d/dx(x2)
2x
limx->2(x2-4)
0
int(x) dx
x2/2 + C
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?
d/dx(lnx2)
2/x
d/dx(5x4 + 18x3 + 2x2 + 50x)
20x3 + 54x2 + 4x
limx->2((x2-x-2)/(x2-2x))
3/2 or 1.5
int(sin(x)) dx
-cos(x) + C
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
limx->inf.(-7x2-2x-13)
-inf.
d/dx(-(x+1)/(4x+7))
-3/(4x+7)^2
limx-2((x2-4)/(x-2))
4
int(1/(x2+4)) dx
1/2arctan(x/2) + C
Let h(x) = x + 3ex−4
Use the equation of the tangent line to h(x) at x = 4 to approximate h(5).
11
int(xe2x) dx
(xe2x)/2-(e2x)/4+C
d/dx(arcsin(5x2))
10x/sqrt(1-25x4)
limx->3((1/(x2-1))-(2/(x4-1)))
1/10 or 0.1
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.
f(x) = 4x2-12x, what is f''(x)?
f''(x)=8
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))
limx->-1((x3+x2+x+1)/(x4+x2-2))
-1/3 or -0.333
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
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)
limx->33((sqrt(x+3)-6)/(x-33))
1/12