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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #194 : Directional Derivatives

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

Possible Answers:

${-546x^{2}y^{2}z^{2}cos(x^{3})sin(z^{3})}$

${0.0675y^{2}cos(z^{3})sin(x^{3}) - 2.94y^{3}z^{2}sin(x^{3})sin(z^{3})}$

${60.7y^{2}cos(z^{3})sin(x^{3}) + 60.7x^{2}y^{3}cos(x^{3})cos(z^{3}) - 60.7y^{3}z^{2}sin(x^{3})sin(z^{3})}$

${- 0.45y^{2}cos(z^{3})sin(x^{3}) - 2.97y^{3}z^{2}sin(x^{3})sin(z^{3})}$

Correct answer:

${- 0.45y^{2}cos(z^{3})sin(x^{3}) - 2.97y^{3}z^{2}sin(x^{3})sin(z^{3})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-3)^2+(20)^2}=20.22\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{20.22}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-3}{20.22}=-0.15\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{20}{20.22}=0.99\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(3x^{2}y^{3}cos(x^{3})cos(z^{3}))+(-0.15)(3y^{2}cos(z^{3})sin(x^{3}))+(0.99)(-3y^{3}z^{2}sin(x^{3})sin(z^{3}))\\&D_{\overrightarrow{u}}(x,y,z)=- 0.45y^{2}cos(z^{3})sin(x^{3}) - 2.97y^{3}z^{2}sin(x^{3})sin(z^{3})\end{align*}$

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Example Question #3622 : Calculus 3

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2^y\cdot 2^zcos(x)\\&\text{In the direction of }\overrightarrow{v}=(-15,0,-14).\end{align*}$

Possible Answers:

${28.4\cdot 2^y\cdot 2^zcos(x) - 20.5\cdot 2^y\cdot 2^zsin(x)}$

${0.73\cdot 2^y\cdot 2^zsin(x) - 0.471\cdot 2^y\cdot 2^zcos(x)}$

${-9.86\cdot 2^y\cdot 2^zsin(x)}$

${0.321\cdot 2^y\cdot 2^zcos(x) - 0.533\cdot 2^y\cdot 2^zsin(x)}$

Correct answer:

${0.73\cdot 2^y\cdot 2^zsin(x) - 0.471\cdot 2^y\cdot 2^zcos(x)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-15)^2+(0)^2+(-14)^2}=20.52\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-15}{20.52}=-0.73\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{20.52}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-14}{20.52}=-0.68\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.73)(-1\cdot 2^y\cdot 2^zsin(x))+(0)(0.693\cdot 2^y\cdot 2^zcos(x))+(-0.68)(0.693\cdot 2^y\cdot 2^zcos(x))\\&D_{\overrightarrow{u}}(x,y,z)=0.73\cdot 2^y\cdot 2^zsin(x) - 0.471\cdot 2^y\cdot 2^zcos(x)\end{align*}$

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Example Question #1261 : Partial Derivatives

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

Possible Answers:

${-\frac{(29.3\cdot 3^{(z^{2})}ze^{(x)})}{y}}$

${- 3\cdot 3^{(z^{2})}e^{(x)}ln(y) - 28.6\cdot 3^{(z^{2})}ze^{(x)}ln(y)}$

${-\frac{ (13.3\cdot 3^{(z^{2})}e^{(x)})}{y}- 13.3\cdot 3^{(z^{2})}e^{(x)}ln(y) - 29.3\cdot 3^{(z^{2})}ze^{(x)}ln(y)}$

${- 0.22\cdot 3^{(z^{2})}e^{(x)}ln(y) - 2.13\cdot 3^{(z^{2})}ze^{(x)}ln(y)}$

Correct answer:

${- 0.22\cdot 3^{(z^{2})}e^{(x)}ln(y) - 2.13\cdot 3^{(z^{2})}ze^{(x)}ln(y)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(3)^2+(0)^2+(13)^2}=13.34\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{3}{13.34}=0.22\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{13.34}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{13}{13.34}=0.97\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.22)(-1\cdot 3^{(z^{2})}e^{(x)}ln(y))+(0)(-\frac{(1\cdot 3^{(z^{2})}e^{(x)})}{y})+(0.97)(-2.2\cdot 3^{(z^{2})}ze^{(x)}ln(y))\\&D_{\overrightarrow{u}}(x,y,z)=- 0.22\cdot 3^{(z^{2})}e^{(x)}ln(y) - 2.13\cdot 3^{(z^{2})}ze^{(x)}ln(y)\end{align*}$

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Example Question #1262 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-2^{(x^{3})}y^{4}z^{2}\\&\text{In the direction of }\overrightarrow{v}=(-19,0,-8).\end{align*}$

Possible Answers:

${- 0.304\cdot 2^{(x^{3})}y^{4}z - 1.76\cdot 2^{(x^{3})}x^{2}y^{4}z^{2}}$

${-343\cdot 2^{(x^{3})}x^{2}y^{3}z}$

${0.78\cdot 2^{(x^{3})}y^{4}z + 1.91\cdot 2^{(x^{3})}x^{2}y^{4}z^{2}}$

${- 41.2\cdot 2^{(x^{3})}y^{4}z - 82.5\cdot 2^{(x^{3})}y^{3}z^{2} - 42.9\cdot 2^{(x^{3})}x^{2}y^{4}z^{2}}$

Correct answer:

${0.78\cdot 2^{(x^{3})}y^{4}z + 1.91\cdot 2^{(x^{3})}x^{2}y^{4}z^{2}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-19)^2+(0)^2+(-8)^2}=20.62\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-19}{20.62}=-0.92\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{20.62}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-8}{20.62}=-0.39\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.92)(-2.08\cdot 2^{(x^{3})}x^{2}y^{4}z^{2})+(0)(-4\cdot 2^{(x^{3})}y^{3}z^{2})+(-0.39)(-2\cdot 2^{(x^{3})}y^{4}z)\\&D_{\overrightarrow{u}}(x,y,z)=0.78\cdot 2^{(x^{3})}y^{4}z + 1.91\cdot 2^{(x^{3})}x^{2}y^{4}z^{2}\end{align*}$

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Example Question #1263 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-ln(x^{3})ln(y^{4})\\&\text{In the direction of }\overrightarrow{v}=(14,0,-20).\end{align*}$

Possible Answers:

${-\frac{ (97.6ln(x^{3}))}{y}-\frac{ (73.2ln(y^{4}))}{x}}$

${-\frac{(0.975ln(y^{4}))}{x}}$

${-\frac{(1.71ln(y^{4}))}{x}}$

{0}

Correct answer:

${-\frac{(1.71ln(y^{4}))}{x}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(14)^2+(0)^2+(-20)^2}=24.41\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{14}{24.41}=0.57\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{24.41}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-20}{24.41}=-0.82\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.57)(-\frac{(3ln(y^{4}))}{x})+(0)(-\frac{(4ln(x^{3}))}{y})+(-0.82)(0.0)\\&D_{\overrightarrow{u}}(x,y,z)=-\frac{(1.71ln(y^{4}))}{x}\end{align*}$

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Example Question #1264 : Partial Derivatives

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

Possible Answers:

${21.2yz^{2}sin(z^{3})e^{(x^{2})} - 7.07cos(z^{3})e^{(x^{2})} - 14.1xycos(z^{3})e^{(x^{2})}}$

${- 0.504cos(z^{3})e^{(x^{2})} - 1.01xycos(z^{3})e^{(x^{2})}}$

${42.4xz^{2}sin(z^{3})e^{(x^{2})}}$

${0.71cos(z^{3})e^{(x^{2})} + 1.42xycos(z^{3})e^{(x^{2})}}$

Correct answer:

${0.71cos(z^{3})e^{(x^{2})} + 1.42xycos(z^{3})e^{(x^{2})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-5)^2+(-5)^2+(0)^2}=7.07\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-5}{7.07}=-0.71\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-5}{7.07}=-0.71\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{7.07}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.71)(-2xycos(z^{3})e^{(x^{2})})+(-0.71)(-1cos(z^{3})e^{(x^{2})})+(0)(3yz^{2}sin(z^{3})e^{(x^{2})})\\&D_{\overrightarrow{u}}(x,y,z)=0.71cos(z^{3})e^{(x^{2})} + 1.42xycos(z^{3})e^{(x^{2})}\end{align*}$

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Example Question #1265 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=4^{(z^{3})}x^{2}y^{4}\\&\text{In the direction of }\overrightarrow{v}=(13,0,12).\end{align*}$

Possible Answers:

${26\cdot 4^{(z^{3})}xy^{4} + 49.9\cdot 4^{(z^{3})}x^{2}y^{4}z^{2}}$

${35.4\cdot 4^{(z^{3})}xy^{4} + 70.8\cdot 4^{(z^{3})}x^{2}y^{3} + 73.6\cdot 4^{(z^{3})}x^{2}y^{4}z^{2}}$

${1.46\cdot 4^{(z^{3})}xy^{4} + 2.83\cdot 4^{(z^{3})}x^{2}y^{4}z^{2}}$

${589\cdot 4^{(z^{3})}xy^{3}z^{2}}$

Correct answer:

${1.46\cdot 4^{(z^{3})}xy^{4} + 2.83\cdot 4^{(z^{3})}x^{2}y^{4}z^{2}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(13)^2+(0)^2+(12)^2}=17.69\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{13}{17.69}=0.73\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{17.69}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{12}{17.69}=0.68\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.73)(2\cdot 4^{(z^{3})}xy^{4})+(0)(4\cdot 4^{(z^{3})}x^{2}y^{3})+(0.68)(4.16\cdot 4^{(z^{3})}x^{2}y^{4}z^{2})\\&D_{\overrightarrow{u}}(x,y,z)=1.46\cdot 4^{(z^{3})}xy^{4} + 2.83\cdot 4^{(z^{3})}x^{2}y^{4}z^{2}\end{align*}$

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Example Question #1266 : Partial Derivatives

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

Possible Answers:

${163x^{3}ze^{(x^{4})}e^{(z^{2})}}$

${20.4e^{(x^{4})}e^{(z^{2})} + 81.6x^{3}ye^{(x^{4})}e^{(z^{2})} + 40.8yze^{(x^{4})}e^{(z^{2})}}$

${1.96yze^{(x^{4})}e^{(z^{2})} - 0.8x^{3}ye^{(x^{4})}e^{(z^{2})}}$

${0.16x^{3}ye^{(x^{4})}e^{(z^{2})} + 1.92yze^{(x^{4})}e^{(z^{2})}}$

Correct answer:

${1.96yze^{(x^{4})}e^{(z^{2})} - 0.8x^{3}ye^{(x^{4})}e^{(z^{2})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-4)^2+(0)^2+(20)^2}=20.4\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-4}{20.4}=-0.2\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{20.4}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{20}{20.4}=0.98\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.2)(4x^{3}ye^{(x^{4})}e^{(z^{2})})+(0)(e^{(x^{4})}e^{(z^{2})})+(0.98)(2yze^{(x^{4})}e^{(z^{2})})\\&D_{\overrightarrow{u}}(x,y,z)=1.96yze^{(x^{4})}e^{(z^{2})} - 0.8x^{3}ye^{(x^{4})}e^{(z^{2})}\end{align*}$

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Example Question #1267 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-\frac{(e^{(x^{2})}cos(y))}{z}\\&\text{In the direction of }\overrightarrow{v}=(9,-16,0).\end{align*}$

Possible Answers:

${-\frac{ (16e^{(x^{2})}sin(y))}{z}-\frac{ (18xe^{(x^{2})}cos(y))}{z}}$

${\frac{(18.4e^{(x^{2})}cos(y))}{z^{2}}+\frac{ (18.4e^{(x^{2})}sin(y))}{z}-\frac{ (36.7xe^{(x^{2})}cos(y))}{z}}$

${-\frac{ (0.87e^{(x^{2})}sin(y))}{z}-\frac{ (0.98xe^{(x^{2})}cos(y))}{z}}$

${-\frac{(36.7xe^{(x^{2})}sin(y))}{z^{2}}}$

Correct answer:

${-\frac{ (0.87e^{(x^{2})}sin(y))}{z}-\frac{ (0.98xe^{(x^{2})}cos(y))}{z}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(9)^2+(-16)^2+(0)^2}=18.36\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{9}{18.36}=0.49\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-16}{18.36}=-0.87\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{18.36}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.49)(-\frac{(2xe^{(x^{2})}cos(y))}{z})+(-0.87)(\frac{(e^{(x^{2})}sin(y))}{z})+(0)(\frac{(e^{(x^{2})}cos(y))}{z^{2}})\\&D_{\overrightarrow{u}}(x,y,z)=-\frac{ (0.87e^{(x^{2})}sin(y))}{z}-\frac{ (0.98xe^{(x^{2})}cos(y))}{z}\end{align*}$

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Example Question #203 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-y^{2}cos(z^{4})sin(x^{4})\\&\text{In the direction of }\overrightarrow{v}=(0,-18,-12).\end{align*}$

Possible Answers:

${86.5y^{2}z^{3}sin(x^{4})sin(z^{4}) - 86.5x^{3}y^{2}cos(x^{4})cos(z^{4}) - 43.3ycos(z^{4})sin(x^{4})}$

${1.66ycos(z^{4})sin(x^{4}) - 2.2y^{2}z^{3}sin(x^{4})sin(z^{4})}$

${36ycos(z^{4})sin(x^{4}) - 48y^{2}z^{3}sin(x^{4})sin(z^{4})}$

${692x^{3}yz^{3}cos(x^{4})sin(z^{4})}$

Correct answer:

${1.66ycos(z^{4})sin(x^{4}) - 2.2y^{2}z^{3}sin(x^{4})sin(z^{4})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-18)^2+(-12)^2}=21.63\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{21.63}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-18}{21.63}=-0.83\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-12}{21.63}=-0.55\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-4x^{3}y^{2}cos(x^{4})cos(z^{4}))+(-0.83)(-2ycos(z^{4})sin(x^{4}))+(-0.55)(4y^{2}z^{3}sin(x^{4})sin(z^{4}))\\&D_{\overrightarrow{u}}(x,y,z)=1.66ycos(z^{4})sin(x^{4}) - 2.2y^{2}z^{3}sin(x^{4})sin(z^{4})\end{align*}$

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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #194 : Directional Derivatives

Example Question #3622 : Calculus 3

Example Question #1261 : Partial Derivatives

Example Question #1262 : Partial Derivatives

Example Question #1263 : Partial Derivatives

Example Question #1264 : Partial Derivatives

Example Question #1265 : Partial Derivatives

Example Question #1266 : Partial Derivatives

Example Question #1267 : Partial Derivatives

Example Question #203 : Directional Derivatives

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