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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #1361 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=5ln(x^{2})ln(y^{2}) + 2^{(z^{2})}xy^{2}\\&\text{In the direction of }\overrightarrow{v}=(18,20,0).\end{align*}$

Possible Answers:

${74.6\cdot 2^{(z^{2})}yz}$

${\frac{(4.49ln(y^{2}))}{x}+\frac{ (5.48ln(x^{2}))}{y}+ 0.449\cdot 2^{(z^{2})}y^{2} + 1.1\cdot 2^{(z^{2})}xy}$

${\frac{(6.7ln(y^{2}))}{x}+\frac{ (7.4ln(x^{2}))}{y}+ 0.67\cdot 2^{(z^{2})}y^{2} + 1.48\cdot 2^{(z^{2})}xy}$

${\frac{(269ln(y^{2}))}{x}+\frac{ (269ln(x^{2}))}{y}+ 26.9\cdot 2^{(z^{2})}y^{2} + 53.8\cdot 2^{(z^{2})}xy + 37.3\cdot 2^{(z^{2})}xy^{2}z}$

Correct answer:

${\frac{(6.7ln(y^{2}))}{x}+\frac{ (7.4ln(x^{2}))}{y}+ 0.67\cdot 2^{(z^{2})}y^{2} + 1.48\cdot 2^{(z^{2})}xy}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(18)^2+(20)^2+(0)^2}=26.91\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{18}{26.91}=0.67\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{20}{26.91}=0.74\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{26.91}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.67)(\frac{(10ln(y^{2}))}{x}+ 2^{(z^{2})}y^{2})+(0.74)(\frac{(10ln(x^{2}))}{y}+ 2\cdot 2^{(z^{2})}xy)+(0)(1.39\cdot 2^{(z^{2})}xy^{2}z)\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(6.7ln(y^{2}))}{x}+\frac{ (7.4ln(x^{2}))}{y}+ 0.67\cdot 2^{(z^{2})}y^{2} + 1.48\cdot 2^{(z^{2})}xy\end{align*}$

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Example Question #291 : Directional Derivatives

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

Possible Answers:

${\frac{(12.6ln(y^{3})sin(z^{2}))}{x^{3}}-\frac{ (12.6cos(y^{3})sin(x^{3}))}{z^{2}}-\frac{ (19sin(z^{2}))}{(x^{2}y)}-\frac{ (12.6zcos(z^{2})ln(y^{3}))}{x^{2}}+\frac{ (37.9x^{2}cos(x^{3})cos(y^{3}))}{z}-\frac{ (37.9y^{2}sin(x^{3})sin(y^{3}))}{z}}$

${\frac{(12cos(y^{3})sin(x^{3}))}{z^{2}}+\frac{ (4ln(y^{3})sin(z^{2}))}{x^{3}}+\frac{ (12zcos(z^{2})ln(y^{3}))}{x^{2}}+\frac{ (12x^{2}cos(x^{3})cos(y^{3}))}{z}}$

${\frac{(75.8zcos(z^{2}))}{(x^{3}y)}+\frac{ (114x^{2}y^{2}cos(x^{3})sin(y^{3}))}{z^{2}}}$

${\frac{(1.9cos(y^{3})sin(x^{3}))}{z^{2}}+\frac{ (0.64ln(y^{3})sin(z^{2}))}{x^{3}}+\frac{ (1.9zcos(z^{2})ln(y^{3}))}{x^{2}}+\frac{ (1.92x^{2}cos(x^{3})cos(y^{3}))}{z}}$

Correct answer:

${\frac{(1.9cos(y^{3})sin(x^{3}))}{z^{2}}+\frac{ (0.64ln(y^{3})sin(z^{2}))}{x^{3}}+\frac{ (1.9zcos(z^{2})ln(y^{3}))}{x^{2}}+\frac{ (1.92x^{2}cos(x^{3})cos(y^{3}))}{z}}$

Explanation:

$\begin{align*}\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{6.32}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-6}{6.32}=-0.95\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.32)(\frac{(2ln(y^{3})sin(z^{2}))}{x^{3}}+\frac{ (6x^{2}cos(x^{3})cos(y^{3}))}{z})+(0)(-\frac{ (3sin(z^{2}))}{(x^{2}y)}-\frac{ (6y^{2}sin(x^{3})sin(y^{3}))}{z})+(-0.95)(-\frac{ (2cos(y^{3})sin(x^{3}))}{z^{2}}-\frac{ (2zcos(z^{2})ln(y^{3}))}{x^{2}})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(1.9cos(y^{3})sin(x^{3}))}{z^{2}}+\frac{ (0.64ln(y^{3})sin(z^{2}))}{x^{3}}+\frac{ (1.9zcos(z^{2})ln(y^{3}))}{x^{2}}+\frac{ (1.92x^{2}cos(x^{3})cos(y^{3}))}{z}\end{align*}$

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Example Question #291 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=yln(x^{2})ln(z^{4})\\&\text{In the direction of }\overrightarrow{v}=(0,13,-16).\end{align*}$

Possible Answers:

${13ln(x^{2})ln(z^{4}) -\frac{ (64yln(x^{2}))}{z}}$

${20.6ln(x^{2})ln(z^{4}) +\frac{ (82.5yln(x^{2}))}{z}+\frac{ (41.2yln(z^{4}))}{x}}$

${0.63ln(x^{2})ln(z^{4}) -\frac{ (3.12yln(x^{2}))}{z}}$

${\frac{165.0}{(xz)}}$

Correct answer:

${0.63ln(x^{2})ln(z^{4}) -\frac{ (3.12yln(x^{2}))}{z}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(13)^2+(-16)^2}=20.62\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{20.62}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{13}{20.62}=0.63\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-16}{20.62}=-0.78\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{(2yln(z^{4}))}{x})+(0.63)(ln(x^{2})ln(z^{4}))+(-0.78)(\frac{(4yln(x^{2}))}{z})\\&D_{\overrightarrow{u}}(x,y,z)=0.63ln(x^{2})ln(z^{4}) -\frac{ (3.12yln(x^{2}))}{z}\end{align*}$

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Example Question #1364 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3ln(z^{2})tan(x) + x^{2}y^{3}ln(z^{2})\\&\text{In the direction of }\overrightarrow{v}=(0,-13,4).\end{align*}$

Possible Answers:

${\frac{(24.0tan(x))}{z}- 39x^{2}y^{2}ln(z^{2}) +\frac{ (8x^{2}y^{3})}{z}}$

${\frac{(163xy^{2})}{z}}$

${\frac{(81.6tan(x))}{z}+ 40.8ln(z^{2})\cdot (tan(x)^{2} + 1) + 40.8x^{2}y^{2}ln(z^{2}) +\frac{ (27.2x^{2}y^{3})}{z}+ 27.2xy^{3}ln(z^{2})}$

${\frac{(1.74tan(x))}{z}- 2.88x^{2}y^{2}ln(z^{2}) +\frac{ (0.58x^{2}y^{3})}{z}}$

Correct answer:

${\frac{(1.74tan(x))}{z}- 2.88x^{2}y^{2}ln(z^{2}) +\frac{ (0.58x^{2}y^{3})}{z}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-13)^2+(4)^2}=13.6\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{13.6}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-13}{13.6}=-0.96\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{4}{13.6}=0.29\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(3ln(z^{2})\cdot (tan(x)^{2} + 1) + 2xy^{3}ln(z^{2}))+(-0.96)(3x^{2}y^{2}ln(z^{2}))+(0.29)(\frac{(6.0tan(x))}{z}+\frac{ (2x^{2}y^{3})}{z})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(1.74tan(x))}{z}- 2.88x^{2}y^{2}ln(z^{2}) +\frac{ (0.58x^{2}y^{3})}{z}\end{align*}$

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Example Question #292 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-\frac{(x^{4}cos(z))}{y}\\&\text{In the direction of }\overrightarrow{v}=(0,20,-7).\end{align*}$

Possible Answers:

${\frac{(21.2x^{4}cos(z))}{y^{2}}-\frac{ (84.8x^{3}cos(z))}{y}+\frac{ (21.2x^{4}sin(z))}{y}}$

${\frac{(0.94x^{4}cos(z))}{y^{2}}-\frac{ (0.33x^{4}sin(z))}{y}}$

${-\frac{(84.8x^{3}sin(z))}{y^{2}}}$

${\frac{(20x^{4}cos(z))}{y^{2}}-\frac{ (7x^{4}sin(z))}{y}}$

Correct answer:

${\frac{(0.94x^{4}cos(z))}{y^{2}}-\frac{ (0.33x^{4}sin(z))}{y}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(20)^2+(-7)^2}=21.19\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{21.19}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{20}{21.19}=0.94\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-7}{21.19}=-0.33\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-\frac{(4x^{3}cos(z))}{y})+(0.94)(\frac{(x^{4}cos(z))}{y^{2}})+(-0.33)(\frac{(x^{4}sin(z))}{y})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(0.94x^{4}cos(z))}{y^{2}}-\frac{ (0.33x^{4}sin(z))}{y}\end{align*}$

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Example Question #1366 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-4^{(y^{3})}e^{(x^{2})}\\&\text{In the direction of }\overrightarrow{v}=(0,10,-20).\end{align*}$

Possible Answers:

${-0.842\cdot 4^{(y^{3})}y^{2}e^{(x^{2})}}$

${-1.87\cdot 4^{(y^{3})}y^{2}e^{(x^{2})}}$

${- 44.7\cdot 4^{(y^{3})}xe^{(x^{2})} - 93\cdot 4^{(y^{3})}y^{2}e^{(x^{2})}}$

Correct answer:

${-1.87\cdot 4^{(y^{3})}y^{2}e^{(x^{2})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(10)^2+(-20)^2}=22.36\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{22.36}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{10}{22.36}=0.45\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-20}{22.36}=-0.89\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-2\cdot 4^{(y^{3})}xe^{(x^{2})})+(0.45)(-4.16\cdot 4^{(y^{3})}y^{2}e^{(x^{2})})+(-0.89)(0.0)\\&D_{\overrightarrow{u}}(x,y,z)=-1.87\cdot 4^{(y^{3})}y^{2}e^{(x^{2})}\end{align*}$

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Example Question #1367 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=e^{(y^{3})}e^{(x)}sin(z)\\&\text{In the direction of }\overrightarrow{v}=(0,10,-16).\end{align*}$

Possible Answers:

${1.59y^{2}e^{(y^{3})}e^{(x)}sin(z) - 0.85e^{(y^{3})}e^{(x)}cos(z)}$

${0.723e^{(y^{3})}e^{(x)}cos(z) + 0.843y^{2}e^{(y^{3})}e^{(x)}sin(z)}$

${56.6y^{2}e^{(y^{3})}e^{(x)}cos(z)}$

${18.9e^{(y^{3})}e^{(x)}cos(z) + 18.9e^{(y^{3})}e^{(x)}sin(z) + 56.6y^{2}e^{(y^{3})}e^{(x)}sin(z)}$

Correct answer:

${1.59y^{2}e^{(y^{3})}e^{(x)}sin(z) - 0.85e^{(y^{3})}e^{(x)}cos(z)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(10)^2+(-16)^2}=18.87\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{18.87}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{10}{18.87}=0.53\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-16}{18.87}=-0.85\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(e^{(y^{3})}e^{(x)}sin(z))+(0.53)(3y^{2}e^{(y^{3})}e^{(x)}sin(z))+(-0.85)(e^{(y^{3})}e^{(x)}cos(z))\\&D_{\overrightarrow{u}}(x,y,z)=1.59y^{2}e^{(y^{3})}e^{(x)}sin(z) - 0.85e^{(y^{3})}e^{(x)}cos(z)\end{align*}$

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Example Question #1361 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=4zln(y)tan(x) -\frac{ e^{(x)}}{(yz)}\\&\text{In the direction of }\overrightarrow{v}=(0,-16,6).\end{align*}$

Possible Answers:

${\frac{(68.4\cdot (tan(x)^{2} + 1))}{y}-\frac{ (17.1e^{(x)})}{(y^{2}z^{2})}}$

${1.4ln(y)tan(x) -\frac{ (3.76ztan(x))}{y}+\frac{ (0.35e^{(x)})}{(yz^{2})}-\frac{ (0.94e^{(x)})}{(y^{2}z)}}$

${0.49ln(y)tan(x) +\frac{ (3.53ztan(x))}{y}+\frac{ (0.122e^{(x)})}{(yz^{2})}+\frac{ (0.884e^{(x)})}{(y^{2}z)}}$

${68.4ln(y)tan(x) +\frac{ (68.4ztan(x))}{y}+ 68.4zln(y)\cdot (tan(x)^{2} + 1) -\frac{ (17.1e^{(x)})}{(yz)}+\frac{ (17.1e^{(x)})}{(yz^{2})}+\frac{ (17.1e^{(x)})}{(y^{2}z)}}$

Correct answer:

${1.4ln(y)tan(x) -\frac{ (3.76ztan(x))}{y}+\frac{ (0.35e^{(x)})}{(yz^{2})}-\frac{ (0.94e^{(x)})}{(y^{2}z)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-16)^2+(6)^2}=17.09\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{17.09}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-16}{17.09}=-0.94\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{6}{17.09}=0.35\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(4zln(y)\cdot (tan(x)^{2} + 1) -\frac{ (e^{(x)})}{(yz)})+(-0.94)(\frac{(4ztan(x))}{y}+\frac{ e^{(x)}}{(y^{2}z)})+(0.35)(4ln(y)tan(x) +\frac{ e^{(x)}}{(yz^{2})})\\&D_{\overrightarrow{u}}(x,y,z)=1.4ln(y)tan(x) -\frac{ (3.76ztan(x))}{y}+\frac{ (0.35e^{(x)})}{(yz^{2})}-\frac{ (0.94e^{(x)})}{(y^{2}z)}\end{align*}$

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Example Question #301 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=x^{4}ln(z^{4})e^{(y)}\\&\text{In the direction of }\overrightarrow{v}=(0,-5,-12).\end{align*}$

Possible Answers:

${- 0.38x^{4}ln(z^{4})e^{(y)} -\frac{ (3.68x^{4}e^{(y)})}{z}}$

${\frac{(208x^{3}e^{(y)})}{z}}$

${52x^{3}ln(z^{4})e^{(y)} + 13x^{4}ln(z^{4})e^{(y)} +\frac{ (52x^{4}e^{(y)})}{z}}$

${0.144x^{4}ln(z^{4})e^{(y)} +\frac{ (3.39x^{4}e^{(y)})}{z}}$

Correct answer:

${- 0.38x^{4}ln(z^{4})e^{(y)} -\frac{ (3.68x^{4}e^{(y)})}{z}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-5)^2+(-12)^2}=13\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{13}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-5}{13}=-0.38\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-12}{13}=-0.92\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(4x^{3}ln(z^{4})e^{(y)})+(-0.38)(x^{4}ln(z^{4})e^{(y)})+(-0.92)(\frac{(4x^{4}e^{(y)})}{z})\\&D_{\overrightarrow{u}}(x,y,z)=- 0.38x^{4}ln(z^{4})e^{(y)} -\frac{ (3.68x^{4}e^{(y)})}{z}\end{align*}$

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Example Question #1362 : Partial Derivatives

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

Possible Answers:

${16.5x^{4}y^{3}sin(z) - 66x^{3}y^{3}cos(z) - 49.5x^{4}y^{2}cos(z) - 66\cdot 4^{(y^{3})}x^{3}cos(z^{3}) - 68.6\cdot 4^{(y^{3})}x^{4}y^{2}cos(z^{3}) + 49.5\cdot 4^{(y^{3})}x^{4}z^{2}sin(z^{3})}$

${16\cdot 4^{(y^{3})}x^{3}cos(z^{3}) + 16x^{3}y^{3}cos(z) + 48x^{4}y^{2}cos(z) + 66.5\cdot 4^{(y^{3})}x^{4}y^{2}cos(z^{3})}$

${0.96\cdot 4^{(y^{3})}x^{3}cos(z^{3}) + 0.96x^{3}y^{3}cos(z) + 2.91x^{4}y^{2}cos(z) + 4.03\cdot 4^{(y^{3})}x^{4}y^{2}cos(z^{3})}$

${198x^{3}y^{2}sin(z) + 823\cdot 4^{(y^{3})}x^{3}y^{2}z^{2}sin(z^{3})}$

Correct answer:

${0.96\cdot 4^{(y^{3})}x^{3}cos(z^{3}) + 0.96x^{3}y^{3}cos(z) + 2.91x^{4}y^{2}cos(z) + 4.03\cdot 4^{(y^{3})}x^{4}y^{2}cos(z^{3})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-4)^2+(-16)^2+(0)^2}=16.49\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-4}{16.49}=-0.24\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-16}{16.49}=-0.97\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{16.49}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.24)(- 4\cdot 4^{(y^{3})}x^{3}cos(z^{3}) - 4x^{3}y^{3}cos(z))+(-0.97)(- 3x^{4}y^{2}cos(z) - 4.16\cdot 4^{(y^{3})}x^{4}y^{2}cos(z^{3}))+(0)(x^{4}y^{3}sin(z) + 3\cdot 4^{(y^{3})}x^{4}z^{2}sin(z^{3}))\\&D_{\overrightarrow{u}}(x,y,z)=0.96\cdot 4^{(y^{3})}x^{3}cos(z^{3}) + 0.96x^{3}y^{3}cos(z) + 2.91x^{4}y^{2}cos(z) + 4.03\cdot 4^{(y^{3})}x^{4}y^{2}cos(z^{3})\end{align*}$

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Calculus 3 : Partial Derivatives

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