The First Derivative Test

                                                                                                                 Briya Kirksey
We have now started to learn about the derivative application #3 which is Curve Sketching. You can find this section in our calculus book labeled as "Connecting the Graphs of f ', f '', and f''' ". The first part of this section is "The First Derivative Test".

                           The First Derivative Test 

     What are we looking for?

• We are taking a functions equation and determining where it is  increasing and/ or decreasing, and its relative extrema (relative maximum/minimum) on its graph. 

   

  How do we do this?

  1.   Take the derivative of the given function (if its not already given). 
  2.   Set the derivative equal to zero. Why? - Because that is where the slopes of the functions graph changes signs (positive to negative or vise versa). The values we get by doing this is what we know as critical points. 
  3. Make an x-axis number line labeling all of the critical points on it. 
  4. Choose numbers on the number before, after, and in between the critical points, known as testing points. Example: If critical points are x= -1, 0, 3 , then my testing points could be x= -1.5, -.5, 2, 4. 
  5. Plug each of the testing points into the derivative. Why?- To determine the behavior of the functions graph in between its critical points. 
  6. The answers produced by doing step number 5 will give you the x intervals in which the graph is increasing/decreasing and which x values there may be a max/min.  

                 Tip: 

  • When plugging in testing points into a functions derivative, pay attention to the values sign to make it easier in determining whether the slope will be positive or negative.
   Example:  y' = x (x-3)^2   testing point : -1
  
   Without doing any math, one can already tell that the slope will end up being negative. The value in the parenthesis is -4, but the fact that it is then squared makes it positive. However the x value  (-1) being multiplied to the positive value makes the slope  negative.   

  Concavity : Refers to the way Graphs look

  This is the worksheet and notes used to learn about Concavity. Concave up= a u-shape facing up. Concave down= u -shape facing downward. Inflation point= point on graph where cure changes; where y '' = 0. 

   

  

  
  

     

   

Existence Theorems

This week, we learned about the first application of derivatives: Existence Theorems. Basically, the theorems state that a certain point exists if it meets some criteria. Here are the four types of Existence Theorems we learned about.

1. Intermediate Value Theorem (IVT)
     - if a function f(x) is continuous on a closed interval [a,b]
     - and k (a y-value) is any value between f(a) and f(b)
     - then there is at least one number "c" (an x-value) in [a,b] such that f(c)= k

An example question: if f(x)= x^3+2x-1 is there a value on the interval [0,1] where c is 0?
     -First, determine if the function is continuous on the interval: in this case, it is
     -Second, find the values of f(0) and f(1)
          -      f(0)= -1      f(1)= 2
     -Since -1<0<2, a "c" exists on the interval such that f(c)= 0
     -To find this, we simply set the equation equal to 0.
          -      x^3+2x-1=0    if you solve this, and plug it into the calculator, c=.453   

2. Extreme Value Theorem (EVT)
     -if f(x) is continuous on the interval [a,b]
     -then f(x) is guaranteed to have an absolute minimum and an absolute maximum on the interval

An Extreme Value can happen at two places: endpoints and critical points (where the derivative is equal to 0 or is undefined)

An Example Question: Can you apply EVT to the equation x^-1 +4x on the interval [-4, -.25]? If so, what are the absolute maximum and minimum?
     -First, determine if the function is continuous on the interval.
          - x is not defined at 0, but this is not in the interval, so it doesn't matter
     -Second, evaluate f(-4) and f(-.25) to find possible points for extreme values
          - f(-4)= -16.25     f(-.25)= -5
     -Third, find the critical points by finding f'(x) and determining where the derivative= 0 or und.
          - f'(x)= -1    +4
                      x^2   
          - this is undefined at x=0, but isn't on the interval, so it doesn't matter
          - f'(x) equals zero at x= .5 or -.5 (however, .5 is not on our interval so it doesn't matter)
     -Fourth, evaluate f(x) at critical points
          - f(-.5)= -4
     -Finally, we can determine where the absolute maxes and mins are. In this situation, the maximum is -4 at x= -.5   and the minimum is -16.25 at x= -4

3. Mean Value Theorem (MVT)
     - if f(x) is continuous on [a,b] AND
     - if f(x) is differentiable on (a,b)
     - then there exists a number "c" in (a,b) such that
                 f'(c)= f(b)-f(a)
                              b-a

An Example Question: Is MVT applicable to the equation x+1/x on the interval [.5, 2]? If so, find the value of c.                                                                            
     - First, determine if the function is continuous and differentiable on the interval. In this case, it is, so MVT applies.
     - Second, evaluate f(.5) and f(2)
            -  f(.5)= 3    f(2)= 1.5
     - Third, plug these values into f(b)-f(a)
                                                          b-a
               In this case, it will simplify to -1
     - Fourth, find f'(x) and set it equal to -1.
          -   -1= -1  
                     x^2
          - x= 1, -1 (however, -1 is not on the interval, so we disregard this.) c=1

4. Rolle's Theorem (RT)
     - if f(x) is continuous on [a,b] AND
     - f(x) is differentiable on (a,b) AND
     - f(a)=f(b)
     - then there exists some number "c" on the interval (a,b) such that f'(c)=0

An Example Question: Is RT applicable to the equation f(x)=sin2x on the interval [π/2, 3π/8]? If so, find the value of c.
     - First, determine if the function is continuous and differentiable on the interval.
 In this case, it is.
     - Second, determine what f(π/2) is and f(3π/8) is.
          - f(π/2)= √2/2  f(3π/8)= √2/2 so yes, they are equal and MVT applies.
     - Third, find f'(x) and set it equal to 0 to find c.
          - f'(x)= 2cos2x 
               - 2cos2x= 0           x=π/4, 3π/4 (however, 3π/4 is not on the interval, so it is not included. c=π/4


That is what we did in class this week! Existence Theorems are so fun just like calculus yay!
 

L'Hopital's Rule- Sean Dandeneau

L’Hopital’s Rule

We recently learned how to use L’Hopital’s rule when we must do so. This situation appears whenever we try to evaluate limits that produce indeterminate forms when we try direct substitution.

To use L’Hopital’s rule, you must find the derivatives of the numerator and the denominator. Once this is found, you plug in c for  x and solve for the limit as x->c.

Lim of f(x)/g(x) as x->c = lim of f’(x)/g’(x) as x->c = lim of f’’(x)/g’’(x) as x->c

Lim 1-cos(x)/x as x->0

= 0/0

    So we must use L’Hopital’s Rule…

Lim 0-(-sin(x))/1 as x->0

= lim sin(x) as x->0

= sin(0)

= 0

When doing L’Hopital’s Rule, we must keep in mind that all derivative rules apply. We can find as many derivatives as we need to in order to solve for the limit at c.

Another example:

f(x)=lnx^5/x

Lim x->infinity lnx^5/x

    = lim x->infinity 5lnx/x

    = lim x->infinity (5/x)/1

    = lim x-> infinity 5/x

    = lim x-> infinity 0/1

    = 0

Continuity and Discontinuity

Over the weekend, our homework was to complete a survey evaluating limits through the topic of roads and bridges. On Monday, we began our discussion on continuity by thinking in these roads and bridges terms:

Continuity- a function f(x) is continuous at x=a if

1. the roads meet
2. there is a bridge
3. the roads meet at the bridge

After conceptually understanding what continuity is, we decided to translate this definition into calculus lingo.

Continuity- a function f(x) is continuous at x=a if

1. the limit of x as it approaches a exists
2. f(a) exists
3. the limit of x as it approaches a is equal to f(a)

In other words

• If you can draw it without lifting your pencil, it’s continuous.
• f(x) is continuous at x=a if
• the limit from the left equals the limit from the right (or the roads come to the same spot)
• a y-value exists at x=a (bridge exists; there isn’t a hole or asymptote there)
• the y-value, the limit from the left, and the limit from the right are all the same number

Discontinuity

• A function has a point of discontinuity at x=a if it fails one or more of the four criteria listed below

The Four Types of Discontinuities and How to Find Them:

1. Removable
1. Piecewise function
2. Hole (a factor cancels)
2. Jumps
1. Absolute value division function
2. Greatest integer function
3. Piecewise function
3. Infinite
1. y=tan(x) (infinite discontinuities at odd integers of pi/2)
2. Reciprocal (vertical asymptote- cannot cancel an x in the denominator)
4. Restricted
1. Square root (continuous for all x-values in its domain, but not outside its domain)
2. Logarithms (only continuous for x>0, unless it is transformed)

The worksheet for discontinuity:

Based on the definition of continuity and the four types of discontinuity, we were able to solve the following problems in class:

To find a value that makes a function continuous, plug in the the boundary (in this case, the boundary is 1) into the x of each part of the piecewise function and solve. Then, set up the answers from each part of the piecewise function and save for a, as shown above. 

Here are answers to the homework problems!

Differentiability

Differentiability: f(x) is not differentiable at x=a (f’(x) does not exist) when f’(x) is not continuous at x=a.

For example:

The Four Ways a Function can Fail to be Differentiable at x=a:

1. A corner
1. Different one sided derivatives causes the graph to jump (ex. absolute value graph)
2. A vertical tangent
1. One-sided derivatives go toward infinity (ex. y=x^⅓)
3. A cusp
1. One-sided derivatives go toward opposite infinities (one toward positive infinity and the other toward negative infinity)
4. Any type of discontinuity
1. Any of the four types of discontinuity explained in the previous section!

The worksheet for differentiability:

Remember:

• Continuity does not imply differentiability, but
• Differentiability does imply continuity.

In other words:

• f’(c)=lim x as it approaches c [f(x)-f(c)]/(x-c) must exist if f is differentiable at x=c

Based on these notes of differentiability, we were able to complete the following in class problems:

To make it differentiable, you find the derivative of the top part of the piecewise function and the derivative of the bottom of the piecewise function, set them equal to one another, and then solve. In this case, you also had to make it continuous, which is explained in a problem above. 

Here are the homework answers!

Reminder

Don’t forget to study for the assessment on tuesday!

Blog for November 4

Overview:
This week we began learning about limits. We had seen limits before, at the beginning of the derivatives unit. The limit definition of the derivative of a function at a point helped us find the slope of the tangent line at x=c. The limit definition of the derivative of a whole function was used to find the slope of a secant line between any point on f(x) and another point that is h units away from it.
As we move away from the topic of derivatives, we focused on limits as x approaches any point in a function.

Graphical Approach
There are three main approaches to limits. We began with the graphical approach. This entails using graphs to find the limit at a point.


We were asked to find the limits from the left and right of each point pictured, as well as the actual limit for each value.


We learned that the limit does not exist if the limits approaching from the left and right sides are different from each other. The limit also does not exist if either of these values is infinity or negative infinity. In other words...


We then did some practice finding limits using graphs.



Numerical Approach
The next approach is the numerical approach. For this approach, we use a table to determine the limits approaching a point from the left and right. As the value gets closer to x, the limit gets closer to what it should be. For example:

In this example, the limits on either side seem to be approaching the same number, .333. This means the limit as x is approaching 2 is .333. If the limits on  either side are not approaching the same number, or they are approaching infinity or negative infinity, the limit does not exist.


Analytical Approach
The last approach is the analytical approach. This is when you use direct substitution to find the limit. You plug in the value, c, into the function, f(x). For normal functions, if the limit as x approaches c exists, it should be f(c).
 
In the last example, the limit ends up being 0/0. This means that there is a hole in the graph at this point.



We were also introduced to the properties of limits in class, which are listed below.


The last topic we covered this week was limits at infinity. This involves plugging infinity into functions as the x-value,  which would result in finding the end behavior of functions. To find the end behavior or a horizontal asymptote, you need to find the limit as x approaches both infinity and negative infinity.
 

Reminder: Monday's class will be a review for our quiz on Thursday.

Blog for October 23

Overview of the week of 10/17-10/21:

Monday: We reviewed derivatives of exponential and logarithmic functions

Tuesday: PSAT day

Wednesday: NO class

Thursday: Chain rule/derivatives of logarithmic and exponential functions assessment

Friday: Started covering our last type of derivative (!!!), implicit differentiation (derivatives of non-functions)

Implicit differentiation- writing the derivative when y isn't alone, or when the equation can not be solved for y. An example of when implicit differentiation must be used is x2+y2=1

All year we have been finding the derivative of functions in respect to x (without knowing it)

When finding the derivative of an equation like y=x+1 we would find y'=1 or dy/dx= 1, where dy/dx is the derivative of y and 1 is the derivative of x+1.

Now that y is not always alone, and can be on both sides of the equal sign dy/dx has a much bigger role in our derivatives.

Now that we are finding derivatives of y terms that are more than just y (ex. y2, 3y, 17y2) dy/dx is always part of our equations, because it's part of the derivative of y.

When using implicit differentiation it is important to remember to always write dy/dx because it is part of the derivative of the y term

example: The derivative of x=4y is 1= 4(dy/dx)

• 1 is the derivative of x
• 4(dy/dx) is the derivative of 4y

Example from class:

The steps we took when solving this were:

1. Take the derivative
2. Collect dy/dx terms on one side of the = (anything else should be moved to the other side)
3. Factor out dy/dx
4. Divide to get dy/dx alone on one side of the =

**Helpful hint: distributing almost always makes these problems easier

Here is another example from class. This problem involves more than just finding the derivative of an equation: 

Implicit differentiation can also be used to find the second derivative of an equation.

1. Find the first derivative normally, using implicit derivation 
2. Find the second derivative (almost always involves the quotient rule)

Here is an example from class:

**Don't forget: when writing the second derivative dy/dx becomes d2y/dx2

Blog for October 16

Overview
This week's material focused on the eighth derivative rule regarding derivatives of exponential and logarithmic functions with introduction to the number e. The material also introduced derivatives of inverse trigonometric functions.

RULE #8 - Derivatives of Exponential and Logarithmic Functions

Derivatives: "e" as an Exponential Function
In the exponential section of derivatives, we were introduced to the numerical value e. The derivative of e in an exponential function will always equal itself, as written below.

• dy/dx (ex) = ex

Applying "e" to Several Derivative Rules

*Product Rule


*Quotient Rule


*Chain Rule


--------------------------

Derivatives: Exponential Functions with Values Other than "e"
The derivative of an exponential function that is not ex (i.e. ax) is:

• dy/dx (ax) = ln (a) * ax

*Proving the Derivative Formula


"a" is only representative of a value. Numerical values can be substituted for this variable, and the same steps can be followed to find the derivative. An example of this process with numerical values is listed below.

*Derivative of an Exponential Function Other Than "e"

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Derivatives: Natural Log Functions
The derivative for natural log functions is given and proven as listed below:

• dy/dx [ln(x)] = 1/x

*Proving the Derivative Formula

Now we can apply this formula to natural logs of numerical values opposed to only working with variables.

*Derivative of a Natural Log Function

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Derivatives: Logarithmic Functions

Logarithmic functions have a derivative formula that follows the format listed below:

• dy/dx [loga(x)] = 1/ln (a) * x

*Derivative of a Logarithmic Function

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Derivatives of Inverse Trig Functions

Aside from working with exponential and logarithmic functions, this week's material encompassed an introduction to derivatives of inverse trig functions for sine, cosine, and tangent. Each of these trig functions has a specific derivative formula.

Derivatives: Inverse Sine

The derivative formula for finding the derivative of inverse sine is provided below with an example highlighting how the formula can be applied. 

• dy/dx [sin-1(x)] = 1/√(1-x2)

*Derivative of Inverse Sine Function

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Derivatives: Inverse Cosine

The formula for the derivative of inverse cosine varies slightly in comparison to the formula for the derivative of inverse sine. One important thing to note is that inverse cosine is negative and inverse sine is positive. Remember this to avoid using the wrong formula. 

• dy/dx [cos-1(x)] = -1/√(1-x2)

*Derivative of Inverse Cosine Function

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Derivatives: Inverse Tangent
Finally, the derivative formula for the inverse of tangent follows:

• d/dx [tan-1(x)] = 1/1+x2

*Derivative of Inverse Tangent

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FINAL TIPS

Don't forget several log properties that are important to use:

• ln(ab) = ln(a) + ln(b)
• ln(a/b) = ln(a) - ln (b)
• ln(ax) = x * ln(a)

Blog for October 10

Overview-
This week in class we focused primarily on the Chain Rule. It is the 7th rule of derivatives and is one of the most important.

Equation-

Application-
Every time there is a function nested inside of another function the chain rule is extremely useful. It is used to link parts of equations together and for differentiating complicated equations. It allows people to take derivatives of much more complex problems.

Explanation-
To take the derivative you have to look at the most outside function, usually an exponent. Just pretend that the inside is just an x as usual and take the derivative and leave the inside portion unchanged, but still copy it down. Then multiply that answer by the derivative of the stuff on the inside that you previously ignored. That is the basics of the rule, but this process may have to be repeated depending on the extent of the problem.

Practice Problems- 

Using the Rule Twice in Same Problem-

  








Finding dr/d(theta)

More Complicated Problem-

Using Values From a Table-

Tips-

- Always work from the outside in

- Anytime there is a nested function, think to use the chain rule