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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #1201 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-3^{(y^2)}\cdot3^zln{(x)}\\&\text{In the direction of }\overrightarrow{v}=(-20,15,0).\end{align*}$

Possible Answers:

${-\frac{ {(25\cdot3^z\cdot3^{(y^2)})}}{x} - 27.5\cdot3^z\cdot3^{(y^2)}ln{(x)} - 54.9\cdot3^z\cdot3^{(y^2)}yln{(x)}}$

${-\frac{ {(0.64\cdot3^z\cdot3^{(y^2)})}}{x} - 0.791\cdot3^z\cdot3^{(y^2)}yln{(x)}}$

${\frac{{(0.8\cdot3^z\cdot3^{(y^2)})}}{x} - 1.32\cdot3^z\cdot3^{(y^2)}yln{(x)}}$

${\frac{-{(60.3\cdot3^z\cdot3^{(y^2)}y)}}{x}}$

Correct answer:

${\frac{{(0.8\cdot3^z\cdot3^{(y^2)})}}{x} - 1.32\cdot3^z\cdot3^{(y^2)}yln{(x)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-20)^2+(15)^2+(0)^2}=25\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-20}{25}=-0.8\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{15}{25}=0.6\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{25}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.8)(\frac{-{(1\cdot3^z\cdot3^{(y^2)})}}{x})+(0.6)(-2.2\cdot3^z\cdot3^{(y^2)}yln{(x)})+(0)(-1.1\cdot3^z\cdot3^{(y^2)}ln{(x)})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{{(0.8\cdot3^z\cdot3^{(y^2)})}}{x} - 1.32\cdot3^z\cdot3^{(y^2)}yln{(x)}\end{align*}$

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Example Question #1202 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{-{(cos{(x^3)}cos{(y^2)})}}{z}\\&\text{In the direction of }\overrightarrow{v}=(0,1,15).\end{align*}$

Possible Answers:

${\frac{{(15cos{(x^3)}cos{(y^2)})}}{z^2} +\frac{ {(30.1ycos{(x^3)}sin{(y^2)})}}{z} +\frac{ {(45.1x^2cos{(y^2)}sin{(x^3)})}}{z}}$

${\frac{{(cos{(x^3)}cos{(y^2)})}}{z^2} +\frac{ {(0.0098ycos{(x^3)}sin{(y^2)})}}{z}}$

${\frac{{(cos{(x^3)}cos{(y^2)})}}{z^2} +\frac{ {(0.14ycos{(x^3)}sin{(y^2)})}}{z}}$

${\frac{{(90.2x^2ysin{(x^3)}sin{(y^2)})}}{z^2}}$

Correct answer:

${\frac{{(cos{(x^3)}cos{(y^2)})}}{z^2} +\frac{ {(0.14ycos{(x^3)}sin{(y^2)})}}{z}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(1)^2+(15)^2}=15.03\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{15.03}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{1}{15.03}=0.07\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{15}{15.03}=1\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{{(3x^2cos{(y^2)}sin{(x^3)})}}{z})+(0.07)(\frac{{(2ycos{(x^3)}sin{(y^2)})}}{z})+(1)(\frac{{(cos{(x^3)}cos{(y^2)})}}{z^2})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{{(cos{(x^3)}cos{(y^2)})}}{z^2} +\frac{ {(0.14ycos{(x^3)}sin{(y^2)})}}{z}\end{align*}$

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Example Question #1203 : Partial Derivatives

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

Possible Answers:

${\frac{{(30.5\cdot4^{(z^2)}zcos{(x)})}}{y^2} -\frac{ {(8\cdot4^{(z^2)}cos{(x)})}}{y^3}}$

${\frac{{(64.9\cdot4^{(z^2)}zsin{(x)})}}{y^3}}$

${\frac{{(32.4\cdot4^{(z^2)}zcos{(x)})}}{y^2} -\frac{ {(11.7\cdot4^{(z^2)}sin{(x)})}}{y^2} -\frac{ {(23.4\cdot4^{(z^2)}cos{(x)})}}{y^3}}$

${\frac{{(2.61\cdot4^{(z^2)}zcos{(x)})}}{y^2} -\frac{ {(0.68\cdot4^{(z^2)}cos{(x)})}}{y^3}}$

Correct answer:

${\frac{{(2.61\cdot4^{(z^2)}zcos{(x)})}}{y^2} -\frac{ {(0.68\cdot4^{(z^2)}cos{(x)})}}{y^3}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(4)^2+(11)^2}=11.7\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{11.7}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{4}{11.7}=0.34\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{11}{11.7}=0.94\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{-{(1\cdot4^{(z^2)}sin{(x)})}}{y^2})+(0.34)(\frac{-{(2\cdot4^{(z^2)}cos{(x)})}}{y^3})+(0.94)(\frac{{(2.77\cdot4^{(z^2)}zcos{(x)})}}{y^2})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{{(2.61\cdot4^{(z^2)}zcos{(x)})}}{y^2} -\frac{ {(0.68\cdot4^{(z^2)}cos{(x)})}}{y^3}\end{align*}$

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Example Question #1204 : Partial Derivatives

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

Possible Answers:

${- 1.16zcos{(y^2)}cos{(z^2)}ln{(x)} - 1.62ysin{(y^2)}sin{(z^2)}ln{(x)}}$

${\frac{-{(34.4yzcos{(z^2)}sin{(y^2)})}}{x}}$

${0.673zcos{(y^2)}cos{(z^2)}ln{(x)} - 1.31ysin{(y^2)}sin{(z^2)}ln{(x)}}$

${\frac{{(8.6cos{(y^2)}sin{(z^2)})}}{x} + 17.2zcos{(y^2)}cos{(z^2)}ln{(x)} - 17.2ysin{(y^2)}sin{(z^2)}ln{(x)}}$

Correct answer:

${- 1.16zcos{(y^2)}cos{(z^2)}ln{(x)} - 1.62ysin{(y^2)}sin{(z^2)}ln{(x)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(7)^2+(-5)^2}=8.6\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{8.6}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{7}{8.6}=0.81\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-5}{8.6}=-0.58\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{{(cos{(y^2)}sin{(z^2)})}}{x})+(0.81)(-2ysin{(y^2)}sin{(z^2)}ln{(x)})+(-0.58)(2zcos{(y^2)}cos{(z^2)}ln{(x)})\\&D_{\overrightarrow{u}}(x,y,z)=- 1.16zcos{(y^2)}cos{(z^2)}ln{(x)} - 1.62ysin{(y^2)}sin{(z^2)}ln{(x)}\end{align*}$

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Example Question #1205 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=y^3sin{(x^2)}sin{(z)}\\&\text{In the direction of }\overrightarrow{v}=(0,-17,-3).\end{align*}$

Possible Answers:

${- 3y^3sin{(x^2)}cos{(z)} - 51y^2sin{(x^2)}sin{(z)}}$

${17.3y^3sin{(x^2)}cos{(z)} + 51.8y^2sin{(x^2)}sin{(z)} + 34.5xy^3cos{(x^2)}sin{(z)}}$

${104xy^2cos{(x^2)}cos{(z)}}$

${- 0.17y^3sin{(x^2)}cos{(z)} - 2.94y^2sin{(x^2)}sin{(z)}}$

Correct answer:

${- 0.17y^3sin{(x^2)}cos{(z)} - 2.94y^2sin{(x^2)}sin{(z)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-17)^2+(-3)^2}=17.26\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{17.26}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-17}{17.26}=-0.98\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-3}{17.26}=-0.17\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(2xy^3cos{(x^2)}sin{(z)})+(-0.98)(3y^2sin{(x^2)}sin{(z)})+(-0.17)(y^3sin{(x^2)}cos{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=- 0.17y^3sin{(x^2)}cos{(z)} - 2.94y^2sin{(x^2)}sin{(z)}\end{align*}$

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Example Question #1206 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-2^{(z^2)}\cdot3^xy^2\\&\text{In the direction of }\overrightarrow{v}=(-10,0,-15).\end{align*}$

Possible Answers:

${-54.9\cdot2^{(z^2)}\cdot3^xyz}$

${- 0.332\cdot2^{(z^2)}\cdot3^xy^2 - 0.955\cdot2^{(z^2)}\cdot3^xy^2z}$

${0.604\cdot2^{(z^2)}\cdot3^xy^2 + 1.15\cdot2^{(z^2)}\cdot3^xy^2z}$

${- 36.1\cdot2^{(z^2)}\cdot3^xy - 19.8\cdot2^{(z^2)}\cdot3^xy^2 - 25\cdot2^{(z^2)}\cdot3^xy^2z}$

Correct answer:

${0.604\cdot2^{(z^2)}\cdot3^xy^2 + 1.15\cdot2^{(z^2)}\cdot3^xy^2z}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-10)^2+(0)^2+(-15)^2}=18.03\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-10}{18.03}=-0.55\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{18.03}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-15}{18.03}=-0.83\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.55)(-1.1\cdot2^{(z^2)}\cdot3^xy^2)+(0)(-2\cdot2^{(z^2)}\cdot3^xy)+(-0.83)(-1.39\cdot2^{(z^2)}\cdot3^xy^2z)\\&D_{\overrightarrow{u}}(x,y,z)=0.604\cdot2^{(z^2)}\cdot3^xy^2 + 1.15\cdot2^{(z^2)}\cdot3^xy^2z\end{align*}$

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Example Question #1207 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-z^3sin{(y^3)}sin{(x)}\\&\text{In the direction of }\overrightarrow{v}=(0,19,11).\end{align*}$

Possible Answers:

${- 1.5z^2sin{(y^3)}sin{(x)} - 2.61y^2z^3cos{(y^3)}sin{(x)}}$

${- 0.75z^2sin{(y^3)}sin{(x)} - 2.27y^2z^3cos{(y^3)}sin{(x)}}$

${-198y^2z^2cos{(y^3)}cos{(x)}}$

${- 22z^3sin{(y^3)}cos{(x)} - 65.8z^2sin{(y^3)}sin{(x)} - 65.8y^2z^3cos{(y^3)}sin{(x)}}$

Correct answer:

${- 1.5z^2sin{(y^3)}sin{(x)} - 2.61y^2z^3cos{(y^3)}sin{(x)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(19)^2+(11)^2}=21.95\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{21.95}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{19}{21.95}=0.87\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{11}{21.95}=0.5\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-1z^3sin{(y^3)}cos{(x)})+(0.87)(-3y^2z^3cos{(y^3)}sin{(x)})+(0.5)(-3z^2sin{(y^3)}sin{(x)})\\&D_{\overrightarrow{u}}(x,y,z)=- 1.5z^2sin{(y^3)}sin{(x)} - 2.61y^2z^3cos{(y^3)}sin{(x)}\end{align*}$

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Example Question #141 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-4^{(y^2)}x^2z^2\\&\text{In the direction of }\overrightarrow{v}=(0,17,4).\end{align*}$

Possible Answers:

${- 0.46\cdot4^{(y^2)}x^2z - 2.69\cdot4^{(y^2)}x^2yz^2}$

${-194\cdot4^{(y^2)}xyz}$

${- 0.106\cdot4^{(y^2)}x^2z - 2.61\cdot4^{(y^2)}x^2yz^2}$

${- 34.9\cdot4^{(y^2)}xz^2 - 34.9\cdot4^{(y^2)}x^2z - 48.4\cdot4^{(y^2)}x^2yz^2}$

Correct answer:

${- 0.46\cdot4^{(y^2)}x^2z - 2.69\cdot4^{(y^2)}x^2yz^2}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(17)^2+(4)^2}=17.46\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{17.46}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{17}{17.46}=0.97\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{4}{17.46}=0.23\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-2\cdot4^{(y^2)}xz^2)+(0.97)(-2.77\cdot4^{(y^2)}x^2yz^2)+(0.23)(-2\cdot4^{(y^2)}x^2z)\\&D_{\overrightarrow{u}}(x,y,z)=- 0.46\cdot4^{(y^2)}x^2z - 2.69\cdot4^{(y^2)}x^2yz^2\end{align*}$

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Example Question #142 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-2^yzcos{(x)}\\&\text{In the direction of }\overrightarrow{v}=(0,-9,-11).\end{align*}$

Possible Answers:

${9.85\cdot2^ysin{(x)}}$

${14.2\cdot2^yzsin{(x)} - 9.85\cdot2^yzcos{(x)} - 14.2\cdot2^ycos{(x)}}$

${0.77\cdot2^ycos{(x)} + 0.437\cdot2^yzcos{(x)}}$

${11\cdot2^ycos{(x)} + 6.24\cdot2^yzcos{(x)}}$

Correct answer:

${0.77\cdot2^ycos{(x)} + 0.437\cdot2^yzcos{(x)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-9)^2+(-11)^2}=14.21\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{14.21}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-9}{14.21}=-0.63\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-11}{14.21}=-0.77\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(2^yzsin{(x)})+(-0.63)(-0.693\cdot2^yzcos{(x)})+(-0.77)(-1\cdot2^ycos{(x)})\\&D_{\overrightarrow{u}}(x,y,z)=0.77\cdot2^ycos{(x)} + 0.437\cdot2^yzcos{(x)}\end{align*}$

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Example Question #143 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-2^xyz^4\\&\text{In the direction of }\overrightarrow{v}=(-2,-9,0).\end{align*}$

Possible Answers:

${-25.6\cdot2^xz^3}$

${- 9.22\cdot2^xz^4 - 36.9\cdot2^xyz^3 - 6.39\cdot2^xyz^4}$

${0.98\cdot2^xz^4 + 0.152\cdot2^xyz^4}$

${- 0.96\cdot2^xz^4 - 0.0335\cdot2^xyz^4}$

Correct answer:

${0.98\cdot2^xz^4 + 0.152\cdot2^xyz^4}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-2)^2+(-9)^2+(0)^2}=9.22\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-2}{9.22}=-0.22\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-9}{9.22}=-0.98\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{9.22}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.22)(-0.693\cdot2^xyz^4)+(-0.98)(-1\cdot2^xz^4)+(0)(-4\cdot2^xyz^3)\\&D_{\overrightarrow{u}}(x,y,z)=0.98\cdot2^xz^4 + 0.152\cdot2^xyz^4\end{align*}$

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #1201 : Partial Derivatives

Example Question #1202 : Partial Derivatives

Example Question #1203 : Partial Derivatives

Example Question #1204 : Partial Derivatives

Example Question #1205 : Partial Derivatives

Example Question #1206 : Partial Derivatives

Example Question #1207 : Partial Derivatives

Example Question #141 : Directional Derivatives

Example Question #142 : Directional Derivatives

Example Question #143 : Directional Derivatives

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