Calculus I Unit 3 - 4 Notations Flashcards
How do you find higher order derivatives?
Taking the derivative of the derivative of the derivative…..
Explain the Solution to:
Limit as x → 0 of sin(ax) / x
Since there’s a constant in the sin operator, we will start by multiplying the function by the constant/constant: a/a
(sin(ax) / x) ⋅ a/a → asin(ax) / ax
Remove the a in the numerator which leaves us with sin(ax)/ax which one of our limit identities (limit x → 0 of sin(x)/x = 1)
This becomes:
a ⋅ limit x → 0 of sin(ax)/ ax
a ⋅ 1 = a
What are the two trig limit identities ?
limit x → 0 of sinx/x = 1
limit x → 0 of 1 - cosx / x = 0
Explain logic behind:
limit x → 0 of sinx/x
The Squeeze Theorem takes our unit circle triangles of:
tanx/2 ≥ x/2 ≥ sinx/2.
Multiplying each side by 2/sinx and flipping the reciprocal of each value will give us
cosx ≥ sinx/x ≥ 1
@ x = 0, cos(0) = 1 giving us
1 > sinx/x > 1 which makes our limit:
Limit x → 0 of sinx/x = 1
Explain logic behind:
limit x → 0 of 1 - cosx / x
Understanding that Limit x → 0 of sinx/x = 1, we start by trying to get the x out of the denominator by multiplying our function by the conjugate of 1- cosx:
1 - cosx/x ⋅ (1 + cosx)/(1 + cosx) → 1 - cos²x/x(1 + cosx)
We then use trig identities to convert 1 - cos²x into sin²x then split the function into two digestible parts:
sin²x/x(1 + cosx) → sinx/x ⋅ sinx/(1 + cosx)
Applying the limit operator, we can now find the limit of both side of the function:
Limit x → 0 of sinx/x = 1
Limit x → 0 of sinx/(1 + cosx) = 0/1+1 = 0
1 ⋅ 0 = 0
Explain Solution to:
limit x → 0 of sinax / sinbx
Start by using the procedure for
limit x → 0 sinax /x, where we learned this is a ⋅ 1
perform this procedure on both the top and bottom of the function to get the fraction:
(a ⋅ 1)/(b ⋅ 1) which will equate to a/b
Explain how to find values of x when the below function has a horizontal tangent line ?
ƒ(x) = 5x - 10sinx
Start by finding the derivative of ƒ(x):
ƒ’(x) = 5 - 10cosx
Solve for when ƒ’(x) = 0, or 0 = 5 - 10cosx
5/10 = cosx
Note that cosx = 1/2 when @ cos(60°) and cos(300°) or cos(π/3) and cos(5π/3).
Now consider that in order to gain either value, the graph must be multiplied by at least 2π
Therefore x = π/3 + 2kπ or 5π/3 + 2kπ where k is integer.
Explain how to find values of x when the below function has a horizontal tangent line ?
ƒ(x) = 6x - 12cosx
Solve for ƒ’(x) @ x = 0
ƒ’(x) @ x = 0 → 0 = 6 + 12sinx
Solve for x:
-1/2 = sinx → sin(210°) and sin(330°) = -1/2
Since the graph requires a full revolution in order to regain these values:
x = 7π/6 + 2kπ or 11π/6 + 2kπ where k is integer.
Explain how the derivative of h(x) = ƒ(g(x)) found?
Using the Composite Chain Rule:
h’(x) = ƒ’(g(x)) ⋅ g’(x)
How to prove:
d/dx[tanx] = sec²x
Using the Quotient Rule
Start by using our fundamental trig identity: tanx = sinx/cosx
Apply the derivative operator then solve the derivative using the Quotient Rule:
d/dx[sinx/cosx] =
(cosx)(cosx) - (-sinx)(sinx) / (cosx)² →
(cos²x) + (sin²x) / cos²x →
Using trig identities, convert sin²x to (1-cos²x) then subtract the cos²x values:
(cos²x) + (1 - cos²x) / cos²x →
1/cos²x → sec²x
Proving d/dx[tanx] = sec²x
Implicit Differentiation:
d/dx[y²]
2y ⋅ dy/dx
Implicit Differentiation:
d/dx[4xy]
Using Chain Rule d/dx[4xy] =
(4)(y) + (4x)(dy/dx)
Implicit Differentiation:
d/dx[sin(xy²)]
Using Chain Rule d/dx[sin(xy²)] =
cos(xy²) ⋅ (y² + 2xy⋅dy/dx)
Implicit Differentiation:
d/dx[y³]
3y² ⋅ dy/dx
Implicit Differentiation:
d/dx[x³y]
3x²y + x³⋅dy/dx
Implicit Differentiation:
d/dx[x²y⁴]
2xy⁴ + x²⋅4y³⋅dy/dx
Implicit Differentiation:
d/dx[x²√(y)]
2x√(y) + x²/2√(y) ⋅ dy/dx
How to do logarithmic differentiation for:
y = √((x² + 1)(x - 2)²)
Start by taking the natural log of both sides and carrying the ¹/² exponent:
ln(y) = 1/2ln[(x² + 1)(x - 2)²]
Then distribute the natural log into the interior brackets:
ln(y) = 1/2[ln(x² + 1) + ln(x - 2)²]
Distribute the exponent on the right hand:
ln(y) = 1/2[ln(x² + 1) + 2ln(x - 2)] → ln(y) = 1/2ln(x² + 1) + ln(x - 2)
Now since the equation is now simplified, we can now differentiate both sides:
d/dx[ln(y)] = d/dx[1/2ln(x² + 1) + ln(x - 2)]
1/y ⋅ dy/dx = x/(x² + 1) + 1/(x - 2) → dy/dx = y[ x/(x² + 1) + 1/(x - 2)]
Substitute our previous value of y into the equation for our answer:
dy/dx = √((x² + 1)(x - 2)²) ⋅ [ x/(x² + 1) + 1/(x - 2) ]
Explain Solution to:
y = 5 ⋅ 4ˣ
Treat the 5 as a constant then use the Base to the x Power property -
d/dx[bˣ] = bˣln(b):
5 ⋅ d/dx[4ˣ] = 5 ⋅ 4ˣln(4) → 20ˣln(4)
Explain Solution to:
y = (sinx)ˣ²
Start by taking the natural log of both sides of the function and bringing the exponent to the back:
ln(y) = x²ln(sinx)
Now apply the derivative operator and solve using product and chain rule:
d/dx[y] = d/dx[x²ln(sinx)] → 1/y ⋅ dy/dx = 2xln(sinx) + (1/sinx ⋅ cosx)(x²)
Use algebra to solve for dy/dx:
dy/dx = y ⋅ [2xln(sinx) + (x²cotx)]
Since we know y = (sinx)ˣ², we can substitute this into our y value:
dy/dx = (sinx)ˣ²[2xln(sinx) + (x²cotx)]
Explain Derivative of:
y = (1 + x²)/(cot⁻¹x)
Identify the derivative of cot⁻¹x as -1/(x² + 1) then solve using the quotient rule:
dy/dx = [2x(cot⁻¹x) - (-1/(x² + 1) ⋅ (1 + x²))] / (cot⁻¹x)²
Simplify the equation to see that our subtraction symbol becomes addition and the right side of the numerator cancels out into 1:
dy/dx = [2x(cot⁻¹x) + (1 + x²)/(x² + 1)] / (cot⁻¹x)² →
dy/dx = [2x(cot⁻¹x) + 1] / (cot⁻¹x)²
Explain the Solution to the following problem:
One end of 13ft ladder is on the ground while the other end rests on a vertical wall. The bottom end of the ladder is drawn away from the wall at 3ft/s. How fast is the top of the ladder falling down the wall when the foot of the ladder is 5ft from the wall.
Start by identifying the knowns and and unknowns along with what we are trying to solve:
Known:
Ladder (L) = 13ft
Bottom of Ladder (B) = 5ft
ROC Bottom of Ladder (db/dt) = 3ft/s
Unknown:
Wall Top of Ladder (W)
ROC Wall (dW/dt)
Recognize that the Vertical Wall, Ladder, and ground form a 90° triangle, and solve for W using the Pythagorean Theorem and the given lengths.
L² = W² + B² → W² = L² - B² → W² = 13ft² - 5ft² → W = 12ft
We can then use the derivative of the same equation to find ROC Wall (dW/dt):
d/dt[L²] = d/dt[W² + B²] = 2L(dL/dx) = 2W(dW/dx) + 2B(dB/dx)
Now we can replace the each variable using our known values: Note that since the ladder is fixed at 13ft, the ROC of the Ladder (dB/dx) = 0ft/s
2(13ft)(0ft) = 2(12ft)(dW/dx) + 2(5ft)(3ft/s) → -30ft²/s / 24ft = (dW/dx)
Solve to:
(dW/dt) = -6/4 ft/s
Extreme Value Theorem
Any continuous function on a closed interval [a,b] is guaranteed to have a least one Absolute Max, and one Absolute Min value between [a,b].
What do Critical Points of a function represent and how are they found ?
Critical Points represent a change in direction/a change in the slope of a function from positive → negative or vice-versa.
They are found by setting the derivative of a function to 0 and solving for the independent variable.
What are the conditions of Extremas ?
Critical Points cannot happen @ endpoints
Local Extrema only occur @ Critical Points
Absolute Extrema occurs at both Critical Points and endpoints.
What is the process of finding Local Extrema of a function
Take the first derivative of a function to find the critical points.
Find the critical points @ ƒ’(x) = 0
Plug those points into f(x) (so long as they are not endpoints)
How do you find the interval which a function is decreasing and increasing ?
Start by taking the derivative of the function.
Set the function’s derivative = 0: ƒ’(x) = 0 then solve for the critical points.
Create a numberline using the critical points and values of numbers between the critical points.
Solve for ƒ’(x) at every value between the critical points and jot down whether these points are positive or negative on the timeline.
The critical points note the end of a particular interval so a negative up until reaching the critical point will indicate a decrease over an interval.
What indicates an Inflection ?
When the Second Derivative of ƒ(x): ƒ’‘(x) changes from positive to negative or positive to negative (ƒ’‘(x) = 0).
How do you find the intervals where a function has a Concave up or Concave down ?
Concave up: Where ƒ’‘(x) is positive
Concave down: Where ƒ’‘(x) is negative
Use the Numberline method (solving each at ƒ’‘(x)) to find the exact interval of positivity and negativity
What does the Second Derivative indicate ?
ƒ’‘(x) indicates the rate which the slope is changing.
If this rate is positive we will see an upward U shape ƒ(x). If this is negative we will see a downward ∩ Shape in the graph ƒ(x).
How are Local Extrema found using Second Derivative ?
Start by taking the first derivative and finding the Critical Points using: ƒ’(x) = 0.
Then take the Second Derivative and solve for ƒ’‘(c) where c is the critical points of the function.
Where ƒ’‘(c) > 0, we have a local minimum
Where ƒ’‘(c) < 0, we have a local maximum
