Derivatives
The derivative of ex
What is
ex
L'hopital's Rule
What is
lim as x approaching a (f(x)/g(x))=(f'(x)/g'(x)) when it's ∞/∞ or 0/0
The formula for Integration By Parts
What is
uv-∫vdx
When should you use disk method and shell method?
What is
You should use disk method when y = f(x) is being rotated around the x - axis, and you should use shell method when y = f(x) is rotated around the y - axis
The Sum Law for Limits
What is
lim𝑥→𝑎(𝑓(𝑥)+𝑔(𝑥))=lim𝑥→𝑎𝑓(𝑥)+lim𝑥→𝑎𝑔(𝑥)
The derivative of ln(x)
What is
1/x
f(x) = x2 - 4x [1,3]
Does this function satisfy the conditions of the Rolle's Theorem and find the value where f'(c) = 0
What is
c = 2
1. It is continuous and differentiable, so it does satisfy the Rolle's Theorem
2. Take derivative
= 2x - 4
3. Find where it equals zero
2x - 4 = 0
x = 2
4. c = 2
When to use u-sub
What is
When you see that your function is being multiplied by its derivative
Find the area bounded by the function f(x) = 3x + 5 [0,5]
What is
125/2
1. Set up integral
∫(3x + 5)dx from 0 to 5
2. Split up because of addition
∫3xdx + ∫5dx all from 0 to 5
3. Integrate
(3/2)x2] from 0 to 5 (plug in 5 first then 0 and subtract it)
= 75/2
5x] from 0 to 5
= 25
3. Combine
(75/2) + 25 = 125/2
Evaluate the limit
lim as x approaches 2 of (3x + 4)
What is
10
1. Plug in 2 for x
(3(2) + 4)
2. 10
The derivative of
f(x)=cos4x+sin4(2x)
What is
8sin3(2x)*cos(2x) - 4cos3(x)*sin(x)
1. (d/dx=cos4x) + (d/dx=sin4(2x))
2. (d/dx=cos4x) = -4cos3x*sinx because of the power rule and chain rule
3. (d/dx=sin4(2x)) = 8sin3(2x)*cos(2x) because of the power rule and chain rule
A beach ball has a radius of 12 inches and is being inflated at a rate of 2 inches/sec2. What is the rate that the radius is increasing?
What is
0.001 inches/sec2
1. The volume of a sphere is V=4/3πr2
2. Derived would be V'=4πr2 * r' because of power rule
3. Plug in the values given
(2) = 4π(12)2 * r'
4. Simplify and solve for r'
r' = 0.001 inches/sec2
∫e4sin((x/2)dx
What is
-2e4cos(x/2) + c
1. Use u-sub
u = x/2
du=1/2dx
2e4∫sin(u)dx
2. Integrate
-2e4cos(u)
3. Back sub
-2e4cos(x/2) + c
Find the area of the region that is bound by the line y=0 and the line f(x)=x2-8
What is
(64/3)√2
1. Draw the graph and identify the region we are looking for
2. Find the points of intersection
x2 - 8 = 0
x = −2√2 and 2√2
3. Set up integral to match previous information by splitting it into two parts (where it's symmetrical)
2 * ∫|x2 - 8|dx from 0 to 2√2
2 * ∫(8 - x2)dx from 0 to 2√2
4. Split because subtraction
2 (∫8dx - ∫x2dx) all from 0 to 2√2
5. Integrate
∫8dx from 0 to 2√2
8x] from 0 to 2√2 = 16√2
∫x2dx from 0 to 2√2
x3/3] from 0 to 2√2 = (16/3)√2
6. Combine all of it
2 (16√2 - (16/3)√2)
7. (64/3)√2
Evaluate the limit
lim as x approaches infinity of (5x2 + 3x)/(2x2 + 10)
What is
5/2
1. Simplify by factoring out an x2
(5 + (3/x))/(2+ (10/x2))
2. Plug in infinity for x
(5 + 0)/(2 + 0)
3. 5/2
The derivative of f(x) = (x4 + 53x - 7)2
What is
f'(x) = 2(x4 + 53x - 7) * (4x3 + 53)
1. Recognize that we will be using power rule AND chain rule
= 2(x4 + 53x - 7) * d/dx (x4 + 53x - 7)
2. Continue the chain rule for each bit inside the parentheses
= 2(x4 + 53x - 7) * (4x3 + 53)
3. f'(x) = 2(x4 + 53x - 7) * (4x3 + 53)
The graph of a function is f'(x)=(x-2)(x+1)
Find the local minimas/maximas
What is
x = -1 is local maxima and x = 2 is local minima
1. Find where the slope equals zero to get the critical points
x = 2 and x= -1
2. Test the points around the critical points by plugging it into the function
f'(-2)=((-2)-2)((-2)+1) = 4 which is positive
f'(0)=((0)-2)((0)+1) = -2 which is negative
f'(1)=((1)-2)((1)+1) = -2 which is negative
f'(3)=((3)-2)((3)+1) = 4 which is positive
3. x = -1 is local maxima and x = 2 is local minima
Evaluate ∫xln(x)dx
What is
x2/2 * ln(x) - x2/4 + c
1. Use IBP
u = ln(x) du = (1/x)dx
dv = xdx v = x2/2
x2/2 * ln(x) - 1/2∫xdx when simplified
2. Integrate the last bit
∫xdx = x2/2
3. Put it all together
x2/2 * ln(x) - 1/2 * x2/2
4. x2/2 * ln(x) - x2/4 + c
Find the volume of the region bounded by the functions y = x2, y = 0, and x = 2 rotated around the y - axis using the shell method.
What is.
8π
1. Draw what it looks like
2. Plug in the given information into the shell formula (which is 2π∫rhdx from a to b)
2π∫x3dx from 0 to 2
3. Integrate
2π * x4/4] from 0 to 2
= 2π (4 - 0)
= 8π
f(x) = { (x2 - 4)/(x - 2) if x /= 2
{ 3 if x = 2
Is this function continuous at x =2? If not, what kind of discontinuity is it?
What is
the limit exists but it's not equal, so it's removable discontinuity
1. f(2) = 3, so it is defined and fulfills the first requirement
2. limit as x approaches 2 of (x2 - 4)/(x - 2) to see if it fulfills the second requirement
x /= 2 for this, so factor out (x - 2)
(x - 2)(x + 2)/(x - 2) = x + 2
= 4
3. Compare
4 /= 3, so it is not continuous
4. the limit exists but it's not equal, so it's removable discontinuity
The derivative of f(x) = ex^7
What is
f'(x) = ex^7 * 7x6
1. Use chain rule
= ex^7 * d/dx x7
= ex^7 * 7x6
2. f'(x) = ex^7 * 7x6
Two rich people want to build a rectangular mansion, and they also want to create a moat surrounding their property. A mountain is directly behind their mansion, but for the other sides, they want the moat to be about 265 feet total. What dimensions would create the maximum area of the mansion?
What is
x = 132.5 and y = 66.25
1. Create an equation to represent this (and draw it out but I can't do it without paying)
2y + x = 265
where y represents the two widths and the x represents the side parallel to the mountain
and Area = xy
2. Move Things Around and Substitute
x = 265 - 2y
Then plug it into the area formula
Area = 265y - 2y2
3. Derive to find the max
(Area)' = 265 - 4y
4. Now plug in zero to find the CPs
(Area)' = 265 - 4y = 0
y = 66.25
5. Now plug y back into the previous equation to find x
x = 265 - 2(66.25)
x = 132.5
6. x = 132.5 and y = 66.25
Evaluate this integral
∫dx/(√(25-9x2))
What is
1/3arcsin(3x/5) + c
1. Recognize that this is an inverse trig function
∫du/(√(a2-u2)) = arcsin(u/a) + c
2. Use trig-sub
u=3x and du=3dx
1/3∫dx/(√(5)2-(u)2)) where a=5
3. Now plug in each value for arcsin(u/a) + c
1/3arcsin(3x/5) + c
Find the surface area of f(x) = x2 [0,1] around the x-axis
What is
21.32
1. Surface area formula (2π∫f(x)√(1 + f'(x)2)dx) needs us to find the derivative and square it
f'(x) = 2x
(f'(x))2=4x2
2. Plug information into formula
2π∫(x2)√(1 + (4x2))dx from 0 to 1
3. Integrate
∫x2 = x3/3
= 2/3π∫√(1 + (4x2))dx from 0 to 1
4. U-sub
u = 1 + 4x2
du = 8x
2/3π∫√(u)dx from 0 to 1
= 2/3π * u3/2] from 0 to 1
=
2/3π * (1 + 4x2)3/2] from 0 to 1
5. 21.32
Prove whether this limit exists or not
limit as x approaches 0 of (x/(|x| +x))
What is
The negative side is undefined, so the limit does not exist
1. Examine 0 from the positive and negative side
x approaches 0+ , x > 0, so |x| = x
x approaches 0- , x < 0, so |x| = - x
2. Substitute
x/2x = 1/2
x/ (-x + x) = x/0
3. The negative side is undefined, so the limit does not exist