I had another enjoyable week working through Derivatives: Definitions and Basic Rules and managed to get a good amount of work done though I probably (once again) only worked for about 5-6 hours in total. I was able to grasp around ~90% of what I went through this week without too much difficulty, all of which were brand new concepts to me so I felt like I learned a lot. A lot of what I did this week involved using algebra to solve equations which I actually enjoy doing. I’m definitely getting better at using BEDMAS and exponent rules properly which makes using algebra a lot easier and makes me feel like a genius when I do it without making mistakes. 👨🏻🔬
The first thing I learned this week had to do with finding the derivative at x-values along the sin(x) and cos(x) functions. What I learned gave me a miniature ‘mind-blown’ moment. As a quick reminder, the derivative of a function at a certain (x, y) point is the slope (m) of the function at that exact point. Keeping that in mind, here’s a graph of the concept that the derivative of sin(x) equals cos(x):
As you can see, the function sin(x) is drawn in red and cos(x) drawn in blue. At the point (0, 0) on the function sin(x), the slope of the function equals 1 (a.k.a. the ‘rise-over-run’ at that exact point is 1/1). You’ll notice that value of cos(x) at (0, 0) equals 1. The slope at every other point along the function sin(x) is equal to the y-value of cos(x) at the same x-value. For example, when plugging in π/2 (a.k.a. 90°) into sin(x) you get sin(π/2) = 1 and can see the slope equals 0. If you then plug π/2 into cos(x), you see that cos(π/2) also equals 0.
After going through this concept, Sal then explained how the derivative of cos(x) equals –(sin(x)):
Sal went through the proof of why d/dx[cos(x)] = –(sin(x)) which I was able to follow along with and, even though I understand the general concept of how it works, it made much less intuitive sense to me than d/dx[sin(x)] = cos(x). When talking about these concepts, I found it useful to pronounce them out loud:
- d/dx[sin(x)] = cos(x)
- “The derivative of Sine-of-x, with respect to x, equals Cosine-of-x.”
- d/dx[cos(x)] = –(sin(x))
- “The derivative of Cosine-of-x, with respect to x, equals negative Sine-of-x.”
The next thing I learned about gave me another ‘mind-blown’ type of moment, which is that the derivative of ex equals ex:
If you don’t have a solid grasp on derivatives, this will be a hard concept to understand. The general idea of this is, on the function f(x) = ex, the slope at any point along the function IS the y-value. This means that whatever x-value you plug into the function, the resulting y-value equals the slope at that same point. 🤯 This is highlighted on the table I wrote beside the function on the left side of the picture where each value under the f(x) column is the same as the value under the m column.
This exact same concept is also true for d/dx[ln(x)] = 1/x (“The derivative of the natural log-of-x equals 1 over x.”):
This is the one thing I learned this week that I really had a hard time understanding. Again, I understand the general concept of what’s going on (i.e. the slope at any point along the function ln(x) equals 1/x) but I found the proof that Sal went through really tough to follow.
The second last thing I learned this week is what’s called the Derivative Product Rule which is essentially a formula that lets you find the derivative of the product of two functions multiplied by each other. (I’m having a hard time putting this concept into words and don’t know if what I just wrote made sense…) The formula and it’s pronunciation is:
- d/dx[f(x)g(x)] = f’(x)g(x) + f(x)g’(x)
- “The derivative of the functions ‘F’-of-x multiplied by ‘G’-of-x equals ‘F’-prime-of-x times ‘G’-of-x plus ‘F’-of-x times ‘G’-prime-of-x.”
When Sal went through the proof of this rule, I was happy that I understood it completely:
Lastly, on Saturday I briefly learned about the Derivative Quotient Rule which is essentially the same thing as the product rule but is used when dividing two functions. The formula for this rule is:
- d/dx[f(x)/g(x)] = (f’(x)g(x) – f(x)g’(x))/(g(x))2
This formula is very similar to the product rule except that you subtract the second term from the first term in the numerator and divide both terms by the divisor function, squared.
All in all, as I mentioned at the top, I was happy with how this week went and felt like I learned quite a bit. I’m now 67% of the way through Derivatives: Definitions and Basic Rules (1680/2500 M.P.) and only have 5 more videos to watch and 4 exercises to go through before the unit test. My guess is I’ll get to the unit test by Wednesday or Thursday this coming week which hopefully won’t take me too many attempts to pass. Once I get through the unit test, the following unit is called Derivatives: Chain Rule and Other Advanced Topics. I have a strong feeling that unit is going to be off the chaaaaain. 🤙🏼🤙🏼🤙🏼 😬