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Optimization and Hyperbolic Functions // Reflection #5

As it is the last topic for the last quarter, I was expecting this concepts to be more challenging, but it wasn’t like that at all. Although it wasn’t that easy, I could say that I understood it faster than I thought. With this finished, all I have to do is prepare for the second long test, and I will already be set to a hopingly stress-free vacation.

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Rates of Change, Rectilinear Motion, and Related Rates // Reflection #4

It's all fun and games until you arrive in related rates - or so I think.

It was intimidating to start learning about rates of change and rectilinear motion because it reeks of Physics, but it was surprisingly easy when I actually started reading about it.

I thought it would be the same with related rates, but it wasn't. It was confusing and I am usually clueless as to what equations to use, but somehow, I survived through it.

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Sketching Graphs of Functions // Reflection #3

To be honest, I'm not sure if I'm doing things right. The rules on sketching in the previous Mathematics subjects have been easy to understand; however, the rules on sketching graphs using derivatives were confusing to me. To this point, I don't think I completely understand how to sketch graphs of functions using derivatives.

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Increasing and Decreasing Functions, First Derivative Test, Concavity Test, and Second Derivative Test // Reflection #2

At first, I was not able to process the concept properly in my head. It really was confusing the first time I studied it, but it was actually pretty easy. All we had to do is apply what we had learned in the previous quarter.

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Extrema of Functions // Reflection #1

Personally I found this topic easy to understand since it's about derivatives, which I understood properly from the previous quarter. I think I did better than before and now I should probably try to understand the concept of increasing and decreasing functions.

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Higher Order Derivatives // Reflection #6

Higher Order Derivatives

Determine the second derivative of the following functions:

For both problems, since we are required to solve for the second derivative, we just solve for the derivative of the function and then solve for the derivative of that derivative.

Reflection #6

For this entry, we solved higher order derivatives, which is just like the previous problems but repetitive. In this topic, we applied all the previous topics we learned to solve the derivatives. In my opinion, since it is basically the same as the previous lessons, I think I did well in this topic.

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Implicit Differentiation // Reflection #5

Implicit Differentiation

Use implicit differentiation to find 𝑦’:

For problems like this, we have to solve for the derivatives of both sides and use transposition depending on what we need to find.

Reflection #5

The chain rule and implicit differentiation are both important because only by learning them do we make our understanding of derivatives deeper. Solving chain rule and implicit differentiation problems are similar to the other problems because we continuously apply what we learned in the previous problems. However, I think I'm still slightly confused whenever I solve implicit differentiation problems and the only way to resolve that is to continue practicing solving problems.

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Chain Rule

To solve the problem above, we use chain rule, rules on differentiation, and the derivative the inverse trigonometric function.

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Derivatives of Transcendental Functions // Reflection #4

Derivatives of Transcendental Functions

Find the derivative of:

In solving for the derivatives of both functions, we have to keep in mind the equivalent derivative of every trigonometric and inverse trigonometric functions. Using them along with the different rules on differentiation will let us solve these problems.

Reflection #4

Solving for the derivatives of transcendental functions is important because it lets us have a deeper understanding of derivatives. For the past entries, we have been solving for derivatives; however, this time, the trigonometric and inverse trigonometric functions have their own unique derivatives. In my opinion, I still did well in this topic, although not as good as my performance in rules on differentiation. As always, I have to keep practicing solving for the derivatives of transcendental functions as it is the only way to improve my solving skills.

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Rules on Differentiation (3/3) // Reflection #3

Rules on Differentiation

Differentiate the following:

Find the equation of the tangent line of the following:

We simply use the quotient rule and solve for the slope and the value of b afterwards. With these, we can find the equation of the tangent line.

Reflection #3

The rules on differentiation was an easy topic compared to the previous ones. It is important to learn this so that we can solve for the derivatives of functions more efficiently. This is the first time that I took on this topic, which is why it was very helpful. Doing the activities made me think that every rule is important to master as they could be combined to solve for the derivatives functions. In my opinion, I think I did well in these topics and I should now master each of them as they could be applied in the coming days.

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Rules on Differentiation (2/3)

Find the derivative of the function:

We first use the product rule to get the expanded form of the derivative of the function and simplify it using the other rules of differentiation.

If 𝑘(𝑥) = 𝑓(𝑥) ∙ 𝑔(𝑥) and 𝑓(2) = 3, 𝑓’(2) = −4, 𝑔(2) = 1, and 𝑔’(2) = 5, find 𝑘’(2).

Since the values and derivatives of both functions are already given, all we have to do is use the product rule.

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Rules on Differentiation (1/3)

Find the derivatives of the following functions:

In the problems above, we make use of the derivative of a constant function, the power rule, the constant multiple rule, the sum rule, and the difference rule. Combining all of these and our previous knowledge on tangent line and derivatives, we could solve for the derivative of the functions easily as seen in the photo above.

In item 1.a, we simply use the constant multiple rule, the power rule, and the addition rule to get the derivative of the function.

Item 1.b is similar to item 1.a, except that we have to convert each terms first so that they are in their exponential forms. We then use the constant multiple rule, the power rule, the addition rule, and the difference rule to get the derivative of the function.

In item 2, we have to find the slope of the normal line of the function at (1, -1). After solving for the slope of the tangent line using the rules on differentiation, we solve for the slope of the normal line by getting the negative reciprocal of the tangent line.

Lastly, in item 3, we have to solve for the equation of the tangent line at (-1, -3/2). First, we solve for the derivative of the function as it is equal to the slope of the tangent line. By substitution, we would get the slope and use it along with the other given constants to solve for the value of b. With this, we can get the equation of the tangent line in the form 𝑦 = 𝑚𝑥 + 𝑏.

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The Tangent Line and Derivative of a Function (2/2) // Reflection #2

The Tangent Line and Derivative of a Function

Determine the derivative of the function.

After substituting the values, we rationalized the equation and combined like terms. We then cancelled Δ𝑥 from the numerator and the denominator and substituted Δ𝑥 with 0. Lastly, we simplified the fraction and arrived at the derivate as seen in the picture above.

Reflection #2

Getting the derivative of a function is basically the same as getting its tangent line. It is important that we learn this because it could be helpful in the coming lessons. In my opinion, I did well in this topic compared to the previous ones and the next thing I should do is to practice more on solving the derivatives of functions.

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The Tangent Line and Derivative of a Function (1/2)

Find the equation of the tangent and normal lines of 𝑓(𝑥) = 3𝑥² + 5𝑥 + 1 at 𝑥 = 1.

Tangent Line:

Normal Line:

We can get the slope of the normal line by getting the negative reciprocal of the slope of the tangent line.

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Continuity of a Function on an Interval // Reflection #1

Continuity of a Function on an Interval

Find the intervals on which the function is continuous.

Continuity at 𝑥 = 1:

It is continuous since 5 - (𝑥 - 2)² is a polynomial.

Continuity at 𝑥 = 3:

The limits as 𝑥 approaches 3 from the right and from the left are equal; thus, the limit exists and the function is continuous at 𝑥 = 3.

Continuity at [1, 6]:

h(𝑥) is continuous in open interval (1, 6) and from the right at 1.

Therefore, the function is continuous at interval [1, 6).

Reflection #1

Checking if a function is discontinuous and getting the continuity of a function on an interval is a continuation of the last topic from the previous quarter. It is important to learn this so that we can identify the difference between continuity and differentiability. In my opinion, I did not do that well in these topics because of my shallow understanding of limits. As always, the only thing left to do is to keep practicing.

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Types of Discontinuity

Determine if the given function is continuous at the indicated number 𝑥 = 𝑐.

The limit of the function as 𝑥 approaches 3 does not exist because the limits as 𝑥 approaches 3 from the left and from the right are not equal. Therefore, there is a jump essential discontinuity at 𝑥 = 3.

Just like the previous item, there is a jump essential discontinuity at 𝑥 = 1 since the limits as 𝑥 approaches 1 from the left and from the right are not equal.

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Continuity of a Function at a Number // Reflection #4

Continuity of a Function at a Number

Reflection #4

Unlike some of the previous topics, limits of trigonometric functions and continuity of a function at a given number seemed to be easy since the concepts from previous lessons could be applied to solve these kind of problems. Doing this was important to assess if we truly understood the previous concepts.

Although I could never be totally confident, I think I understood these lessons easily. Finishing these series of lessons just means that I should prepare now to learn more and possibly harder mathematical concepts in the coming days.