Last week I said that I should be able to get through this unit, Derivatives: Chain Rule and Other Advanced Topics, before the end of June without much difficulty as long as nothing comes up that stumps me for too long. Well, I definitely jinxed myself and spent about ~95% of this week trying to understand one formula. The good news is that I eventually figured it out (I think), but I didn’t get nearly as far through the unit as I hoped I would. As bummed as I am about my lack of progress, I’m happy with the effort I put in this week trying to understand this particular formula. I had to watch 3 or 4 videos from other YouTube channels, read a few articles from math websites, and I even asked the question on the r/Mathhelp subreddit. When I finally started to understand it, it was really gratifying. Woo!!

Before I get into the formula I worked through this week, I made a brief note on the difference between differentiable functions and continuous functions. I learned about the difference between these two functions a few weeks ago but Sal reviewed them this week and I found them to be a bit easier to understand this time around:

From what I understand, for a function to be differentiable at a given point, you must be able to find the slope (i.e. the derivative) of that point on the function. This means that there cannot be any “sharp turns”, as Sal says, or any discontinuities. In the first function, you can see that there’s a line discontinuity at the point x which means the function is neither continuous nor differentiable at that point. In the second function, there is a ‘sharp point’ meaning there is no defined slope at that point and therefore it’s not differentiable at that point (but is continuous!). In the third function, you can see that there is no point discontinuity or ‘sharp turn’ at the point in question so therefore the function is both differentiable and continuous at that point.

The equation I had a hard time working through this week had to do with changing a base number, for example a in a^x, to (e^ln(a))^x:

First off, the reason why you would want to change a base number to e^ln(a) is because e to the power of anything is the derivative of itself. I’ve been trying to figure out how to phrase the expression b^logb(a) = a for about 10 minutes and the best I can come up with is, “a base to the power of a logarithm of the same base of any argument is equal to the argument”:

• 2⁴ = 16
• Log₂(16) = 4
  • Since we know that Log₂(16) is the exact same value as 4, we can swap it with the exponent 4 in 2⁴:
• 2⁴ = 2^Log2(16) = 16
  • Note: You can choose ANY base greater than 0 and this expression will still work. The base doesn’t have to be a perfect root of the number being replaced.
• Example:
  • 10^x = (2^log2(10))^x
    • = (100^log100(10))^x
    • = (e^ln(10))^x
  • 354^x = (14^log14(354))^x
    • = (56^log56(354))^x

After learning how to swap a base number with e^ln(a), I was then shown how to find the derivative of any base a raised to the power of x:

The thing that took me that longest to understand this week was that the expression e^ln(a)x is a composite function where e^x is the outside expression and ln(a)x is the inside function. Since it’s a composite function, to find it’s derivative you must apply the chain rule:

• F(x) = f(g(x)) = 10^x
  • 10^x = e^ln(10)x
    • f(x) = e^x
    • f’(x) = e^x
    • g(x) = ln(10)x
    • g’(x) = ln(10)
      • (Note: you use the power rule to find g’(x))
• F’(x) = f’(g(x)) * g’(x)
  • = e^ln(10)x * ln(10)
    • (Note: f(g(x)) = f’(g(x)) = e^ln(10)x since e raised to the power of any value equals the derivative of itself.)
  • = ln(10) * (e^ln(10))^x
    • (Note: Sal always seems to put the natural log to the front of the expression. I think he does this because it’s essentially a coefficient and it’s standard form to put the coefficient at the front of the expression.)
  • = ln(10)*10^x
    • (Note: You can finish of the equation by switching e^ln(10) back to 10.)

The thing that took me the longest time to understand was that the outer function in the above example is 10, i.e. e^ln(10), and the inner function is ln(10)x. It also took me awhile to realize that e^zx, where z is any constant, is the derivative of itself. I kept thinking I needed to use the power rule and do something like d/dx[e^zx] = (zx)e^zx-1. In any case, after about 3 days of looking through other videos about this equation I finally figured out how and why the equation works.

The last thing I was able to get through this week was finding the derivative of log_a(x):

This equation is somewhat similar to finding the derivative of a^x in the sense that both equations utilize e to find the derivative of any other number, a. The key thing to remember to find derivative of log_a(x) is that:

• Log_a(b) = log_c(b)/log_c(a)

I know that this is known as the logarithm quotient rule but I forget how and why it works. Using this rule, you can find the derivative of log_a(x) as follows:

• d/dx[log_a(x)] = d/dx[ln(x)/ln(a)]
  • = d/dx[ln(x) * 1/ln(a)]
  • = 1/ln(a) * d/dx[ln(x)]
  • = 1/ln(a) * 1/x
  • = 1/ln(a)x

This equation didn’t take me quite as long to understand as it did to find the derivative of a^x but it still took me the better part of a day to figure out. Using both of these formulas, I worked through the 2 exercises that were given and finished them on Saturday but those were the only exercises I managed to get through this week. I didn’t feel like I had a strong grasp on what I was doing in the exercises so I’ll need more practice working through these types of questions to feel more confident with them. This feels similar to when I worked through the unit circle in Trig and had no clue what I was doing it but eventually it all clicked for me and became clear. I hope it’s the same anyways!

This coming week I’m hoping to get through 11 videos and 2 exercises. If I do, that would get me just under 50% of the way through this unit, Derivatives: Chain Rule and Other Advanced Topics (400/1600 M.P.). Given that I have 3 weeks before the end of the month, as long as I don’t run into anything else that’s as tricky to understand, I think I’m still in good shape to get through the unit before the month turns over. Fingers crossed!