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Multivariable Calculus : Partial Differentiation

Study concepts, example questions & explanations for Multivariable Calculus

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Example Questions

Example Question #1 : Differentiation Rules

Find $\frac{\partial f}{dz}$ .

$f(x,y,z)=xyz^{10}+\frac{xy}{\ln(yz)}+10e^z+e^{xy}$

Possible Answers:

$\frac{\partial f}{dz}=10xyz^9$

$\frac{\partial f}{dz}=\frac{xy}{z\ln^2(yz)}$

$\frac{\partial f}{dz}=10xyz^9+10e^z$

$\frac{\partial f}{dz}=10xyz^9-\frac{xy}{z\ln^2(yz)}+10e^z$

$\frac{\partial f}{dz}=10e^z$

Correct answer:

$\frac{\partial f}{dz}=10xyz^9-\frac{xy}{z\ln^2(yz)}+10e^z$

Explanation:

In order to find $\frac{\partial f}{dz}$ , we need to take the derivative of in respect to , and treat , and as constants. We also need to remember what the derivatives of natural log, exponential functions and power functions are for single variables.

Natural Log:

$f(x)=\ln(g(x))$

$f'(x)=\frac{g'(x)}{g(x)}$

Exponential Functions:

$f(x)=e^{g(x)}$

$f'(x)=g'(x)e^{g(x)}$

Power Functions:

f(x)=x^n

$f'(x)=nx^{n-1}$

$\frac{\partial f}{dz}=10xyz^9-xy(\ln(yz))^{-2}(\frac{y}{yz})+10e^z+0$

$\frac{\partial f}{dz}=10xyz^9-\frac{xy}{z\ln^2(yz)}+10e^z$

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Example Question #2 : Differentiation Rules

Find $\frac{\partial f}{dx}$ .

$f(x,y,z)=xyz^{10}+\frac{xy}{\ln(yz)}+10e^z+e^{xy}$

Possible Answers:

$\frac{\partial f}{dx}=\frac{y}{\ln(yz)}+ye^{xy}$

$\frac{\partial f}{dx}=xz^{10}+\frac{1}{\ln(yz)}+xe^{xy}$

$\frac{\partial f}{dx}=yz^{10}+\frac{y}{\ln(yz)}+ye^{xy}$

$\frac{\partial f}{dx}=yz^{10}+\frac{y}{\ln(yz)}$

$\frac{\partial f}{dx}=xyz^{10}+\frac{xy}{\ln(yz)}+ye^{xy}$

Correct answer:

$\frac{\partial f}{dx}=yz^{10}+\frac{y}{\ln(yz)}+ye^{xy}$

Explanation:

In order to find $\frac{\partial f}{dx}$ , we need to take the derivative of in respect to , and treat , and as constants. We also need to remember what the derivatives of natural log, exponential functions and power functions are for single variables.

Natural Log:

$f(x)=\ln(g(x))$

$f'(x)=\frac{g'(x)}{g(x)}$

Exponential Functions:

$f(x)=e^{g(x)}$

$f'(x)=g'(x)e^{g(x)}$

Power Functions:

f(x)=x^n

$f'(x)=nx^{n-1}$

$\frac{\partial f}{dx}=yz^{10}+\frac{y}{\ln(yz)}+0+ye^{xy}$

$\frac{\partial f}{dx}=yz^{10}+\frac{y}{\ln(yz)}+ye^{xy}$

Report an Error

Example Question #3 : Differentiation Rules

Find $\frac{\partial f}{dz}$ .

$f(x,y,z)=xyz^{10}+\frac{xy}{\ln(yz)}+10e^z+e^{xy}$

Possible Answers:

$\frac{\partial f}{dz}=10xyz^9+10e^z$

$\frac{\partial f}{dz}=\frac{xy}{z\ln^2(yz)}$

$\frac{\partial f}{dz}=10xyz^9$

$\frac{\partial f}{dz}=10xyz^9-\frac{xy}{z\ln^2(yz)}+10e^z$

$\frac{\partial f}{dz}=10e^z$

Correct answer:

$\frac{\partial f}{dz}=10xyz^9-\frac{xy}{z\ln^2(yz)}+10e^z$

Explanation:

Natural Log:

$f(x)=\ln(g(x))$

$f'(x)=\frac{g'(x)}{g(x)}$

Exponential Functions:

$f(x)=e^{g(x)}$

$f'(x)=g'(x)e^{g(x)}$

Power Functions:

f(x)=x^n

$f'(x)=nx^{n-1}$

$\frac{\partial f}{dz}=10xyz^9-xy(\ln(yz))^{-2}(\frac{y}{yz})+10e^z+0$

$\frac{\partial f}{dz}=10xyz^9-\frac{xy}{z\ln^2(yz)}+10e^z$

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Example Question #1 : Arc Length

Determine the length of the curve $\vec{r}(t)=\left \langle t,4\sin(2t),4\cos(2t)\right \rangle$ , on the interval $0\leq t\leq 2\pi$ .

Possible Answers:

$\pi$

$2\pi$

$2\pi\sqrt{65}$

$\sqrt{65}$

Correct answer:

$2\pi\sqrt{65}$

Explanation:

First we need to find the tangent vector, and find its magnitude.

$\vec{v}(t)=\vec{r}\ '(t)=\left \langle 1, 8\cos(2t), -8\sin(2t)\right \rangle$

$\left \| \vec{v}(t)\right \|=\sqrt{1^2+(8\cos(2t))^2+(-8\sin(2t))^2}$

$\left \| \vec{v}(t)\right \|=\sqrt{1+64\cos^2(2t)+64\sin^2(2t)}$

$\left \| \vec{v}(t)\right \|=\sqrt{1+64(\cos^2(2t)+\sin^2(2t))}$

$\left \| \vec{v}(t)\right \|=\sqrt{1+64}=\sqrt{65}$

Now we can set up our arc length integral

$L=\int_{a}^{b}\left \| \vec{v}(t)\right \|dt$

$L=\int_{0}^{2\pi} \sqrt{65}\ dt=t\sqrt{65}\Big|_{0}^{2\pi}=2\pi\sqrt{65}$

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Example Question #11 : Multivariable Calculus

Determine the length of the curve $\vec{r}(t)=\left \langle t,4\sin(2t),4\cos(2t)\right \rangle$ , on the interval $0\leq t\leq 2\pi$ .

Possible Answers:

$2\pi$

$2\pi\sqrt{65}$

$\pi$

$\sqrt{65}$

Correct answer:

$2\pi\sqrt{65}$

Explanation:

First we need to find the tangent vector, and find its magnitude.

$\vec{v}(t)=\vec{r}\ '(t)=\left \langle 1, 8\cos(2t), -8\sin(2t)\right \rangle$

$\left \| \vec{v}(t)\right \|=\sqrt{1^2+(8\cos(2t))^2+(-8\sin(2t))^2}$

$\left \| \vec{v}(t)\right \|=\sqrt{1+64\cos^2(2t)+64\sin^2(2t)}$

$\left \| \vec{v}(t)\right \|=\sqrt{1+64(\cos^2(2t)+\sin^2(2t))}$

$\left \| \vec{v}(t)\right \|=\sqrt{1+64}=\sqrt{65}$

Now we can set up our arc length integral

$L=\int_{a}^{b}\left \| \vec{v}(t)\right \|dt$

$L=\int_{0}^{2\pi} \sqrt{65}\ dt=t\sqrt{65}\Big|_{0}^{2\pi}=2\pi\sqrt{65}$

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Multivariable Calculus : Partial Differentiation

Study concepts, example questions & explanations for Multivariable Calculus

All Multivariable Calculus Resources

Example Questions

Example Question #1 : Differentiation Rules

Example Question #2 : Differentiation Rules

Example Question #3 : Differentiation Rules

Example Question #1 : Arc Length

Example Question #11 : Multivariable Calculus

All Multivariable Calculus Resources

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