Calculus Teacher vs. Power Rule Student

The definition of derivative is the foundation of differential calculus. However, many students tend to ignore it and want to use the "shortcuts" (differentiation rules such as the power rule). Not only do we need the definition of derivative whenever we need to determine the derivative of a function, but I also give a calculus 1 problem where using the definition of derivative is easier than using the differentiation rules!

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[space_engineer]

Math teacher: spends 1 hour explaining a method
Math teacher: and, there is an easy method too

[mikevids]

I felt so smart in calc when I figured out the power rule on my own while doing the homework in my class but then the next day my teacher had a lesson on it and everyone else learned it too.

[haakon]

Impressive! I know this too, you know

[earl]

5:55 I laughed at that facial expression, "can we use L'hopital's rule?"

[chloemccarthy]

Honestly Calculus is what made me want to be a mathmetician. I cannot explain why, but integrals were so seemingly magical to me as well as differentiation. Hope I'm smart enough to get my PhD in applied mathematics after my Bachelors!

[OverlordOfDarkness]

Funnily enough, when I first encountered the formal definition of a derivative in a real analysis textbook at university, my first question was why I was required to spend so much time in high school memorising the rules when just working from the definition was intuitively so much easier. Hindsight is often very ironic like that.

[johnchessant]

You can solve the (2^x-1)...(2^x-10) with the product rule too, namely (fgh)' = (f')gh + f(g')h + fg(h'), etc. In this case all terms except the first one vanish and the first term is -(9!)log(2).

[monke]

The fact that he just marked that timestamp as "final words"!!😆

[michaelroyer]

As a former math tutor, I find that with the "harder" derivatives, I break this out and it allows me to solve the problem with ease. It also helps me to better understand limits.

[MonsterIsABlock]

Inaccurate, calculus teachers actually just throw a bunch of formulas at you, bprp is just built different.

[JasonOvalles]

Student wants to use the power rule? "Oh that's brilliant! How did you come up with that?"

If they don't know how it's derived - "that's why we're practicing using the definition of the derivative, so we can use the definition to prove interesting patterns we might notice."

If they do know how it's derived - "that's great that you've already learned some of this! I don't want to jump ahead so we can give other students time to think and process. But maybe consider yourself a resource for other students in this first unit!"

[RobsMiscellania]

To grow in mathematical maturity, it is necessary to look well beyond the equations. They fall naturally into place from the structure we are exploring: the mathematical scaffolding of the equations themselves. When I was 16, I was in such a hurry. I could certainly have been described as THIS student, and I just wanted the formula. Today, I struggle to make certain formulas stick in my memory, aside from the ones I learned much earlier in my youth. It leads to occasional moments of embarrassment in front of scholars smarter than I (by that I mean all of them); I mainly shrug it off, figuring I can derive them again from the landscape I've already explored, or look it up in a book. It's no problem for me. The equations come naturally from a deeper understanding of the structure I've studied. And more often than I realize, when I wander back through well-studied material, I find new perspectives to look at old things. What is old truly is new.

[TheCodeSatan]

Nice video!

Try f(x) = 0 if x = 0 and f(x) = x²sin(1/x) for x not equal to 0

The derivative for this function isn't continuous and hence normal differentiating techniques don't work...we have to go by the definition of the derivative

[biffwebster]

It's 2:11 AM. I haven't taken a math class in roughly 15 years. I've never taken a calculus class. I have no idea what you're talking about. I have no idea why YouTube brought me here. With all this being said for whatever reason...I LOVE IT. I guess the tube knows me better than I know myself.

[sploofmcsterra]

In physics we also "discover" derivatives. E.g. If you are considering the pressure on a horizontal slice of fluid in a vertical tank, you end up with something like P(z+h) - P(z) = gravity*density*h. Then you know that if you divide both sides by h, the left hand side is just dP/Dz, so the pressure is the integral of gravity * density!

[bobsanchez]

Your example can still be solved just as easily with the product rule.

Let g(x) = 2^x - 1, and let h(x) = (2^x - 2) ... (2^x - 9).

Then f(x) = g(x) h(x). So f'(0) = g'(0) h(0) + g(0) h'(0).

Since g(0) = 0, the quantity is just g'(0) h(0).

[nicolastorres]

Q: Is it possible to have a question so that it is actually easier to solve with the definition of derivative rather than the differentiation rules, which limit is actually easier to solve using the epsilon-delta definition rather than the limit rules?

[tsawy]

Man, that limit of 2^h-1/h is pretty crazy man. Pretty triv w/ L'hopitals, but seeing that you used it in your search for derivs in your other video (and abusing the fact that you can get derivs of all the exponentials from just e^x) I tried to explicitly prove the case for e. Surprisingly tricky, and naturally one of those hard to search sorta math problems. Eventually managed it using epsilon delta and the logarithm definition of the exponential (ln(x+1) as integral from 0 to x of 1/t+1, actually the first rigorous def I was given in uni!) and the taylor expansion around 1+x of 1/x [which is gettable naively]). This was really helpful in taking care of the extra 1 in the limit numerator.

Pretty cool problem!

[harshvardhanpandey]

Solution using general differentiation techniques (Modified Product Rule):

derivative of f(x) * g(x) * h(x) = f'(x) * g(x) * h(x) + f(x) * g'(x) * h(x) + f(x )*g(x) * h'(x) [This kind of pattern goes on for any number functions multiplied together]

2^0 = 1
So, all the terms that retain (2^x-1) will just become 0.


Hence, derivative of (2^x-1) * (2^x-2) * ... *(2^x-10) at 0 = (2^x-2) * (2^x-3)*...*(2^x-10) * [derivative of (2^x-1)].

[samuelatienzo]

3:42 😂😂😂 he was just like “you get my point, there’s a shitload of rules, no need to list more”