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Calculus 3 : Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1153 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=x^2cos{(z^3)}sin{(y)} - 4x^4sin{(y^3)}\\&\text{In the direction of }\overrightarrow{v}=(-15,7,0).\end{align*}$

Possible Answers:

${33.1xcos{(z^3)}sin{(y)} - 265x^3sin{(y^3)} - 199x^4y^2cos{(y^3)} + 16.5x^2cos{(z^3)}cos{(y)} - 49.7x^2z^2sin{(z^3)}sin{(y)}}$

${-99.3xz^2sin{(z^3)}cos{(y)}}$

${14.6x^3sin{(y^3)} - 1.82xcos{(z^3)}sin{(y)} - 5.04x^4y^2cos{(y^3)} + 0.42x^2cos{(z^3)}cos{(y)}}$

${240x^3sin{(y^3)} - 30xcos{(z^3)}sin{(y)} - 84x^4y^2cos{(y^3)} + 7x^2cos{(z^3)}cos{(y)}}$

Correct answer:

${14.6x^3sin{(y^3)} - 1.82xcos{(z^3)}sin{(y)} - 5.04x^4y^2cos{(y^3)} + 0.42x^2cos{(z^3)}cos{(y)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-15)^2+(7)^2+(0)^2}=16.55\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-15}{16.55}=-0.91\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{7}{16.55}=0.42\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{16.55}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.91)(2xcos{(z^3)}sin{(y)} - 16x^3sin{(y^3)})+(0.42)(x^2cos{(z^3)}cos{(y)} - 12x^4y^2cos{(y^3)})+(0)(-3x^2z^2sin{(z^3)}sin{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=14.6x^3sin{(y^3)} - 1.82xcos{(z^3)}sin{(y)} - 5.04x^4y^2cos{(y^3)} + 0.42x^2cos{(z^3)}cos{(y)}\end{align*}$

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Example Question #1154 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=4^{(y^2)}sin{(z^3)} - 3ln{(x^2)}e^{(z^2)}e^{(y)}\\&\text{In the direction of }\overrightarrow{v}=(15,0,20).\end{align*}$

Possible Answers:

${60\cdot4^{(y^2)}z^2cos{(z^3)} -\frac{ {(90e^{(z^2)}e^{(y)})}}{x} - 120zln{(x^2)}e^{(z^2)}e^{(y)}}$

${\frac{-{(300ze^{(z^2)}e^{(y)})}}{x}}$

${2.4\cdot4^{(y^2)}z^2cos{(z^3)} -\frac{ {(3.6e^{(z^2)}e^{(y)})}}{x} - 4.8zln{(x^2)}e^{(z^2)}e^{(y)}}$

${69.3\cdot4^{(y^2)}ysin{(z^3)} -\frac{ {(150e^{(z^2)}e^{(y)})}}{x} - 75ln{(x^2)}e^{(z^2)}e^{(y)} + 75\cdot4^{(y^2)}z^2cos{(z^3)} - 150zln{(x^2)}e^{(z^2)}e^{(y)}}$

Correct answer:

${2.4\cdot4^{(y^2)}z^2cos{(z^3)} -\frac{ {(3.6e^{(z^2)}e^{(y)})}}{x} - 4.8zln{(x^2)}e^{(z^2)}e^{(y)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(15)^2+(0)^2+(20)^2}=25\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{15}{25}=0.6\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{25}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{20}{25}=0.8\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.6)(\frac{-{(6e^{(z^2)}e^{(y)})}}{x})+(0)(2.77\cdot4^{(y^2)}ysin{(z^3)} - 3ln{(x^2)}e^{(z^2)}e^{(y)})+(0.8)(3\cdot4^{(y^2)}z^2cos{(z^3)} - 6zln{(x^2)}e^{(z^2)}e^{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=2.4\cdot4^{(y^2)}z^2cos{(z^3)} -\frac{ {(3.6e^{(z^2)}e^{(y)})}}{x} - 4.8zln{(x^2)}e^{(z^2)}e^{(y)}\end{align*}$

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Example Question #1155 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=sin{(x^2)}e^{(y)}cos{(z)} - 4y^2cos{(z^2)}\\&\text{In the direction of }\overrightarrow{v}=(5,7,0).\end{align*}$

Possible Answers:

${0.81sin{(x^2)}e^{(y)}cos{(z)} - 6.48ycos{(z^2)} + 1.16xcos{(x^2)}e^{(y)}cos{(z)}}$

${-17.2xcos{(x^2)}e^{(y)}sin{(z)}}$

${8.6sin{(x^2)}e^{(y)}cos{(z)} - 68.8ycos{(z^2)} - 8.6sin{(x^2)}e^{(y)}sin{(z)} + 68.8y^2zsin{(z^2)} + 17.2xcos{(x^2)}e^{(y)}cos{(z)}}$

${7sin{(x^2)}e^{(y)}cos{(z)} - 56ycos{(z^2)} + 10xcos{(x^2)}e^{(y)}cos{(z)}}$

Correct answer:

${0.81sin{(x^2)}e^{(y)}cos{(z)} - 6.48ycos{(z^2)} + 1.16xcos{(x^2)}e^{(y)}cos{(z)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(5)^2+(7)^2+(0)^2}=8.6\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{5}{8.6}=0.58\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{7}{8.6}=0.81\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{8.6}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.58)(2xcos{(x^2)}e^{(y)}cos{(z)})+(0.81)(sin{(x^2)}e^{(y)}cos{(z)} - 8ycos{(z^2)})+(0)(8y^2zsin{(z^2)} - sin{(x^2)}e^{(y)}sin{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=0.81sin{(x^2)}e^{(y)}cos{(z)} - 6.48ycos{(z^2)} + 1.16xcos{(x^2)}e^{(y)}cos{(z)}\end{align*}$

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Example Question #1156 : Partial Derivatives

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

Possible Answers:

${7.07e^{(x)}sin{(y)} - 7.07e^{(x)}cos{(y)}}$

{0}

${-5e^{(x)}cos{(y)}}$

${-0.71e^{(x)}cos{(y)}}$

Correct answer:

${-0.71e^{(x)}cos{(y)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(5)^2+(0)^2+(5)^2}=7.07\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{5}{7.07}=0.71\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{7.07}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{5}{7.07}=0.71\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.71)(-1e^{(x)}cos{(y)})+(0)(e^{(x)}sin{(y)})+(0.71)(0.0)\\&D_{\overrightarrow{u}}(x,y,z)=-0.71e^{(x)}cos{(y)}\end{align*}$

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Example Question #1157 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=y^3sin{(z^3)} - y^2z\\&\text{In the direction of }\overrightarrow{v}=(0,1,-2).\end{align*}$

Possible Answers:

${3y^2sin{(z^3)} - 2yz + 2y^2 - 6y^3z^2cos{(z^3)}}$

${6.72y^2sin{(z^3)} - 4.48yz - 2.24y^2 + 6.72y^3z^2cos{(z^3)}}$

{0}

${1.35y^2sin{(z^3)} - 0.9yz + 0.89y^2 - 2.67y^3z^2cos{(z^3)}}$

Correct answer:

${1.35y^2sin{(z^3)} - 0.9yz + 0.89y^2 - 2.67y^3z^2cos{(z^3)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(1)^2+(-2)^2}=2.24\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{2.24}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{1}{2.24}=0.45\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-2}{2.24}=-0.89\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(0.0)+(0.45)(3y^2sin{(z^3)} - 2yz)+(-0.89)(3y^3z^2cos{(z^3)} - y^2)\\&D_{\overrightarrow{u}}(x,y,z)=1.35y^2sin{(z^3)} - 0.9yz + 0.89y^2 - 2.67y^3z^2cos{(z^3)}\end{align*}$

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Example Question #91 : Directional Derivatives

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

Possible Answers:

${208\cdot3^{(x^2)}xy^2z}$

${90xe^{(y^2)} + 10\cdot3^{(x^2)}y^3z - 33\cdot3^{(x^2)}xy^3z^2}$

${31.6\cdot3^{(x^2)}y^3z - 94.9xe^{(y^2)} + 47.4\cdot3^{(x^2)}y^2z^2 - 94.9x^2ye^{(y^2)} + 34.7\cdot3^{(x^2)}xy^3z^2}$

${5.7xe^{(y^2)} + 0.64\cdot3^{(x^2)}y^3z - 2.09\cdot3^{(x^2)}xy^3z^2}$

Correct answer:

${5.7xe^{(y^2)} + 0.64\cdot3^{(x^2)}y^3z - 2.09\cdot3^{(x^2)}xy^3z^2}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-15)^2+(0)^2+(5)^2}=15.81\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-15}{15.81}=-0.95\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{15.81}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{5}{15.81}=0.32\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.95)(2.2\cdot3^{(x^2)}xy^3z^2 - 6xe^{(y^2)})+(0)(3\cdot3^{(x^2)}y^2z^2 - 6x^2ye^{(y^2)})+(0.32)(2\cdot3^{(x^2)}y^3z)\\&D_{\overrightarrow{u}}(x,y,z)=5.7xe^{(y^2)} + 0.64\cdot3^{(x^2)}y^3z - 2.09\cdot3^{(x^2)}xy^3z^2\end{align*}$

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Example Question #92 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=4^{(z^2)}cos{(x^3)}\\&\text{In the direction of }\overrightarrow{v}=(0,-19,18).\end{align*}$

Possible Answers:

{0}

${49.9\cdot4^{(z^2)}zcos{(x^3)}}$

${72.6\cdot4^{(z^2)}zcos{(x^3)} - 78.5\cdot4^{(z^2)}x^2sin{(x^3)}}$

${1.91\cdot4^{(z^2)}zcos{(x^3)}}$

Correct answer:

${1.91\cdot4^{(z^2)}zcos{(x^3)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-19)^2+(18)^2}=26.17\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{26.17}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-19}{26.17}=-0.73\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{18}{26.17}=0.69\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-3\cdot4^{(z^2)}x^2sin{(x^3)})+(-0.73)(0.0)+(0.69)(2.77\cdot4^{(z^2)}zcos{(x^3)})\\&D_{\overrightarrow{u}}(x,y,z)=1.91\cdot4^{(z^2)}zcos{(x^3)}\end{align*}$

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Example Question #93 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=4^{(z^2)}cos{(x^2)}\\&\text{In the direction of }\overrightarrow{v}=(-7,16,0).\end{align*}$

Possible Answers:

${0.8\cdot4^{(z^2)}xsin{(x^2)}}$

${14\cdot4^{(z^2)}xsin{(x^2)}}$

{0}

${48.4\cdot4^{(z^2)}zcos{(x^2)} - 34.9\cdot4^{(z^2)}xsin{(x^2)}}$

Correct answer:

${0.8\cdot4^{(z^2)}xsin{(x^2)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-7)^2+(16)^2+(0)^2}=17.46\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-7}{17.46}=-0.4\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{16}{17.46}=0.92\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{17.46}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.4)(-2\cdot4^{(z^2)}xsin{(x^2)})+(0.92)(0.0)+(0)(2.77\cdot4^{(z^2)}zcos{(x^2)})\\&D_{\overrightarrow{u}}(x,y,z)=0.8\cdot4^{(z^2)}xsin{(x^2)}\end{align*}$

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Example Question #94 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=x^3sin{(z^2)}sin{(y)} - 5yz^4\\&\text{In the direction of }\overrightarrow{v}=(0,-7,-13).\end{align*}$

Possible Answers:

${17.6yz^3 + 2.35z^4 - 0.47x^3sin{(z^2)}cos{(y)} - 1.76x^3zcos{(z^2)}sin{(y)}}$

${260yz^3 + 35z^4 - 7x^3sin{(z^2)}cos{(y)} - 26x^3zcos{(z^2)}sin{(y)}}$

${88.6x^2zcos{(z^2)}cos{(y)}}$

${14.8x^3sin{(z^2)}cos{(y)} - 73.8z^4 - 295yz^3 + 44.3x^2sin{(z^2)}sin{(y)} + 29.5x^3zcos{(z^2)}sin{(y)}}$

Correct answer:

${17.6yz^3 + 2.35z^4 - 0.47x^3sin{(z^2)}cos{(y)} - 1.76x^3zcos{(z^2)}sin{(y)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-7)^2+(-13)^2}=14.76\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{14.76}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-7}{14.76}=-0.47\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-13}{14.76}=-0.88\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(3x^2sin{(z^2)}sin{(y)})+(-0.47)(x^3sin{(z^2)}cos{(y)} - 5z^4)+(-0.88)(2x^3zcos{(z^2)}sin{(y)} - 20yz^3)\\&D_{\overrightarrow{u}}(x,y,z)=17.6yz^3 + 2.35z^4 - 0.47x^3sin{(z^2)}cos{(y)} - 1.76x^3zcos{(z^2)}sin{(y)}\end{align*}$

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Example Question #95 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=cos{(z^3)}e^{(x)} - 3xz^3cos{(y)}\\&\text{In the direction of }\overrightarrow{v}=(0,16,-18).\end{align*}$

Possible Answers:

${217z^2sin{(y)}}$

${2.25z^2sin{(z^3)}e^{(x)} + 6.75xz^2cos{(y)} + 1.98xz^3sin{(y)}}$

${54z^2sin{(z^3)}e^{(x)} + 162xz^2cos{(y)} + 48xz^3sin{(y)}}$

${24.1cos{(z^3)}e^{(x)} - 72.2z^3cos{(y)} - 72.2z^2sin{(z^3)}e^{(x)} - 217xz^2cos{(y)} + 72.2xz^3sin{(y)}}$

Correct answer:

${2.25z^2sin{(z^3)}e^{(x)} + 6.75xz^2cos{(y)} + 1.98xz^3sin{(y)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(16)^2+(-18)^2}=24.08\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{24.08}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{16}{24.08}=0.66\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-18}{24.08}=-0.75\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(cos{(z^3)}e^{(x)} - 3z^3cos{(y)})+(0.66)(3xz^3sin{(y)})+(-0.75)(- 3z^2sin{(z^3)}e^{(x)} - 9xz^2cos{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=2.25z^2sin{(z^3)}e^{(x)} + 6.75xz^2cos{(y)} + 1.98xz^3sin{(y)}\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1153 : Partial Derivatives

Example Question #1154 : Partial Derivatives

Example Question #1155 : Partial Derivatives

Example Question #1156 : Partial Derivatives

Example Question #1157 : Partial Derivatives

Example Question #91 : Directional Derivatives

Example Question #92 : Directional Derivatives

Example Question #93 : Directional Derivatives

Example Question #94 : Directional Derivatives

Example Question #95 : Directional Derivatives

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