.34\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{14.87}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{14}{14.87}=0.94\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.34)(-\frac{ (2z^{4}e^{(y)})}{x}- 4x^{3}cos(y^{3})e^{(x^{4})}cos(z))+(0)(3y^{2}sin(y^{3})e^{(x^{4})}cos(z) - z^{4}ln(x^{2})e^{(y)})+(0.94)(cos(y^{3})e^{(x^{4})}sin(z) - 4z^{3}ln(x^{2})e^{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=0.94cos(y^{3})e^{(x^{4})}sin(z) - 3.76z^{3}ln(x^{2})e^{(y)} +\frac{ (0.68z^{4}e^{(y)})}{x}+ 1.36x^{3}cos(y^{3})e^{(x^{4})}cos(z)\end{align*}$

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Example Question #1355 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- 4y^{3}cos(z^{3}) -\frac{ (cos(x^{2})ln(z^{2}))}{y}\\&\text{In the direction of }\overrightarrow{v}=(6,0,11).\end{align*}$

Possible Answers:

${10.6y^{3}z^{2}sin(z^{3}) -\frac{ (1.76cos(x^{2}))}{(yz)}+\frac{ (0.96xln(z^{2})sin(x^{2}))}{y}}$

${132y^{3}z^{2}sin(z^{3}) -\frac{ (22cos(x^{2}))}{(yz)}+\frac{ (12xln(z^{2})sin(x^{2}))}{y}}$

${-\frac{(50.1xsin(x^{2}))}{(y^{2}z)}}$

${150y^{3}z^{2}sin(z^{3}) -\frac{ (25.1cos(x^{2}))}{(yz)}- 150y^{2}cos(z^{3}) +\frac{ (12.5cos(x^{2})ln(z^{2}))}{y^{2}}+\frac{ (25.1xln(z^{2})sin(x^{2}))}{y}}$

Correct answer:

${10.6y^{3}z^{2}sin(z^{3}) -\frac{ (1.76cos(x^{2}))}{(yz)}+\frac{ (0.96xln(z^{2})sin(x^{2}))}{y}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(6)^2+(0)^2+(11)^2}=12.53\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{6}{12.53}=0.48\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{12.53}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{11}{12.53}=0.88\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.48)(\frac{(2xln(z^{2})sin(x^{2}))}{y})+(0)(\frac{(cos(x^{2})ln(z^{2}))}{y^{2}}- 12y^{2}cos(z^{3}))+(0.88)(12y^{3}z^{2}sin(z^{3}) -\frac{ (2cos(x^{2}))}{(yz)})\\&D_{\overrightarrow{u}}(x,y,z)=10.6y^{3}z^{2}sin(z^{3}) -\frac{ (1.76cos(x^{2}))}{(yz)}+\frac{ (0.96xln(z^{2})sin(x^{2}))}{y}\end{align*}$

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Example Question #1356 : Partial Derivatives

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

Possible Answers:

${0.74zcos(x^{2})sin(z^{2})cos(y) -\frac{ (1.86e^{(x^{2})}sin(z))}{y^{2}}- 0.93cos(x^{2})cos(z^{2})sin(y) -\frac{ (0.74e^{(x^{2})}cos(z))}{y}}$

${-\frac{ (21.6xe^{(x^{2})}cos(z))}{y^{2}}- 21.6xzsin(x^{2})sin(z^{2})sin(y)}$

${\frac{(0.274e^{(x^{2})}cos(z))}{y}-\frac{ (1.73e^{(x^{2})}sin(z))}{y^{2}}- 0.865cos(x^{2})cos(z^{2})sin(y) - 0.274zcos(x^{2})sin(z^{2})cos(y)}$

${\frac{(10.8e^{(x^{2})}cos(z))}{y}-\frac{ (10.8e^{(x^{2})}sin(z))}{y^{2}}- 5.39cos(x^{2})cos(z^{2})sin(y) - 10.8xcos(z^{2})sin(x^{2})cos(y) - 10.8zcos(x^{2})sin(z^{2})cos(y) +\frac{ (21.6xe^{(x^{2})}sin(z))}{y}}$

Correct answer:

${0.74zcos(x^{2})sin(z^{2})cos(y) -\frac{ (1.86e^{(x^{2})}sin(z))}{y^{2}}- 0.93cos(x^{2})cos(z^{2})sin(y) -\frac{ (0.74e^{(x^{2})}cos(z))}{y}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(5)^2+(-2)^2}=5.39\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{5.39}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{5}{5.39}=0.93\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-2}{5.39}=-0.37\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{(4xe^{(x^{2})}sin(z))}{y}- 2xcos(z^{2})sin(x^{2})cos(y))+(0.93)(-\frac{ (2e^{(x^{2})}sin(z))}{y^{2}}- cos(x^{2})cos(z^{2})sin(y))+(-0.37)(\frac{(2e^{(x^{2})}cos(z))}{y}- 2zcos(x^{2})sin(z^{2})cos(y))\\&D_{\overrightarrow{u}}(x,y,z)=0.74zcos(x^{2})sin(z^{2})cos(y) -\frac{ (1.86e^{(x^{2})}sin(z))}{y^{2}}- 0.93cos(x^{2})cos(z^{2})sin(y) -\frac{ (0.74e^{(x^{2})}cos(z))}{y}\end{align*}$

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Example Question #1357 : Partial Derivatives

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

Possible Answers:

${13.9\cdot 2^{(y^{4})}y^{3}cos(x^{3})ln(z) -\frac{ (15\cdot 2^{(y^{4})}cos(x^{3}))}{z}}$

${\frac{(132\cdot 2^{(y^{4})}x^{2}y^{3}sin(x^{3}))}{z}}$

${47.4\cdot 2^{(y^{4})}x^{2}sin(x^{3})ln(z) - 43.8\cdot 2^{(y^{4})}y^{3}cos(x^{3})ln(z) -\frac{ (15.8\cdot 2^{(y^{4})}cos(x^{3}))}{z}}$

${0.887\cdot 2^{(y^{4})}y^{3}cos(x^{3})ln(z) -\frac{ (0.95\cdot 2^{(y^{4})}cos(x^{3}))}{z}}$

Correct answer:

${0.887\cdot 2^{(y^{4})}y^{3}cos(x^{3})ln(z) -\frac{ (0.95\cdot 2^{(y^{4})}cos(x^{3}))}{z}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-5)^2+(15)^2}=15.81\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{15.81}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-5}{15.81}=-0.32\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{15}{15.81}=0.95\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(3\cdot 2^{(y^{4})}x^{2}sin(x^{3})ln(z))+(-0.32)(-2.77\cdot 2^{(y^{4})}y^{3}cos(x^{3})ln(z))+(0.95)(-\frac{(1\cdot 2^{(y^{4})}cos(x^{3}))}{z})\\&D_{\overrightarrow{u}}(x,y,z)=0.887\cdot 2^{(y^{4})}y^{3}cos(x^{3})ln(z) -\frac{ (0.95\cdot 2^{(y^{4})}cos(x^{3}))}{z}\end{align*}$

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Example Question #1358 : Partial Derivatives

Determine the directional derivative of the following function in the direction of $\vec{v}=\left \langle 1, 3, 1\right \rangle$ :

f(x, y, z)=zx+xy+z^2y^2

Possible Answers:

$\frac{1}{\sqrt{11}}(4x+ z+y+8yz^2)$

$\frac{1}{\sqrt{11}}(x+ z+y+6yz^2+2zy^2)$

$\frac{1}{\sqrt{11}}(4x+ z+y+6yz^2+2zy^2)$

2x+ z+y+6yz^2+2zy^2

4x+ z+y+6yz^2+2zy^2

Correct answer:

$\frac{1}{\sqrt{11}}(4x+ z+y+6yz^2+2zy^2)$

Explanation:

The directional derivative is given by

$D_{\vec{u}}f(x, y, z)=f_x(x, y, z)a+f_y(x, y, z)b+f_c(x, y, z)c$ , where $\left \langle a, b, c\right \rangle$ is the unit directional vector.

Now, we must find the gradient of f, $\bigtriangledown f=\left \langle f_x, f_y, f_z\right \rangle$ :

f_x=z+y , f_y=x+2yz^2 , f_z=x+2zy^2

The partial derivatives were found by keeping the other variables constant, and using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Next, we must find the unit directional vector, which is given by

$\frac{\vec{v}}{\left | \vec{v}\right |}=\frac{\left \langle 1, 3, 1\right \rangle}{\sqrt{1^2+3^2+1^2}}=\left \langle \frac{1}{\sqrt{11}}, \frac{3}{\sqrt{11}}, \frac{1}{\sqrt{11}} \right \rangle$

Plugging all of this into the above formula, and after simplifying, we get

$\frac{1}{\sqrt{11}}(4x+ z+y+6yz^2+2zy^2)$

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Example Question #3721 : Calculus 3

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=y^{3}cos(z^{4})e^{(x^{4})}\\&\text{In the direction of }\overrightarrow{v}=(0,-15,-12).\end{align*}$

Possible Answers:

${48y^{3}z^{3}sin(z^{4})e^{(x^{4})} - 45y^{2}cos(z^{4})e^{(x^{4})}}$

${-922x^{3}y^{2}z^{3}sin(z^{4})e^{(x^{4})}}$

${2.48y^{3}z^{3}sin(z^{4})e^{(x^{4})} - 2.34y^{2}cos(z^{4})e^{(x^{4})}}$

${57.6y^{2}cos(z^{4})e^{(x^{4})} + 76.8x^{3}y^{3}cos(z^{4})e^{(x^{4})} - 76.8y^{3}z^{3}sin(z^{4})e^{(x^{4})}}$

Correct answer:

${2.48y^{3}z^{3}sin(z^{4})e^{(x^{4})} - 2.34y^{2}cos(z^{4})e^{(x^{4})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-15)^2+(-12)^2}=19.21\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{19.21}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-15}{19.21}=-0.78\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-12}{19.21}=-0.62\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(4x^{3}y^{3}cos(z^{4})e^{(x^{4})})+(-0.78)(3y^{2}cos(z^{4})e^{(x^{4})})+(-0.62)(-4y^{3}z^{3}sin(z^{4})e^{(x^{4})})\\&D_{\overrightarrow{u}}(x,y,z)=2.48y^{3}z^{3}sin(z^{4})e^{(x^{4})} - 2.34y^{2}cos(z^{4})e^{(x^{4})}\end{align*}$

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Example Question #1360 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3^{(y^{4})}cos(x^{4})cos(z^{3}) - sin(x^{3})e^{(y^{2})}e^{(z^{2})}\\&\text{In the direction of }\overrightarrow{v}=(-12,0,6).\end{align*}$

Possible Answers:

${59\cdot 3^{(y^{4})}y^{3}cos(x^{4})cos(z^{3}) - 53.7\cdot 3^{(y^{4})}x^{3}cos(z^{3})sin(x^{4}) - 40.3\cdot 3^{(y^{4})}z^{2}cos(x^{4})sin(z^{3}) - 26.8ysin(x^{3})e^{(y^{2})}e^{(z^{2})} - 26.8zsin(x^{3})e^{(y^{2})}e^{(z^{2})} - 40.3x^{2}cos(x^{3})e^{(y^{2})}e^{(z^{2})}}$

${3.56\cdot 3^{(y^{4})}x^{3}cos(z^{3})sin(x^{4}) - 1.35\cdot 3^{(y^{4})}z^{2}cos(x^{4})sin(z^{3}) - 0.9zsin(x^{3})e^{(y^{2})}e^{(z^{2})} + 2.67x^{2}cos(x^{3})e^{(y^{2})}e^{(z^{2})}}$

${48\cdot 3^{(y^{4})}x^{3}cos(z^{3})sin(x^{4}) - 18\cdot 3^{(y^{4})}z^{2}cos(x^{4})sin(z^{3}) - 12zsin(x^{3})e^{(y^{2})}e^{(z^{2})} + 36x^{2}cos(x^{3})e^{(y^{2})}e^{(z^{2})}}$

${708\cdot 3^{(y^{4})}x^{3}y^{3}z^{2}sin(x^{4})sin(z^{3}) - 161x^{2}yzcos(x^{3})e^{(y^{2})}e^{(z^{2})}}$

Correct answer:

${3.56\cdot 3^{(y^{4})}x^{3}cos(z^{3})sin(x^{4}) - 1.35\cdot 3^{(y^{4})}z^{2}cos(x^{4})sin(z^{3}) - 0.9zsin(x^{3})e^{(y^{2})}e^{(z^{2})} + 2.67x^{2}cos(x^{3})e^{(y^{2})}e^{(z^{2})}}$

Explanation:

$\begin{align*}\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{13.42}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{6}{13.42}=0.45\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.89)(- 4\cdot 3^{(y^{4})}x^{3}cos(z^{3})sin(x^{4}) - 3x^{2}cos(x^{3})e^{(y^{2})}e^{(z^{2})})+(0)(4.39\cdot 3^{(y^{4})}y^{3}cos(x^{4})cos(z^{3}) - 2ysin(x^{3})e^{(y^{2})}e^{(z^{2})})+(0.45)(- 3\cdot 3^{(y^{4})}z^{2}cos(x^{4})sin(z^{3}) - 2zsin(x^{3})e^{(y^{2})}e^{(z^{2})})\\&D_{\overrightarrow{u}}(x,y,z)=3.56\cdot 3^{(y^{4})}x^{3}cos(z^{3})sin(x^{4}) - 1.35\cdot 3^{(y^{4})}z^{2}cos(x^{4})sin(z^{3}) - 0.9zsin(x^{3})e^{(y^{2})}e^{(z^{2})} + 2.67x^{2}cos(x^{3})e^{(y^{2})}e^{(z^{2})}\end{align*}$

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