I hit my goal this week by getting 50% of the way through the unit Derivatives: Definitions and Basic Rules. I enjoyed what I went through this week as a lot of what I did was use algebra to solve equations, plus I didn’t have to learn too much theory which I always find a bit dull. I had hoped to do ~7.5 hours of KA this week which I definitely fell short of but, regardless, I’m happy with how the week played out, overall.

The first thing I went through this week is what’s called differentiability which is in regards to a point on a function. The concept of differentiability is something I’m still struggling to fully understand, but it has to do with whether or not you’re able to find the derivative (i.e. the slope) of a point on a function:

As far as I understand, if you’re able to find the slope at a point on a function then that point is considered differentiable. If there is a discontinuity at that point or if there’s, in Sal’s words, a “sharp turn”, then you won’t be able to find the slope/derivative at that point and it is considered non-differentiable. It’s important to note that a function MUST be continuous at the point in question for it to be differentiable but it does NOT have to be differentiable at a given point for it to also be continuous at that point.

I was shown/given a number of questions this week that switched back and forth between Lagrange’s derivative notation, f’(x), and Leibniz’s derivative notation, d/dx. I found that literally saying the notation out loud as I read the questions helped me to better understand what was being asked.

Next, I learned about what turned out to be a very important rule/formula for the remainder of the week which is what’s known as the Power Rule of derivatives. By applying this rule/formula to each term in a polynomial, you can come up with the expression that will give you the value of the derivative at every point along the function. (That definitely sounds confusing.)

• Power Rule
  • f(x) = xⁿ
    • f’(x) = n * x^n-1

I really started to understand this rule once Sal went through the following example:

As you can see from the example, the function f’(x) = x² looks like a standard parabola. The line running across the parabola represents the value of the slope at every x-value along the function f(x). This means that when f(3) = 3² = 9, then f’(3) = 2(3) = 6 meaning the derivative (i.e. the slope) at the point (3, 9) equals 6. It’s interesting how this still works for negative values of x. For example, f(-4) = 4² = 16 and therefore f’(-4) = 2(-4) = -8 meaning the slope at the point (-4, 16) equals -8. It’s a bit mind blowing to me that this actually works. Sal went through the justifications for the power rule and they made sense but were tough for me to understand and I definitely don’t know them well enough to explain here myself.

The last concept I learned about this week was what Sal referred to as Derivative Rules:

• Constant Rule
  • States that the derivative of a constant (a.k.a. a ‘stand-alone’ number) is equal to 0.
    • Ex. If f(x) = 4 then d/dx (“any x-value”) = 0
  • This is because when function equals a constant, this means that function is a horizontal line that runs through that value on the y-axis and is parallel to the x-axis which means there is no slope and therefore the derivative of every single point along the function equals 0.
    
• Constant Multiple Rule
  • States that if a term in a function is being multiplied by a constant, you can factor out the constant, apply the power rule to the variable, and then multiply the derivative by the constant.
    • Ex. If f(x) = 5x³ then 
      • d/dx f(x) = d/dx(5x³)
        • = (5)d/dx(x³)
          • (**Apply power rule**)
        • = (5) (3x^3-1)
        • = (5) (3x²)
        • = 15x²
          
• Sum and Difference Rule
  • States that when finding the derivative of a polynomial function, you can find the derivative of each individual term of the polynomial separately using the power rule and then add/subtract them afterwards.
    • Ex. If f(x) = 2x – 1  and g(x) = x⁴ then
      • d/dx [f(x) + g(x)] = d/dx f(x) + d/dx g(x)
        • = d/dx(2x – 1) + d/dx (x⁴)
          • (**Apply Power Rule**)
        • = 2*1x^1-1 + 4x^4-1
        • = 2 + 4x³

These rules may seem confusing but they’re very straightforward once you get the hang of them. I spent most of Friday, all of Saturday, and a bit of time on Sunday morning working through questions that had me apply all four derivative rules to functions and come up with the equations for the derivatives and for the tangent line at different points along the functions. I enjoyed working on these types of questions since it was all algebra based which felt like solving little puzzles. It was frustrating at times, however, since most of the questions had multiple steps to them meaning they took 3-4 minutes each and if I made a single mistake on any question I would have to redo the entire exercise, each of which contained 4 questions. Overall though, it was good practice and fairly enjoyable.

I’m happy with the progress I’ve been making so far working through Derivatives: Definitions and Basic Rules(1280/2500 M.P.). Looking ahead, there’s a chance I may be able to finish off this unit this week but it may not happen until next week which would be fine. Either way, considering my goal was to get through a unit a month, I’ll likely get through the unit <2/3 of the way through May which I’ll be happy about. I know it’s a ways away, but I’m pumped to eventually get through the course Calculus 1 just because I think it will be dope to be able to say I’ve done it lol. So hopefully I can keep up this pace! 💪🏼