Partial Derivatives - AP Calculus BC

Card 0 of 7240

Question

Find the differential of the following function:

$f(x, y, z)=x^5+\tan(y^2z)$

Answer

The differential of a function is given by

So, we must find the partial derivatives of the function. To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$f_y=2yz\sec^2(y^2z)$

$f_z=y^2\sec^2(y^2z)$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

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Question

Find the differential of the function

$f(x, y, z)=ze^y+\ln(x^2)$

Answer

The differential of a function is given by

The partial derivatives of the function are

$f_x=\frac{2}{x}$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \ln(u)=\frac{1}{u}\frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

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Question

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

Answer

$\begin{align}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\&\text{First make sure that }\overrightarrow{v}\&\text{is in unit vector form. Find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(8)^2+(19)^2+(0)^2}=20.62\&\text{Unit vector components are:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{8}{20.62}=0.39\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{19}{20.62}=0.92\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{20.62}=0\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(x,y,z)=(0.39)(4x^{3}cos(z^{4})ln(y^{4})sin(x^{4}))+(0.92)(-\frac{(4cos(x^{4})cos(z^{4}))}{y})+(0)(4z^{3}cos(x^{4})ln(y^{4})sin(z^{4}))\&D_{\overrightarrow{u}}(x,y,z)=1.56x^{3}cos(z^{4})ln(y^{4})sin(x^{4}) -\frac{ (3.68cos(x^{4})cos(z^{4}))}{y}\end{align}$

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Question

Evaluate:

$\lim_{(x,y)\rightarrow (0,1)} \frac{x^2+xy}{y}$

Answer

Since we won't have any divided by zero problems, we can just evaluate at the point.

$\lim_{(x,y)\rightarrow (0,1)} \frac{x^2+xy}{y}=\frac{0^2+0\cdot1}{1}=0$

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Question

Calculate the following limit.

$\small \lim_{(x,y)\to (1,0)} x^2+2xy+y^2$

Answer

We can calculate the limit

$\small \lim_{(x,y)\to (1,0)} x^2+2xy+y^2$

by simply plugging in $\small (x,y)=(1,0)$ because the value is defined at $\small (1,0)$ and around that point.

So we get

$\small \small \lim_{(x,y)\to (1,0)} x^2+2xy+y^2=1^2+2\cdot 0\cdot 1+0^2=1$ .

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Question

Calculate the following limit.

$\small \small \lim_{(x,y)\to (2,1)} 2x-5y$

Answer

We can calculate the limit

$\small \small \lim_{(x,y)\to (2,1)} 2x-5y$

by simply plugging in $\small \small (x,y)=(2,1)$ because the value is defined at $\small \small (2,1)$ and at every value in $\small \mathbb{R}^2$ .

So we get

$\small \small \small \lim_{(x,y)\to (2,1)} 2x-5y=2\cdot 2-5\cdot1=4-5=-1$ .

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Question

Evaluate $\lim_{(x,y)\to(0,0)}\frac{6x}{x^2+y}$

Answer

If we try evaluating the limit along two different paths, , we have

$\lim_{(x,x)\to(0,0)}\frac{6x}{x^2+x}= \lim_{(x,x)\to(0,0)}\frac{6}{x+1} = 6$

And

$\lim_{(x,2x)\to(0,0)}\frac{6x}{x^2+(2x)} = \lim_{(x,2x)\to(0,0)}\frac{6}{x+2} = 3$

Since evaluating along these two paths yielded diffferent values, the limit does not exist.

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Question

$\begin{align}&\text{Evaluate the following limit:}\&lim_{x\rightarrow13}-\frac{(x - 13)}{(12x - x^{2} + 13)}\end{align}$

Answer

$\begin{align}&\text{We cannot simply plug in the value of }x=13\text{, as that would}\&\text{create a zero value in the denominator. However, note that the expression}\&-\frac{(x - 13)}{(12x - x^{2} + 13)}\&\text{can be factored. This will allow us to find the limit:}\&\frac{(x - 13)}{((x + 1)\cdot (x - 13))}\&\frac{1}{(x + 1)}\&lim_{x\rightarrow13}-\frac{(x - 13)}{(12x - x^{2} + 13)}= 1/14\end{align}$

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Question

$\begin{align}&\text{Evaluate the following limit:}\&lim_{x\rightarrow12}\frac{(7x + x^{2} - 228)}{(1863x - 133x^{2} - 15x^{3} + x^{4} + 1980)}\end{align}$

Answer

$\begin{align}&\text{We cannot simply plug in the value of }x=12\text{, as that would}\&\text{create a zero value in the denominator. However, note that the expression}\&\frac{(7x + x^{2} - 228)}{(1863x - 133x^{2} - 15x^{3} + x^{4} + 1980)}\&\text{can be factored. This will allow us to find the limit:}\&\frac{((x - 12)\cdot (x + 19))}{((x + 1)\cdot (x + 11)\cdot (x - 12)\cdot (x - 15))}\&\frac{(x + 19)}{((x + 1)\cdot (x + 11)\cdot (x - 15))}\&lim_{x\rightarrow12}\frac{(7x + x^{2} - 228)}{(1863x - 133x^{2} - 15x^{3} + x^{4} + 1980)}= -\frac{31}{897}\end{align}$

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Question

$\begin{align}&\text{Evaluate the following limit:}\&lim_{x\rightarrow2}-\frac{(14x + x^{2} - 32)}{(144x + 11x^{2} - x^{3} - 324)}\end{align}$

Answer

$\begin{align}&\text{We cannot simply plug in the value of }x=2\text{, as that would}\&\text{create a zero value in the denominator. However, note that the expression}\&-\frac{(14x + x^{2} - 32)}{(144x + 11x^{2} - x^{3} - 324)}\&\text{can be factored. This will allow us to find the limit:}\&\frac{((x - 2)\cdot (x + 16))}{((x - 2)\cdot (x + 9)\cdot (x - 18))}\&\frac{(x + 16)}{((x + 9)\cdot (x - 18))}\&lim_{x\rightarrow2}-\frac{(14x + x^{2} - 32)}{(144x + 11x^{2} - x^{3} - 324)}= -\frac{9}{88}\end{align}$

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Question

$\begin{align}&\text{Evaluate the following limit:}\&lim_{x\rightarrow13}-\frac{(9x - x^{2} + 52)}{(2x + x^{2} - 195)}\end{align}$

Answer

$\begin{align}&\text{We cannot simply plug in the value of }x=13\text{, as that would}\&\text{create a zero value in the denominator. However, note that the expression}\&-\frac{(9x - x^{2} + 52)}{(2x + x^{2} - 195)}\&\text{can be factored. This will allow us to find the limit:}\&\frac{((x + 4)\cdot (x - 13))}{((x - 13)\cdot (x + 15))}\&\frac{(x + 4)}{(x + 15)}\&lim_{x\rightarrow13}-\frac{(9x - x^{2} + 52)}{(2x + x^{2} - 195)}=\frac{ 17}{28}\end{align}$

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Question

$\begin{align}&\text{Evaluate the following limit:}\&lim_{x\rightarrow-1}-\frac{(x + 1)}{(3576x + 98x^{2} - 23x^{3} - x^{4} + 3456)}\end{align}$

Answer

$\begin{align}&\text{We cannot simply plug in the value of }x=-1\text{, as that would}\&\text{create a zero value in the denominator. However, note that the expression}\&-\frac{(x + 1)}{(3576x + 98x^{2} - 23x^{3} - x^{4} + 3456)}\&\text{can be factored. This will allow us to find the limit:}\&\frac{(x + 1)}{((x + 1)\cdot (x - 12)\cdot (x + 16)\cdot (x + 18))}\&\frac{1}{((x - 12)\cdot (x + 16)\cdot (x + 18))}\&lim_{x\rightarrow-1}-\frac{(x + 1)}{(3576x + 98x^{2} - 23x^{3} - x^{4} + 3456)}= -\frac{1}{3315}\end{align}$

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Question

$\begin{align}&\text{Evaluate the following limit:}\&lim_{x\rightarrow-18}-\frac{(x - x^{2} + 342)}{(764x + 48x^{2} + x^{3} + 4032)}\end{align}$

Answer

$\begin{align}&\text{We cannot simply plug in the value of }x=-18\text{, as that would}\&\text{create a zero value in the denominator. However, note that the expression}\&-\frac{(x - x^{2} + 342)}{(764x + 48x^{2} + x^{3} + 4032)}\&\text{can be factored. This will allow us to find the limit:}\&\frac{((x + 18)\cdot (x - 19))}{((x + 14)\cdot (x + 16)\cdot (x + 18))}\&\frac{(x - 19)}{((x + 14)\cdot (x + 16))}\&lim_{x\rightarrow-18}-\frac{(x - x^{2} + 342)}{(764x + 48x^{2} + x^{3} + 4032)}= -\frac{37}{8}\end{align}$

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Question

$\begin{align}&\text{Evaluate the following limit:}\&lim_{x\rightarrow-17}\frac{(11x + x^{2} - 102)}{(5x^{3} - 565x^{2} - 1805x + x^{4} + 73644)}\end{align}$

Answer

$\begin{align}&\text{We cannot simply plug in the value of }x=-17\text{, as that would}\&\text{create a zero value in the denominator. However, note that the expression}\&\frac{(11x + x^{2} - 102)}{(5x^{3} - 565x^{2} - 1805x + x^{4} + 73644)}\&\text{can be factored. This will allow us to find the limit:}\&\frac{((x - 6)\cdot (x + 17))}{((x - 12)\cdot (x + 17)\cdot (x - 19)\cdot (x + 19))}\&\frac{(x - 6)}{((x - 12)\cdot (x - 19)\cdot (x + 19))}\&lim_{x\rightarrow-17}\frac{(11x + x^{2} - 102)}{(5x^{3} - 565x^{2} - 1805x + x^{4} + 73644)}= -\frac{23}{2088}\end{align}$

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Question

$\begin{align}&\text{Evaluate the following limit:}\&lim_{x\rightarrow12}\frac{(3x + x^{2} - 180)}{(13x^{3} - 500x^{2} - 3600x + x^{4} + 72000)}\end{align}$

Answer

$\begin{align}&\text{We cannot simply plug in the value of }x=12\text{, as that would}\&\text{create a zero value in the denominator. However, note that the expression}\&\frac{(3x + x^{2} - 180)}{(13x^{3} - 500x^{2} - 3600x + x^{4} + 72000)}\&\text{can be factored. This will allow us to find the limit:}\&\frac{((x - 12)\cdot (x + 15))}{((x - 12)\cdot (x - 15)\cdot (x + 20)^{2})}\&\frac{(x + 15)}{((x - 15)\cdot (x + 20)^{2})}\&lim_{x\rightarrow12}\frac{(3x + x^{2} - 180)}{(13x^{3} - 500x^{2} - 3600x + x^{4} + 72000)}= -\frac{9}{1024}\end{align}$

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Question

$\begin{align}&\text{Evaluate the following limit:}\&lim_{x\rightarrow7}\frac{(x^{2} - 9x + 14)}{(x^{2} - 23x + 112)}\end{align}$

Answer

$\begin{align}&\text{We cannot simply plug in the value of }x=7\text{, as that would}\&\text{create a zero value in the denominator. However, note that the expression}\&\frac{(x^{2} - 9x + 14)}{(x^{2} - 23x + 112)}\&\text{can be factored. This will allow us to find the limit:}\&\frac{((x - 2)\cdot (x - 7))}{((x - 7)\cdot (x - 16))}\&\frac{(x - 2)}{(x - 16)}\&lim_{x\rightarrow7}\frac{(x^{2} - 9x + 14)}{(x^{2} - 23x + 112)}= -\frac{5}{9}\end{align}$

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Question

$\begin{align}&\text{Evaluate the following limit:}\&lim_{x\rightarrow-13}-\frac{(25x + x^{2} + 156)}{(2930x + 103x^{2} - 20x^{3} - x^{4} + 5304)}\end{align}$

Answer

$\begin{align}&\text{We cannot simply plug in the value of }x=-13\text{, as that would}\&\text{create a zero value in the denominator. However, note that the expression}\&-\frac{(25x + x^{2} + 156)}{(2930x + 103x^{2} - 20x^{3} - x^{4} + 5304)}\&\text{can be factored. This will allow us to find the limit:}\&\frac{((x + 12)\cdot (x + 13))}{((x + 2)\cdot (x - 12)\cdot (x + 13)\cdot (x + 17))}\&\frac{(x + 12)}{((x + 2)\cdot (x - 12)\cdot (x + 17))}\&lim_{x\rightarrow-13}-\frac{(25x + x^{2} + 156)}{(2930x + 103x^{2} - 20x^{3} - x^{4} + 5304)}= -\frac{1}{1100}\end{align}$

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Question

$\begin{align}&\text{Evaluate the following limit:}\&lim_{x\rightarrow4}\frac{(x - 4)}{(x - 14x^{2} + x^{3} + 156)}\end{align}$

Answer

$\begin{align}&\text{We cannot simply plug in the value of }x=4\text{, as that would}\&\text{create a zero value in the denominator. However, note that the expression}\&\frac{(x - 4)}{(x - 14x^{2} + x^{3} + 156)}\&\text{can be factored. This will allow us to find the limit:}\&\frac{(x - 4)}{((x + 3)\cdot (x - 4)\cdot (x - 13))}\&\frac{1}{((x + 3)\cdot (x - 13))}\&lim_{x\rightarrow4}\frac{(x - 4)}{(x - 14x^{2} + x^{3} + 156)}= -\frac{1}{63}\end{align}$

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Question

$\begin{align}&\text{Evaluate the following limit:}\&lim_{x\rightarrow-6}-\frac{(33x^{2} - 144x + 14x^{3} + x^{4} - 324)}{(12x - x^{2} + 108)}\end{align}$

Answer

$\begin{align}&\text{We cannot simply plug in the value of }x=-6\text{, as that would}\&\text{create a zero value in the denominator. However, note that the expression}\&-\frac{(33x^{2} - 144x + 14x^{3} + x^{4} - 324)}{(12x - x^{2} + 108)}\&\text{can be factored. This will allow us to find the limit:}\&\frac{((x + 2)\cdot (x - 3)\cdot (x + 6)\cdot (x + 9))}{((x + 6)\cdot (x - 18))}\&\frac{((x + 2)\cdot (x - 3)\cdot (x + 9))}{(x - 18)}\&lim_{x\rightarrow-6}-\frac{(33x^{2} - 144x + 14x^{3} + x^{4} - 324)}{(12x - x^{2} + 108)}= -\frac{9}{2}\end{align}$

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Question

$\begin{align}&\text{Evaluate the following limit:}\&lim_{x\rightarrow-16}-\frac{(12416x + 156x^{2} - 31x^{3} - x^{4} + 97280)}{(31x^{2} - 7200x + 32x^{3} + x^{4} - 57600)}\end{align}$

Answer

$\begin{align}&\text{We cannot simply plug in the value of }x=-16\text{, as that would}\&\text{create a zero value in the denominator. However, note that the expression}\&-\frac{(12416x + 156x^{2} - 31x^{3} - x^{4} + 97280)}{(31x^{2} - 7200x + 32x^{3} + x^{4} - 57600)}\&\text{can be factored. This will allow us to find the limit:}\&\frac{((x + 16)^{2}\cdot (x + 19)\cdot (x - 20))}{((x - 15)\cdot (x + 15)\cdot (x + 16)^{2})}\&\frac{((x + 19)\cdot (x - 20))}{((x - 15)\cdot (x + 15))}\&lim_{x\rightarrow-16}-\frac{(12416x + 156x^{2} - 31x^{3} - x^{4} + 97280)}{(31x^{2} - 7200x + 32x^{3} + x^{4} - 57600)}= -\frac{108}{31}\end{align}$

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