Question 1

$\text { Use implicit differentiation to find } \frac { d y } { d x } \text {. }$ $x ^ { 3 } + 5 \ln y = 4$

Accepted Answer

A)  $- \frac { 3 x ^ { 2 } y } { 5 }$ 
B)  $\frac { 3 x ^ { 2 } y } { 5 }$ 
C)  $\frac { y \left( 4 + 3 x ^ { 2 } \right) y } { 5 }$ 
D)  $\frac { 3 x ^ { 2 } } { 5 y }$ 
E)  $\frac { y \left( 4 - 3 x ^ { 2 } \right) } { 5 }$ 
A)  $- \frac { 3 x ^ { 2 } y } { 5 }$ 
B)  $\frac { 3 x ^ { 2 } y } { 5 }$ 
C)  $\frac { y \left( 4 + 3 x ^ { 2 } \right) y } { 5 }$ 
D)  $\frac { 3 x ^ { 2 } } { 5 y }$ 
E)  $\frac { y \left( 4 - 3 x ^ { 2 } \right) } { 5 }$

Question 2

A projectile is shot upwards from the surface of the earth with an initial velocity of 108 meters per second. The position function is $s ( t ) = - 4.9 t ^ { 2 } + v _ { 0 } t + s _ { 0 }$ What is its velocity after 7 seconds?

Accepted Answer

A)  The velocity after 7 seconds is $181.7$ meters per second. 
B)  The velocity after 7 seconds is $- 142.3$ meters per second. 
C)  The velocity after 7 seconds is $39.4$ meters per second. 
D)  The velocity after 7 seconds is $73.7$ meters per second. 
E)  The velocity after 7 seconds is $- 176.6$ meters per second. 
A)  The velocity after 7 seconds is $181.7$ meters per second. 
B)  The velocity after 7 seconds is $- 142.3$ meters per second. 
C)  The velocity after 7 seconds is $39.4$ meters per second. 
D)  The velocity after 7 seconds is $73.7$ meters per second. 
E)  The velocity after 7 seconds is $- 176.6$ meters per second.

Question 3

Find the rate of change of the distance between the origin and a moving point on the graph of $y = x ^ { 2 } + 7$ if $\frac { d x } { d t } = 6$ centimeters per second.

Accepted Answer

A)  $\frac { d D } { d t } = \frac { 12 x ^ { 3 } + 90 x } { \sqrt { x ^ { 4 } + 15 x ^ { 2 } + 49 } }$ 
B)  $\frac { d D } { d t } = \frac { 6 x ^ { 3 } - 20 x } { x ^ { 3 } + 15 x ^ { 2 } + 49 }$ 
C)  $\frac { d D } { d t } = \frac { 12 x ^ { 3 } - 90 x } { \sqrt { x ^ { 3 } + 15 x ^ { 2 } + 49 } }$ 
D) $\frac { d D } { d t } = \frac { 6 x ^ { 3 } + 90 x } { \sqrt { x ^ { 4 } + 14 x ^ { 2 } + 49 } }$ 
E) $\frac { d D } { d t } = \frac { 12 x ^ { 3 } + 20 x } { x ^ { 4 } + 12 x ^ { 2 } + 49 }$ 
A)  $\frac { d D } { d t } = \frac { 12 x ^ { 3 } + 90 x } { \sqrt { x ^ { 4 } + 15 x ^ { 2 } + 49 } }$ 
B)  $\frac { d D } { d t } = \frac { 6 x ^ { 3 } - 20 x } { x ^ { 3 } + 15 x ^ { 2 } + 49 }$ 
C)  $\frac { d D } { d t } = \frac { 12 x ^ { 3 } - 90 x } { \sqrt { x ^ { 3 } + 15 x ^ { 2 } + 49 } }$ 
D) $\frac { d D } { d t } = \frac { 6 x ^ { 3 } + 90 x } { \sqrt { x ^ { 4 } + 14 x ^ { 2 } + 49 } }$ 
E) $\frac { d D } { d t } = \frac { 12 x ^ { 3 } + 20 x } { x ^ { 4 } + 12 x ^ { 2 } + 49 }$

Question 4

A petrol car is parked 50 feet from a long warehouse (see figure). The revolving light on top of the car turns at a rate of 30 revolutions per minute. How fast is the light beam moving along the wall when the beam makes an angle of $\theta = 45 ^ { \circ }$ with the perpendicular from the light to the wall.

Accepted Answer

A)  $1600 \pi \mathrm { ft } / \mathrm { sec }$ 
B)  $180 \pi \mathrm { ft } / \mathrm { sec }$ 
C)  $2500 \pi \mathrm { ft } / \mathrm { sec }$ 
D)  $400 \pi \mathrm { ft } / \mathrm { sec }$ 
E)  $100 \pi \mathrm { ft } / \mathrm { sec }$ 
A)  $1600 \pi \mathrm { ft } / \mathrm { sec }$ 
B)  $180 \pi \mathrm { ft } / \mathrm { sec }$ 
C)  $2500 \pi \mathrm { ft } / \mathrm { sec }$ 
D)  $400 \pi \mathrm { ft } / \mathrm { sec }$ 
E)  $100 \pi \mathrm { ft } / \mathrm { sec }$

Question 5

an equation of the tangent line to the graph of $y = \arcsin ( 7 x ) \text { at the point }$ $\left( \frac { 1 } { 7 \sqrt { 2 } } , \frac { \pi } { 4 } \right)$

Accepted Answer

A)  $y = \frac { 7 } { \sqrt { 2 } } x + \frac { \pi } { 4 }$ 
B)  $y = 7 \sqrt { 2 } x + \frac { \pi } { 4 }$ 
C)  $y = 7 \sqrt { 2 } x + \frac { \pi } { 4 } - 1$ 
D)  $y = \frac { 7 } { \sqrt { 2 } } x + \frac { \pi } { 4 } - 1$ 
E) $y = 49 \sqrt { 2 } x + \frac { \pi } { 4 } - 7$ 
A)  $y = \frac { 7 } { \sqrt { 2 } } x + \frac { \pi } { 4 }$ 
B)  $y = 7 \sqrt { 2 } x + \frac { \pi } { 4 }$ 
C)  $y = 7 \sqrt { 2 } x + \frac { \pi } { 4 } - 1$ 
D)  $y = \frac { 7 } { \sqrt { 2 } } x + \frac { \pi } { 4 } - 1$ 
E) $y = 49 \sqrt { 2 } x + \frac { \pi } { 4 } - 7$

Question 6

$\text { Find the derivative of the function } f ( x ) = \ln \left( \frac { 3 x } { x ^ { 2 } + 4 } \right) \text {. }$

Accepted Answer

A)  $f ^ { t } ( x ) = \frac { 1 } { x } + \frac { 2 x } { x ^ { 2 } + 4 }$ 
B) $f ^ { t} ( x ) = \frac { 4 - x ^ { 2 } } { x \left( x ^ { 2 } + 4 \right) }$ 
C) $f ^ { t} ( x ) = \frac { 1 } { x } - \frac { 2 x } { x ^ { 2 } + 4 }$ 
D)  $f ^ { t } ( x ) = \frac { 3 } { x } + \frac { 2 x } { x ^ { 2 } + 4 }$ 
E)  $f ^ { t } ( x ) = \frac { 1 } { 3 x } - \frac { 2 x } { x ^ { 2 } + 4 }$ 
A)  $f ^ { t } ( x ) = \frac { 1 } { x } + \frac { 2 x } { x ^ { 2 } + 4 }$ 
B) $f ^ { t} ( x ) = \frac { 4 - x ^ { 2 } } { x \left( x ^ { 2 } + 4 \right) }$ 
C) $f ^ { t} ( x ) = \frac { 1 } { x } - \frac { 2 x } { x ^ { 2 } + 4 }$ 
D)  $f ^ { t } ( x ) = \frac { 3 } { x } + \frac { 2 x } { x ^ { 2 } + 4 }$ 
E)  $f ^ { t } ( x ) = \frac { 1 } { 3 x } - \frac { 2 x } { x ^ { 2 } + 4 }$

Question 7

$\text { Use Newton's Method to approximate the zero(s) of the function } f ( x ) = x ^ { 3 } + x + 1$ accurate to three decimal places.

Accepted Answer

A)  $0.682$ 
B)  $- 0.694$ 
C)  $0.703$ 
D)  $0.649$ 
E)  $- 0.682$ 
A)  $0.682$ 
B)  $- 0.694$ 
C)  $0.703$ 
D)  $0.649$ 
E)  $- 0.682$

Question 8

$\text { Use logarithmic differentiation to find } \frac { d y } { d x } \text {. }$ $y = x ^ { 8 x }$

Accepted Answer

A)  $8 x ^ { 7 x }$ 
B)  $8 x ^ { 8 x - 1 }$ 
C)  $8 x ^ { 8 x } \ln x$ 
D)  $8 ( \ln x + 1 )$ 
E)  $8 x ^ { 8 x } ( \ln x + 1 )$ 
A)  $8 x ^ { 7 x }$ 
B)  $8 x ^ { 8 x - 1 }$ 
C)  $8 x ^ { 8 x } \ln x$ 
D)  $8 ( \ln x + 1 )$ 
E)  $8 x ^ { 8 x } ( \ln x + 1 )$

Question 9

$\text { Find } \frac { d y } { d x } \text { by implicit differentiation given that } \tan ( 4 x + y ) = 4 x \text {. Use the original }$ equation to simplify your answer.

Accepted Answer

A)  $\frac { d y } { d x } = \frac { 4 x } { x ^ { 2 } + 1 }$ 
B)  $\frac { d y } { d x } = - \frac { 4 x ^ { 2 } } { x ^ { 2 } - 1 }$ 
C)  $\frac { d y } { d x } = \frac { 4 x ^ { 2 } } { x ^ { 2 } + 1 }$ 
D)  $\frac { d y } { d x } = - \frac { 4 x ^ { 2 } } { x ^ { 2 } + 1 }$ 
E)  $\frac { d y } { d x } = \frac { 4 x ^ { 2 } } { x ^ { 2 } - 1 }$ 
A)  $\frac { d y } { d x } = \frac { 4 x } { x ^ { 2 } + 1 }$ 
B)  $\frac { d y } { d x } = - \frac { 4 x ^ { 2 } } { x ^ { 2 } - 1 }$ 
C)  $\frac { d y } { d x } = \frac { 4 x ^ { 2 } } { x ^ { 2 } + 1 }$ 
D)  $\frac { d y } { d x } = - \frac { 4 x ^ { 2 } } { x ^ { 2 } + 1 }$ 
E)  $\frac { d y } { d x } = \frac { 4 x ^ { 2 } } { x ^ { 2 } - 1 }$

Question 10

$\text { A buoy oscillates in simple harmonic motion } y = A \cos \text { at as waves move past }$ it. The buoy moves a total of $3.5$ feet (vertically) between its low point and its high point. It returns to its high point every 14 seconds. Write an equation describing the motion of the buoy if it is at its high point at $t = 0$.

Accepted Answer

A)  $y = 3.5 \cos \frac { \pi } { 7 } t$ 
B)  $y = 1.75 \cos 7 \pi t$ 
C)  $y = 1.75 \cos \frac { \pi } { 14 } t$ 
D)  $y = 1.75 \cos \frac { \pi } { 7 } t$ 
E)  $y = 3.5 \cos 14 \pi t$ 
A)  $y = 3.5 \cos \frac { \pi } { 7 } t$ 
B)  $y = 1.75 \cos 7 \pi t$ 
C)  $y = 1.75 \cos \frac { \pi } { 14 } t$ 
D)  $y = 1.75 \cos \frac { \pi } { 7 } t$ 
E)  $y = 3.5 \cos 14 \pi t$

Question 11

$\text { Use implicit differentiation to find } \frac { d y } { d x }$ $7 e ^ { x y } + 5 y ^ { 2 } = 7$

Accepted Answer

A) $\frac { d y } { d x } = - \frac { 7 e ^ { x y } } { 7 e ^ { x y } x - 10 y }$ 
B) 
$\frac { d y } { d x } = - \frac { 7 y e ^ { x y } } { 7 e ^ { x y } x + 10 y }$ 
C) 
$\frac { d y } { d x } = \frac { 7 y e ^ { x y } } { 7 e ^ { x y } - 10 y }$ 
D) 
$\frac { d y } { d x } = - \frac { 7 y e ^ { x y } } { 7 e ^ { x y } x + 5 y }$ 
E) 
$\frac { d y } { d x } = \frac { 7 y e ^ { x y } } { 7 e ^ { x y } + 10 y }$ 
A) $\frac { d y } { d x } = - \frac { 7 e ^ { x y } } { 7 e ^ { x y } x - 10 y }$ 
B) 
$\frac { d y } { d x } = - \frac { 7 y e ^ { x y } } { 7 e ^ { x y } x + 10 y }$ 
C) 
$\frac { d y } { d x } = \frac { 7 y e ^ { x y } } { 7 e ^ { x y } - 10 y }$ 
D) 
$\frac { d y } { d x } = - \frac { 7 y e ^ { x y } } { 7 e ^ { x y } x + 5 y }$ 
E) 
$\frac { d y } { d x } = \frac { 7 y e ^ { x y } } { 7 e ^ { x y } + 10 y }$

Question 12

$\text { Find the derivative of the function } y = \arctan \left( \frac { x } { 7 } \right) + \frac { 4 x - 6 } { 7 \left( x ^ { 2 } + 4 \right) } \text {. }$

Accepted Answer

A) $\frac { d y } { d x } = \frac { 1 } { 7 } \cdot \left( \frac { 1 } { x + 49 } + \frac { 16 - 12 x - 4 x ^ { 2 } } { \left( x ^ { 2 } + 4 \right) ^ { 2 } } \right)$ 
B) $\frac { d y } { d x } = \frac { 1 } { 7 } \cdot \left( \frac { 1 } { 1 + ( x / 7 ) ^ { 2 } } + \frac { 16 + 12 x - 4 x ^ { 2 } } { \left( x ^ { 2 } + 4 \right) ^ { 2 } } \right)$ 
C)  $\frac { d y } { d x } = \frac { 1 } { 7 } \cdot \left( \frac { 1 } { ( x + 7 ) ^ { 2 } } + \frac { 4 x } { x ^ { 2 } + 16 } \right)$ 
D) $\frac { d y } { d x } = \frac { 1 } { 7 } \cdot \left( \frac { 1 } { x + 49 } + \frac { 16 + 12 x - 4 x ^ { 2 } } { \left( x ^ { 2 } + 4 \right) ^ { 2 } } \right)$ 
E) $\frac { d y } { d x } = \frac { 1 } { 7 } \cdot \left( \frac { 1 } { x + 49 } + \frac { 4 x } { \left( x ^ { 2 } + 4 \right) ^ { 2 } } \right)$ 
A) $\frac { d y } { d x } = \frac { 1 } { 7 } \cdot \left( \frac { 1 } { x + 49 } + \frac { 16 - 12 x - 4 x ^ { 2 } } { \left( x ^ { 2 } + 4 \right) ^ { 2 } } \right)$ 
B) $\frac { d y } { d x } = \frac { 1 } { 7 } \cdot \left( \frac { 1 } { 1 + ( x / 7 ) ^ { 2 } } + \frac { 16 + 12 x - 4 x ^ { 2 } } { \left( x ^ { 2 } + 4 \right) ^ { 2 } } \right)$ 
C)  $\frac { d y } { d x } = \frac { 1 } { 7 } \cdot \left( \frac { 1 } { ( x + 7 ) ^ { 2 } } + \frac { 4 x } { x ^ { 2 } + 16 } \right)$ 
D) $\frac { d y } { d x } = \frac { 1 } { 7 } \cdot \left( \frac { 1 } { x + 49 } + \frac { 16 + 12 x - 4 x ^ { 2 } } { \left( x ^ { 2 } + 4 \right) ^ { 2 } } \right)$ 
E) $\frac { d y } { d x } = \frac { 1 } { 7 } \cdot \left( \frac { 1 } { x + 49 } + \frac { 4 x } { \left( x ^ { 2 } + 4 \right) ^ { 2 } } \right)$

Question 13

A manufacturer of digital audio players estimates that the profit for selling a particular model is $P = - 76 x ^ { 3 } + 4830 x ^ { 3 } - 320,000,00 \leq x \leq 60$, where $P$ is the profit in dollars and $x$ is the advertising expense in 10,000 's of dollars (see figure). Find the smaller of two advertising amounts that yield a profit $P$ of $\$ 2,100,000$. Round your answer to the nearest dollar. Advertising expense (in 10,000 s of dollars)

Accepted Answer

A)  $\$ 460,892$ 
B)  $\$ 315,369$ 
C)  $\$ 363,471$ 
D)  $\$ 512,458$ 
E)  $\$ 384,356$ 
A)  $\$ 460,892$ 
B)  $\$ 315,369$ 
C)  $\$ 363,471$ 
D)  $\$ 512,458$ 
E)  $\$ 384,356$

Question 14

$f ( x ) = x - 2 \sqrt { x + 2 }$ Use Newton's Method to approximate the zero(s) of the function accurate to three decimal places.

Accepted Answer

A)  $5.464$ 
B)  $- 5.496$ 
C)  $5.401$ 
D)  $- 5.464$ 
E)  $5.515$ 
A)  $5.464$ 
B)  $- 5.496$ 
C)  $5.401$ 
D)  $- 5.464$ 
E)  $5.515$

Question 15

$\text { If the annual rate of inflation averages } 8 \% \text { over the next } 10 \text { years, the approximate }$ cost $C$ of goods or services during any year in that decade is $C ( t ) = P ( 1.08 ) ^ { t }$ where $t$ is the time in years and $P$ is the present cost. The price of an oil change for your car is presently $\$ 23.65$. Find the rate of change of $C$ with respect to $t$ when $t = 3$. Round your answer to three decimal places.

Accepted Answer

A)  $32.176$ dollars per year 
B)  $1.966$ dollars per year 
C)  $2.293$ dollars per year 
D)  $5.962$ dollars per year 
E)  $32.730$ dollars per year 
A)  $32.176$ dollars per year 
B)  $1.966$ dollars per year 
C)  $2.293$ dollars per year 
D)  $5.962$ dollars per year 
E)  $32.730$ dollars per year

Question 16

$\text { Use the Product Rule to differentiate } f ( t ) = t ^ { - 4 } \sin t \text {. }$

Accepted Answer

A)  $f ( t ) = t ^ { - 4 } \cos t - 4 t ^ { 3 } \sin t$ 
B)  $f ^ { t} ( t ) = - t ^ { - 4 } \cos t + 4 t ^ { - 5 } \sin t$ 
C)  $f ^ { t } ( t ) = - t ^ { - 4 } \cos t - 4 t ^ { - 3 } \sin t$ 
D)  $f ^ { t } ( t ) = t ^ { - 4 } \cos t + 4 t ^ { - 5 } \sin t$ 
E)  $f ^ { t} ( t ) = t ^ { - 4 } \cos t - 4 t ^ { - 5 } \sin t$ 
A)  $f ( t ) = t ^ { - 4 } \cos t - 4 t ^ { 3 } \sin t$ 
B)  $f ^ { t} ( t ) = - t ^ { - 4 } \cos t + 4 t ^ { - 5 } \sin t$ 
C)  $f ^ { t } ( t ) = - t ^ { - 4 } \cos t - 4 t ^ { - 3 } \sin t$ 
D)  $f ^ { t } ( t ) = t ^ { - 4 } \cos t + 4 t ^ { - 5 } \sin t$ 
E)  $f ^ { t} ( t ) = t ^ { - 4 } \cos t - 4 t ^ { - 5 } \sin t$

Question 17

$\text { The radius of a right circular cylinder is } \sqrt { 3 t + 6 } \text { and its height is } t ^ { 5 } \text {, where } t \text { is }$ time in seconds and the dimensions are in inches. Find the rate of change of the volume of the cylinder, V, with respect to time.

Accepted Answer

A)  $\frac { d V } { d t } = \pi t ^ { 4 } ( 30 + 15 t )$ cubic inches per second 
B)  $\frac { d V } { d t } = \pi t ^ { 4 } ( 30 + 18 t )$ cubic inches per second 
C)  $\frac { d V } { d t } = \pi t ^ { 3 } ( 30 + 18 t )$ cubic inches per second 
D)  $\frac { d V } { d t } = \pi t ^ { 4 } ( 6 + 18 t )$ cubic inches per second 
E)  $\frac { d V } { d t } = \pi t ^ { 5 } ( 30 + 18 t )$ cubic inches per second 
A)  $\frac { d V } { d t } = \pi t ^ { 4 } ( 30 + 15 t )$ cubic inches per second 
B)  $\frac { d V } { d t } = \pi t ^ { 4 } ( 30 + 18 t )$ cubic inches per second 
C)  $\frac { d V } { d t } = \pi t ^ { 3 } ( 30 + 18 t )$ cubic inches per second 
D)  $\frac { d V } { d t } = \pi t ^ { 4 } ( 6 + 18 t )$ cubic inches per second 
E)  $\frac { d V } { d t } = \pi t ^ { 5 } ( 30 + 18 t )$ cubic inches per second

Question 18

Find the slope of the graph of the function at the given value. $f ( x ) = - 2 x ^ { 2 } + \frac { 6 } { x ^ { 2 } }$ when $x = 5$

Accepted Answer

A)  $f ^ { t } ( 5 ) = - \frac { 2488 } { 125 }$ 
B)  $f ^ { t } ( 5 ) = \frac { 2488 } { 125 }$ 
C)  $f ^ { t } ( 5 ) = \frac { 2512 } { 125 }$ 
D)  $f ^ { t } ( 5 ) = - \frac { 2512 } { 125 }$ 
E)  $f ^ { t } ( 5 ) = - \frac { 1256 } { 125 }$ 
A)  $f ^ { t } ( 5 ) = - \frac { 2488 } { 125 }$ 
B)  $f ^ { t } ( 5 ) = \frac { 2488 } { 125 }$ 
C)  $f ^ { t } ( 5 ) = \frac { 2512 } { 125 }$ 
D)  $f ^ { t } ( 5 ) = - \frac { 2512 } { 125 }$ 
E)  $f ^ { t } ( 5 ) = - \frac { 1256 } { 125 }$

Question 19

Use the alternative form of the derivative to find the derivative of the function $$f ( x ) = x ^ { 2 } - 9 \text { at } x = 5$$

Accepted Answer

A)  $f ^ { t } ( 5 ) = 1$ 
B)  $f ^ { t } ( 5 ) = 250$ 
C)  $f ^ { t } ( 5 ) = 2$ 
D)  $f ^ {t} ( 5 ) = 125$ 
E)  $f ^ { t } ( 5 ) = 10$ 
A)  $f ^ { t } ( 5 ) = 1$ 
B)  $f ^ { t } ( 5 ) = 250$ 
C)  $f ^ { t } ( 5 ) = 2$ 
D)  $f ^ {t} ( 5 ) = 125$ 
E)  $f ^ { t } ( 5 ) = 10$

Question 20

Given the derivative below find the requested higher-order derivative. $f ^ { tt } ( x ) = 8 x ^ { \frac { 7 } { 5 } } , f ^ { ( i v ) } ( x )$

Accepted Answer

A)  $f ^ { ( i v ) } ( x ) = - ( 56 / 5 ) x ^ { 2 / 5 }$ 
B)  $f ^ { ( i v ) } ( x ) = ( 112 / 25 ) x ^ { 2 / 5 }$ 
C) $f ^ { ( i v ) } ( x ) = ( 112 / 25 ) x ^ { \frac { - 3 } { 5 } }$ 
D)  $f ^ { ( i v ) } ( x ) = ( 56 / 5 ) x ^ { 2 / 5 }$ 
E)  $f ^ { ( i v ) } ( x ) = - ( 112 / 25 ) x ^ { \frac { - 3 } { 5 } }$ 
A)  $f ^ { ( i v ) } ( x ) = - ( 56 / 5 ) x ^ { 2 / 5 }$ 
B)  $f ^ { ( i v ) } ( x ) = ( 112 / 25 ) x ^ { 2 / 5 }$ 
C) $f ^ { ( i v ) } ( x ) = ( 112 / 25 ) x ^ { \frac { - 3 } { 5 } }$ 
D)  $f ^ { ( i v ) } ( x ) = ( 56 / 5 ) x ^ { 2 / 5 }$ 
E)  $f ^ { ( i v ) } ( x ) = - ( 112 / 25 ) x ^ { \frac { - 3 } { 5 } }$

$\text { Use implicit differentiation to find } \frac { d y } { d x } \text {. }$ $x ^ { 3 } + 5 \ln y = 4$

A projectile is shot upwards from the surface of the earth with an initial velocity of 108 meters per second. The position function is $s ( t ) = - 4.9 t ^ { 2 } + v _ { 0 } t + s _ { 0 }$ What is its velocity after 7 seconds?

Find the rate of change of the distance between the origin and a moving point on the graph of $y = x ^ { 2 } + 7$ if $\frac { d x } { d t } = 6$ centimeters per second.

an equation of the tangent line to the graph of $y = \arcsin ( 7 x ) \text { at the point }$ $\left( \frac { 1 } { 7 \sqrt { 2 } } , \frac { \pi } { 4 } \right)$

$\text { Find the derivative of the function } f ( x ) = \ln \left( \frac { 3 x } { x ^ { 2 } + 4 } \right) \text {. }$

$\text { Use Newton's Method to approximate the zero(s) of the function } f ( x ) = x ^ { 3 } + x + 1$ accurate to three decimal places.

$\text { Use logarithmic differentiation to find } \frac { d y } { d x } \text {. }$ $y = x ^ { 8 x }$

$\text { Find } \frac { d y } { d x } \text { by implicit differentiation given that } \tan ( 4 x + y ) = 4 x \text {. Use the original }$ equation to simplify your answer.

$\text { Use implicit differentiation to find } \frac { d y } { d x }$ $7 e ^ { x y } + 5 y ^ { 2 } = 7$

$\text { Find the derivative of the function } y = \arctan \left( \frac { x } { 7 } \right) + \frac { 4 x - 6 } { 7 \left( x ^ { 2 } + 4 \right) } \text {. }$

$f ( x ) = x - 2 \sqrt { x + 2 }$ Use Newton's Method to approximate the zero(s) of the function accurate to three decimal places.

$\text { Use the Product Rule to differentiate } f ( t ) = t ^ { - 4 } \sin t \text {. }$

$\text { The radius of a right circular cylinder is } \sqrt { 3 t + 6 } \text { and its height is } t ^ { 5 } \text {, where } t \text { is }$ time in seconds and the dimensions are in inches. Find the rate of change of the volume of the cylinder, V, with respect to time.

Find the slope of the graph of the function at the given value. $f ( x ) = - 2 x ^ { 2 } + \frac { 6 } { x ^ { 2 } }$ when $x = 5$

Use the alternative form of the derivative to find the derivative of the function $f ( x ) = x ^ { 2 } - 9 \text { at } x = 5$

Given the derivative below find the requested higher-order derivative. $f ^ { tt } ( x ) = 8 x ^ { \frac { 7 } { 5 } } , f ^ { ( i v ) } ( x )$

Graphs and Models

A Preview of Calculus

Extrema on an Interval

Antiderivatives and Indefinite Integration

Slope Fields and Eulers Method

Area of a Region Between Two Curves

Basic Integration Rules

Sequences

Conics and Calculus

Vectors in the Plane

Vector-Valued Functions

Introduction to Functions of Several Variables

Iterated Integrals and Area in the Plane

Vector Fields

Exact First-Order Equations

Filters

Exam 3: The Derivative and the Tangent Line Problem

Use implicit differentiation to find dydx. \text { Use implicit differentiation to find } \frac { d y } { d x } \text {. } Use implicit differentiation to find dxdy​. x3+5ln⁡y=4x ^ { 3 } + 5 \ln y = 4x3+5lny=4

A projectile is shot upwards from the surface of the earth with an initial velocity of 108 meters per second. The position function is s(t)=−4.9t2+v0t+s0s ( t ) = - 4.9 t ^ { 2 } + v _ { 0 } t + s _ { 0 }s(t)=−4.9t2+v0​t+s0​ What is its velocity after 7 seconds?

Find the rate of change of the distance between the origin and a moving point on the graph of y=x2+7y = x ^ { 2 } + 7y=x2+7 if dxdt=6\frac { d x } { d t } = 6dtdx​=6 centimeters per second.

an equation of the tangent line to the graph of y=arcsin⁡(7x) at the point y = \arcsin ( 7 x ) \text { at the point }y=arcsin(7x) at the point (172,π4)\left( \frac { 1 } { 7 \sqrt { 2 } } , \frac { \pi } { 4 } \right)(72​1​,4π​)

Find the derivative of the function f(x)=ln⁡(3xx2+4). \text { Find the derivative of the function } f ( x ) = \ln \left( \frac { 3 x } { x ^ { 2 } + 4 } \right) \text {. } Find the derivative of the function f(x)=ln(x2+43x​).

Use Newton’s Method to approximate the zero(s) of the function f(x)=x3+x+1\text { Use Newton's Method to approximate the zero(s) of the function } f ( x ) = x ^ { 3 } + x + 1 Use Newton’s Method to approximate the zero(s) of the function f(x)=x3+x+1 accurate to three decimal places.

Use logarithmic differentiation to find dydx. \text { Use logarithmic differentiation to find } \frac { d y } { d x } \text {. } Use logarithmic differentiation to find dxdy​. y=x8xy = x ^ { 8 x }y=x8x

Use implicit differentiation to find dydx\text { Use implicit differentiation to find } \frac { d y } { d x } Use implicit differentiation to find dxdy​ 7exy+5y2=77 e ^ { x y } + 5 y ^ { 2 } = 77exy+5y2=7

Find the derivative of the function y=arctan⁡(x7)+4x−67(x2+4). \text { Find the derivative of the function } y = \arctan \left( \frac { x } { 7 } \right) + \frac { 4 x - 6 } { 7 \left( x ^ { 2 } + 4 \right) } \text {. } Find the derivative of the function y=arctan(7x​)+7(x2+4)4x−6​.

f(x)=x−2x+2f ( x ) = x - 2 \sqrt { x + 2 }f(x)=x−2x+2​ Use Newton's Method to approximate the zero(s) of the function accurate to three decimal places.

Use the Product Rule to differentiate f(t)=t−4sin⁡t. \text { Use the Product Rule to differentiate } f ( t ) = t ^ { - 4 } \sin t \text {. } Use the Product Rule to differentiate f(t)=t−4sint.

Find the slope of the graph of the function at the given value. f(x)=−2x2+6x2f ( x ) = - 2 x ^ { 2 } + \frac { 6 } { x ^ { 2 } }f(x)=−2x2+x26​ when x=5x = 5x=5

Use the alternative form of the derivative to find the derivative of the function f(x)=x2−9 at x=5f ( x ) = x ^ { 2 } - 9 \text { at } x = 5f(x)=x2−9 at x=5

Given the derivative below find the requested higher-order derivative. ftt(x)=8x75,f(iv)(x)f ^ { tt } ( x ) = 8 x ^ { \frac { 7 } { 5 } } , f ^ { ( i v ) } ( x )ftt(x)=8x57​,f(iv)(x)

Graphs and Models

A Preview of Calculus

Extrema on an Interval

Antiderivatives and Indefinite Integration

Slope Fields and Eulers Method

Area of a Region Between Two Curves

Basic Integration Rules

Sequences

Conics and Calculus

Vectors in the Plane

Vector-Valued Functions

Introduction to Functions of Several Variables

Iterated Integrals and Area in the Plane

Vector Fields

Exact First-Order Equations

Filters

$\text { Use implicit differentiation to find } \frac { d y } { d x } \text {. }$ $x ^ { 3 } + 5 \ln y = 4$

A projectile is shot upwards from the surface of the earth with an initial velocity of 108 meters per second. The position function is $s ( t ) = - 4.9 t ^ { 2 } + v _ { 0 } t + s _ { 0 }$ What is its velocity after 7 seconds?

Find the rate of change of the distance between the origin and a moving point on the graph of $y = x ^ { 2 } + 7$ if $\frac { d x } { d t } = 6$ centimeters per second.

an equation of the tangent line to the graph of $y = \arcsin ( 7 x ) \text { at the point }$ $\left( \frac { 1 } { 7 \sqrt { 2 } } , \frac { \pi } { 4 } \right)$

$\text { Find the derivative of the function } f ( x ) = \ln \left( \frac { 3 x } { x ^ { 2 } + 4 } \right) \text {. }$

$\text { Use Newton's Method to approximate the zero(s) of the function } f ( x ) = x ^ { 3 } + x + 1$ accurate to three decimal places.

$\text { Use logarithmic differentiation to find } \frac { d y } { d x } \text {. }$ $y = x ^ { 8 x }$

$\text { Use implicit differentiation to find } \frac { d y } { d x }$ $7 e ^ { x y } + 5 y ^ { 2 } = 7$

$\text { Find the derivative of the function } y = \arctan \left( \frac { x } { 7 } \right) + \frac { 4 x - 6 } { 7 \left( x ^ { 2 } + 4 \right) } \text {. }$

$f ( x ) = x - 2 \sqrt { x + 2 }$ Use Newton's Method to approximate the zero(s) of the function accurate to three decimal places.

$\text { Use the Product Rule to differentiate } f ( t ) = t ^ { - 4 } \sin t \text {. }$

Find the slope of the graph of the function at the given value. $f ( x ) = - 2 x ^ { 2 } + \frac { 6 } { x ^ { 2 } }$ when $x = 5$

Use the alternative form of the derivative to find the derivative of the function $f ( x ) = x ^ { 2 } - 9 \text { at } x = 5$

Given the derivative below find the requested higher-order derivative. $f ^ { tt } ( x ) = 8 x ^ { \frac { 7 } { 5 } } , f ^ { ( i v ) } ( x )$