This week was another week where I didn’t get nearly as far through KA as I wanted to but still feel like I made a lot of progress. My goal at the start of this week was to at least get through the exercise Integration Using Completing the Square which I didn’t even end up starting… I spent the entire week doing two things instead; trying to fully understand/memorize the 18 derivatives that I screen-shoted in my last past, and thinking through the FToC and trying to better understand it. By the end of the week I had a MUCH better grasp on the derivatives (specifically the trig derivatives) and also a better grasp on the FToC. I’m certain that both of these things are crucial components of calculus so I’m happy with how this week went even though I didn’t make any progress on getting through the last exercise of Integrals.

Here’s the list of derivatives I found last week when I Googled, “derivatives to memorize for integration”:

I started at the top of the list and covered up the solutions to each derivative and tried to come up with the answer on my own. This is the list I was able to come up with:

I was able to come up with the correct answer for the first 12 derivatives apart from the 6^th one, d/dx[n^x], which I had to review. To be fair, the answer given on the list turns out to be wrong but, regardless, I wouldn’t have been able to figure out the correct answer anyways. I remembered going through this derivative multiple times before but couldn’t remember how to solve it. I looked it up on Google and was shown a KA video that explained how it works. Here’s the page from my notes that explains how it works:

As a recap to my note, to solve this derivative you need to turn a into e and change the function from a^x into a version of e^x since it is the derivative of itself. To do this, you need to use logarithm rules. I always find it easiest to think through log rules using base 2, exponent 3, and argument 8:

• B^e = A
  • 2³ = 8
• Log_B(A) = e
  • Log₂(8) = 3
• B^e = B^log_B(A)
  • 2³ = 2^Log_2(8)
    • = 8

Using this rules, you can change a^x into (e^ln(a))^x where a = e^ln(a). From there, you simply use the chain rule to solve the derivative which you can see how I did at the bottom of my notes above. 

Of that list of 12 derivatives, I had all of them memorized except for the 10^th, 11^th, and 12^th. I was able to solve these ones on my own without having to look them up, however, so I considered them to be derivatives that I already knew. I was stumped once I got to the arc-trig functions, however, and spent the majority of the week learning and working through them. Just below this is how to derive arcsin(x). I’m not going to go over how to derive arccos(x) and arctan(x) since they simply follow the same pattern:

• d/dx[arcsin(x)] = 
  • Step 1 – State that arcsin(x) = y:
    • y = arcsin(x)
  • Step 2 – Use the Sine operation on each side of the equation to isolate x:
    • sin(y) = sin(arcsin(x))
      • sin(y) = x
        • (Note: This works because sine is the inverse function of arcsin meaning that they cancel each other out in the same way that multiplication and division cancel each other out.)
  • Step 3 – Take the derivative with respect to x of each side:
    • d/dx[sin(y)] = d/dx[x]
      • cos(y) * dy/dx = 1
  • Step 4 – Divide both sides by cos(y):
    • (cos(y) * dy/dx)/cos(y) = 1/cos(y)
      • (cos(y) * dy/dx)/cos(y) = 1/cos(y)
      • dy/dx = 1/(cos(y)
  • Step 5 – Use the trig version of Pythag’s theorem to come up with the value of cos(y) in terms of sin(y):
    • a² + b² = c²
    • sin²(y) + cos²(y) = 1²
      • cos(y) = √(1 – sin²(y))
  • Step 6 – Replace cos(y) with √(1 – sin²(y)):
    • dy/dx = 1/(cos(y)
      • dy/dx = 1/√(1 – sin²(y))
        • = 1/√(1 – (sin(y))²)
  • Step 7 – From Step 2, replace sin(y) with x:
    • dy/dx = 1/√(1 – (sin(y))²)
      • d/dx = 1/√(1 – (x)²)
• d/dx[arcsin(x)] = 1/√(1 – (x)²)

The part that confuses me here is that when you replace sin(y) with x in the final step, you change dy/dx to d/dx which ends up being the solution to d/dx[arcsin(x)]. My understanding of why this works is because since y = arcsin(x) which is the same thing as sin(y) = x, if you solve for the derivative of y in sin(y) that’s equivalent to finding the derivative of x in arcsin(x). I’m not sure if that’s accurate though.

You can use the same steps from above to solve the derivatives of arccos(x) and arctan(x):

(Note: One thing I had to review was how to use Pythag’s theorem in terms of tan(x) and sec(x) which you can see that I did at the bottom of the second photo above.)

When I got to the arc-inverse-trig derivatives, arccot(x), arcsec(x) and arccsc(x), I realized that the steps I used above wouldn’t work. I found the following video that explained how to solve arccot(x) which showed me the technique used in order to solve the other arc-inverse-trig functions, as well:

• d/dx[arccot(x)] = 
  • Step 1 – Note that y = arccot(x) and take the cotangent of both sides:
    • y = arccot(x):
      • cot(y) = x
  • Step 2 – Draw a right triangle and label one of the non-90° angles as y:
    • (See photo above.)
  • Step 3 – Knowing that cotangent is the inverse of tangent, label the adjacent and opposite sides of the triangle:
    • cot(y) = adjacent/opposite
      • = x/1
      • (See photo above for labels on triangle.)
  • Step 4 – Use Pyhtag’s theorem to solve for the value of the hypotenuse and label on triangle:
    • hyp² = x² + 1²
      • hyp = √(x² + 1)
      • (Again, see photo above for label on triangle.)
  • Step 5 – Note the value of sin(y) based off sides of triangle:
    • Sin(y) = opp/adj 
      • = 1/√(x² + 1)
  • Step 6 – Find the derivative with respect to x of cot(y) = x and isolate dy/dx:
    • d/dx[cot(y)] = d/dx[x]
      • –csc²(y) * dy/dx = 1
      • dy/dx = 1/–csc²(y)
        • (Side note: Since csc(y) = 1/sin(y), you need to replace dy/dx = 1/–csc²(y) with dy/dx = 1/1/–sin²(y) = –sin²(y).)
      • dy/dx = –sin²(y)
        • = –(sin(y))²
  • Step 7 – Replace sin(y) with 1/√(x² + 1):
    • dy/dx = –(sin(y))²
      • dy/dx = –(1/√(x² + 1))²
        • = –(1²/(√(x² + 1)²)
        • = –1/(x² + 1)
• d/dx[arccot(x)] = –1/(x² + 1)

After learning how to draw out a triangle and label one of the angles as y, you can then use the same process to solve arcsec(x) and arccsc(x):

After having gone over these derivatives multiple times throughout the week, I now have a much better understanding of what arc-functions work and how to solve them.

In the middle of the week, I spent a day and a half trying to think through the FToC using the example f(x) = x² and its derivative f’(x) = 2x. I tried to think through how the FToC works with this particular function and its derivative from the interval x = [–1, 3]. I’m not going to write out the math here, but here are the pages from notes:

My big realization, which you can see that I wrote at the bottom of the last photo, is that if you’re able to find the average slope between two points on a function, because the derivative IS the slope at every point along that function, the average slope of the function is the average height of the derivative.  As far as I understand, this is essentially what the FToC states. (I’m not 100% confident about that though.)

At some point this week I realized that I passed the 2.5-year mark of working on KA at the end of last week. 😳 I find it pretty crazy to think that I’ve been studying math for that long. This coming week I need to get through the exercise Integration Using Completing the Square so that I can finally retry the unit test for Integrals (3,070/3,200 M.P.). I’m getting pretty tired of writing the same thing at the end of each post and really want to finally move onto the next unit. Hopefully I’ll have some good news to report in my next post. 🤞🏼