Derivatives
Limits
Integrals
Applications of Derivatives/ Integrals
Random
100

d/dx(x2)

2x

100

limx->2(x2-4)

0

100

int(x) dx

x2/2 + C

100

The relationship between m and v is defined by the equation 8x3=sqrt(4v). 

At the instant when dv/dt=12 and v=16, what is the value of dx/dt?

1/8
100

d/dx(lnx2)

2/x

200

d/dx(5x4 + 18x3 + 2x2 + 50x)

20x3 + 54x2 + 4x

200

limx->2((x2-x-2)/(x2-2x))

3/2 or 1.5

200

int(sin(x)) dx

-cos(x) + C

200

Let g be the function defined by g(x) = 5 + 4x − x3

What is the minimum value of the function on the interval [−2, 1] ?

1.921

200

limx->inf.(-7x2-2x-13)

-inf.

300

d/dx(-(x+1)/(4x+7))

-3/(4x+7)^2

300

limx-2((x2-4)/(x-2))

4

300

int(1/(x2+4)) dx

1/2arctan(x/2) + C

300

Let h(x) = x + 3ex−4

Use the equation of the tangent line to h(x) at x = 4 to approximate h(5).

11

300

int(xe2x) dx

(xe2x)/2-(e2x)/4+C

400

d/dx(arcsin(5x2))

10x/sqrt(1-25x4)

400

limx->3((1/(x2-1))-(2/(x4-1)))

1/10 or 0.1

400
int(1-tan(x+2)) dx
x+ln|cos(x+2)| + C
400

Bob rides his bike along a straight path. Bob’s velocity can be modeled by the function

v(t) = (t + 2)esqrt(t) meters per minute for 0 ≤ t ≤ 10 minutes.

Find Bob’s acceleration at time t = 4. Indicate units of measure.

At time t=4 minutes, Bob's acceleration is 18.473 meters per minutes squared.

400

f(x) = 4x2-12x, what is f''(x)?

f''(x)=8

500

d/dx((9x2-4x+9)/(2(x2-3)2sqrt(3x-1))

(-27x4+36x3-386x2+180x+57)/(4(x2-3)2(3x-1)sqrt(3x-1))

500

limx->-1((x3+x2+x+1)/(x4+x2-2))

-1/3 or -0.333

500

int(1/(x-sqrt(1-x2))) dx

1/2ln|tan2(1/2arcsin(x))+2tan(1/2arcsin(x))-1|-1/2ln|tan2(1/2arcsin(x))+1|-arctan(tan(1/2arcsin(x)))+C

500

The derivative of the function f is f'(x) = (x3+x)2cos(2.5x)

On what interval(s) in (-1, 1) is the graph of f concave down?

(-0.466, 0) and (0.466, 1)

500

limx->33((sqrt(x+3)-6)/(x-33))

1/12

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