Calc 1 - Exam 2

Unit Circle

A function is continuous at c if

• f(c) is defined

• Lim F(x) exists

• Lim f(x) = f(c)

Ln1, Ln>!, Ln<1

Ln 1 = 0 and Ln >1 = + and Ln < 1 = -

log_axy =

log_a(x/y) =

log_axⁿ =

Derivative of a constant

0

Derivative of A linear (y = mx + b) function

m (slope)

Power Rule

(xⁿ) = nx^n-1or (axⁿ) = (an)x^n-1

Sum and difference rule

The derivative of f + or - g = the sum/diff of the derivative of f and g.

Derivative of e^x

e^x

Product Rule

 [f(x)g(x)] = f’(x)g(x) + f(x)g’(x)

Quotient Rule

(Low)(Dee High) - (high)(dee low) all dived by (low) squared

Use when functions being divided by each other

Chain Rule

• f’(g(x)) ⋅ g’(x)

• Where:

  • f(x) = outer (momma) function

  • g(x) = inner baby function

Equation For Linearization

if f is diff at x = c, then the equation of the tangent line, L(x) = f(c) + f’(c)(x-c)defines the linear approximation of f at c Where:

• x is the number you’re trying to approximate

• c is the number you are using to approximate x, where f(c) and f’(c)

• L(x) is the approximate value of f at x

Steps for linearization

1. Find a function in the problem

   1. Ex. if you’re trying to find ³√1003 or 1003^1/3 the function would be x^1/3

2. Write your goal

   1. i.e. what value you want to approximate

3. Pick a i.e where you’re gonna draw your tangent line

   1. If trying to find 1003^1/3 then your number would be 1000

      1. a should be near the approximation x

      2. you should know f(a)

      3. You should know f’(a)

4. Write the tangent line

   1. Current example: 

      1. L(x) = f(a) + f’(a)(x-a)

         1. L(1003) = f(1000) + f’(1000)(1003-1000)

      2. Find f(1000) 

         1. 1000^1/3 = 10

      3. Find f’(1000)

         1. f’(x) = ⅓x^-⅔ = ⅓/ (1/x^⅔)

         2. f’(1000) = ⅓ (1/1000^2/3) 

         3. f’(1000) = 1/300

      4. Fill in the variables:

         1. L(x) = 10 + (1/300) (x - 1000)

5. Write: f(x) ≈ L(x) for all x near a

   1. 1003^1/3 ≈ 10 + (1/300) (x - 1000) for all x near 1000

6. Do the approximation:

   1. 1003^1/3 ≈ 10 + 1/300 (1003 - 1000) = 10 + .01 = 10.01

How do we know if we over/under approximate the linearization? And do we need the second derivative?

At the value we’re estimating…

If f''(x)<0 → Concave DOWN → Tangent line is ABOVE the curve → OVERESTIMATE

If f''(x)>0 → Concave UP → Tangent line is BELOW the curve → UNDERESTIMATE

U don’t even need the second derivative, just look at graphic concavity

Local Extrema vs Absolute Extrema

Local -

• Relative to nearby points (peakers/valleys/end points only)

• Occurs at critical point(s)

Absolute -

• Over the entire domain or specified interval
  Either at critical points in (x,y) or could be end points in [x.y]

Critical Point

• Points in the domain of f where f’(x) = 0 or where f’ ( x) doesn’t exists, therefore:

• If f has a relative max at x=c, c is critical # of f

To find local Extrema

Use First Derivative Test with Number Line or Second Derivative Test with T Chart

First Derivative Test with Number Line

1. Set f’(x) = 0 or DNE

2. Find critical numbers

3. Draw number line of f’(x) with critical numbers

4. Do test values in between each critical number and see if neg/pos/0

• Since f’(x) pos = increasing and neg = decreasing

5. Where it goes from pos to neg is a local max; where it goes from neg to pos is a local min at c

Second Derivative Test With T Chart

1. Set f’(x) = 0 or DNE

2. Find critical numbers

3. Draw T chart and for each Crit plug in f’’(x)

   1. If f’’(x) > 0, f(x) is concave up meaning there is local min at x = c

   2. If f’’(x) < 0, f(x) is concave down meaning there is local max at x = c

   3. If f’’(x) = 0, inconclusive, use first derivative test with number line instead

Extreme Value Theorem - EVT

If f is a function that is CONTINUOUS on a CLOSED interval, then f must have an an absolute maximum and an absolute minimum

To find Absolute Extrema On A Closed Interval

The Closed Interval Method: First Derivative Test With Closed Interval: T Chart

The Closed Interval Method: First Derivative Test With Closed Interval: T Chart

If f is continuous on a closed interval [a,b], to find the absolute extrema on [a,b]:

1. Find Critical Points

2. Plug critical points and endpoints into f(x)

3. The largest is the absolute max, the smallest is the absolute min

Which test to use when for optimization

1st Derivative Test or 2nd Derivative Test: Only When there is exactly one CP and it is the one you want

Closed interval method: Needs a domain

Optimization Steps For Credit

1. Write a goal

   1. What you are maximizing or minimizing

2. Draw a picture and assign variables

3. Write down an expression for the function in your GOAL that you are trying to optimize

4. Use data from the problem and picture to write the function from step 3 as a function of one variable

5. Find the critical points for the function from step 4

6. Use either 1st or 2nd derivative test, the closed interval method, or a graph

7. Ensure you answer in the context of the question WITH UNITS (Closing Statement)

Trig Derivatives

Implicit Differentiation Steps

1. Differentiate both sides of the equation

2. Bring all dy/dx terms to one side

3. Factor out dy/dx from those terms

4. Solve for dy/dx by dividing both sides by the remaining factor

Derivative of lnx

• Derivative of lnx = 1/x

• Chain rule: ln(u) = 1/u ⋅ (u’)

  • (u’)/u

    • Again, just the same thing times the derivative of the thing inside

Derivaitve of a^u

• Derivative of a^u = a^u ⋅ lna

• Chain Rule: a^u = a^u ⋅ lna ⋅u’

  • Chain rule: a^u ⋅ lna ⋅ u’

    • Again, just the same thing times the derivative of the thing inside 

Derivative of [log_au]

Derivative of [log_au] = 1/(u ⋅ lna)

• Chain rule: 1/(u ⋅ lna) ⋅ u’

  • u’/(u ⋅ lna)

  • Again, just the same thing times the derivative of the thing inside 

Derivative of inverse sin

Derivative of inverse tan

Derivative of inverse cos