I was happy with the amount of work I managed to get through this week. I watched 11 of the 17 videos left in Derivatives: Chain Rule and Other Advanced Topics but only got through 4 of the final 9 exercises. I spent the majority of the week learning about the derivatives of inverse sine, cosine, and tangent with respect to x which I found somewhat tricky to understand. I finished the week by briefly getting started on how to differentiate composite functions that require using a combination of the power, product, quotient, and chain rules. I didn’t get all the way through this section, however, and found it pretty tough but I’m pretty confident that I’ll be able to understand how to do these types of questions with a bit more practice. 🤓

The first few videos I watched this week showed me how to derive the formulas for the derivatives of inverse sine, cosine, and tangent with respect to x. The first one I was shown was the derivative of the inverse of sine:

(As a side note – I don’t fully understand how inverse trig functions work since an x-coordinate maps to multiple y-coordinates, BUT I do know how to use the formulas to solve the questions I was given. 😬)

• To find the derivative of the inverse of sine with respect to x, you start with the equation y = sin^-1x which states, “y equals the inverse of sine (sin^-1) of x.”
  • 1) Given that y = sin^-1x, you can then take the sine of y which equals x.
    • (I don’t really understand how this works.)
  • 2) Take the derivative of both sides.
  • 3) The derivative of sine of y is cos(y) and dy/dx because of the chain rule and because it’s a composite function, and the derivative of x is 1.
  • 4) Divide both sides by cos(y).
    • 4.1 – 4.4) This uses the unit circle definition of Pythag’s theory to solve for cos(y).
  • 5) After substituting √(1 – x²) in for cos(y) you’re left with dy/dx = 1/√(1 – x²) which states, “the derivative of y with respect to x equals 1 over the square root of 1 minus x².”
• As stated at the very beginning, y = sin^-1x so you can substitute sin^-1x back in for y leaving you with d/dx[sin^-1x] = 1/√(1 – x²) which states, “The derivative of the inverse of sine with respect to x equals 1 over the square root of 1 minus x².”

The way you derive the inverse of cosine with respect to x is essentially the same as how you derive the inverse of sine with respect to x except you use Pythag’s theory to solve for sin(y) which leaves you with the final equation being d/dx[cos^-1x] = -1/√(1 – x²) (the only difference is that there’s a negative in front of the right side of the equation). The way you derive the inverse of tangent with respect to x is:

• Start with the same assumption as with the inverse of sine with respect to x that the inverse of tangent with respect to x equals tan(y) = x.
  • 1) Take the derivative of both sides.
  • 2) Using SOH CAH TOA and the unit circle, substitute sin(y)/cos(y) for tan(y) and state that the derivative of x is simply 1.
  • 3) Use the quotient rule and chain rule to differentiate sin(y)/cos(y).
  • 4) Simplify.
  • 5) State that cos²y + sin²y equals 1 (Pythag’s theory and unit circle) and then multiply both sides of the equation by cos²y to leave you with dy/dx = cos²y.
  • 6) Just for funsies throw a 1 underneath cos²y.
  • 7) Replace the 1 with cos²y + sin²y.
  • 8) Multiply both the numerator and denominator by 1/cos²y which you can do because multiplying the numerator and the denominator by the same thing is the same thing as multiplying the term by 1 which doesn’t change it’s value.
  • 9) Simplify.
  • 10) Replace both cos²y/cos²y with 1 and replace sin²y/cos²y with tan²y which is the same thing as [tan(y)]².
• A stated at the very beginning, tan(y) = x so you can replace [tan(y)]² with [x]² leaving you with 1/(1 + x²) = dy/dx = d/dx[tan^-1(x)].

Here are each of the 3 formulas and an example of how to use the inverse of sine with respect to x formula at the bottom:

After spending all of Wednesday and Thursday working on inverse trig functions, on Friday I worked on using different combinations of the derivative rules to differentiate composite functions. I started by writing out all the derivative rules:

The key to solving these types of questions is recognizing the individual functions within the overall composite function, labelling them accordingly, finding their individual derivatives as necessary, and then using the appropriate derivative rules. Here’s an example:

• y = sin³(x²) = [sin(x²)]³
  • f(x) = x³
    • f’(x) = 3x²
  • g(x) = sin(x)
    • g’(x) = cos(x)
  • h(x) = x²
    • h’(x) = 2x
  • (Note: [sin(x²)]³ is a function within a function within a function so, as you can see below, you have to use the chain rule twice.)
• dy/dx = f’(g(h(x))) * g’(h(x)) * h’(x)
  • = 3(sin(x²))² * cos(x²) * 2x
  • = 6x(sin(x²))²cos(x²)

I found solving these types of question pretty difficult. I have a hard time discerning the individual functions that make up the composite function and applying the derivative rules appropriately. As I said at the top, I do think I’ll be able to figure it out with a bit of practice but right now it seems very unintuitive to me.

As a final side note – I had a BEDMAS lightbulb go off for me at the beginning of the week which seemed worth mentioning. I realized that an expression underneath a radical needs to be solved before taking the root of the answer. I think of it now as if there are invisible brackets around the expression underneath the radical. I think the reason I didn’t realize this before was because I’d seen Sal finding the roots of the numerator and denominator of a fraction individually many times so I assumed you could find the roots of terms in an expression individually, as well. After doing some quick examples myself, however, I realized that is not the case: 

I have 10 days remaining before the end of June so I definitely think it’s possible to finish off Derivatives: Chain Rule and Other Advanced Topics (880/1600 M.P.) before the end of the month. There are 3 sections left for me to get through with a combined 6 videos and 5 exercises. I’m hoping to finish all the videos and exercises by the end of Thursday this coming week which would give me 2 days to pass the unit test before the week ends. If that doesn’t happen, I’ll still have 3 days the following week to get through it so, as long as I don’t run into anything insanely difficult, it’s looking pretty good! 💪🏼