"Calculus Is EASIER Than PreCalc"

Do Science And Math Classes Get Easier? Harder? Or Stay The Same As You Make Progress?! #Physics #Chemistry #Math #ScienceIsFun #NicholasGKK #Shorts

[@user-dt7vt3cm2b]

Calculus concepts are far more down to earth. Understanding anything about derivatives was easier that idk, polar functions.

[@devonharvey8414]

Concepts are more abstract than subjects before it, but everyone mind is different, what maybe easy to you may be harder to others, how you like calculus, I’m learning limits and continuity

[@BabyMessi_]

You literally have to find derivatives of polar functions . It is hell

[@Boltclick]

@@BabyMessi_ That's like the end of Calculus II at most places (BC Calculus if you're in American High School). Calculus I ends a little bit after basic integration: it's actually really simple. Calculus II is when you get into the torturous stuff, like finding the volume of areas of rotation, Calculus with polar functions, and limits of summations. (and Calculus III (Multivariable) is when you question life, before finally having a bit of respite in Calculus IV (Differential)) At least, that's how it is at most places: some institutions may do things differently.

[@244295132171]

Bro confused between Analysis and Calculus. 😂

[@Boltclick]

@@244295132171 Take Calculus, and you'll get what he's saying. Calculus concepts are extremely observable. Seeing the relationship between speed and velocity is so much easier than most Algebra concepts. This is all Calculus revolves around: using the slope of equations to figure stuff out. It's legit just finding slope, but on a larger scale with more moving parts.

[@architech5940]

I still think calculus was easier to absorb and retain than precalculus

[@SuperSyntax-s7w]

Amazing distinction to make, well designed video

[@Obecox]

The study of pre-calculus involves the application of various mathematical formulas. Calculus, on the other hand, focuses on determining the slope of a curve and calculating the area under a curve.

[@Garfield_Minecraft]

That's why I'm bad at arithmetic but did well on set theory and calculus

[@imagod4796]

For anyone curious about the bottom problem, the solution is 0

[@HanzCastroyearsago]

what does f^10 notation mean? the 10th derivative? or f(f(f(…))) with 10 f’s?

[@burningtime7746]

​@@HanzCastroyearsago 10th derivative

[@KotikKelTuzada]

It took me 11 seconds I guess I'm not so good in calculus(

[@BenMartin-f5v]

Lmaooo it’s not 0

[@shingiaxiom6650]

You clearly have no idea what you're talking about if you can't figure out the answer for this question ​@@BenMartin-f5v

[@wmpowell8]

(10+3√(5x-2))/√(5x-2)=10/√(5x-2)+3=3+√(10)⇒10/√(5x-2)=√(10)=10/√(10)⇒5x-2=10⇒x=12/5

[@MrYellowAndYacello]

Though I don't have experience in calculus yet, I want to know the solution and how to calculate the bottom expressions.

[@BilalAhmed-on4kd]

0

[@jakegoddard9149]

Integrating an 8th degree polynomial gives you a 9th degree polynomial. Differentiating a 9th degree polynomial 10 times gives you 0.

[@meraldlag4336]

Are we not integrating 10 times though? I’m probably too tired for this

[@jakegoddard9149]

​@@meraldlag4336f(10) means differentiate f 10 times.

[@abehankens7456]

@@meraldlag4336 no, just differentiating 10 times

[@ComposedBySam]

Idk… I would solve the bottom problem tbh

[@Puppeteer_in_the_Void]

I've noticed that honors math classes are less tedious than regular math classes of the same grade and subject

[@UltraHydrationSupreme]

The kids in the honors class typically need to do less to get the same understanding

[@mathematicalworld-x5w]

x² =16² - 250 - 6 = 0

[@zqwas6306]

Thank you ! Needed to be said

[@Dubs913]

I’m in calculus right now, and can’t understand derivatives to save the life of me. And I’m sure that’s a problem, because I can only imagine the derivatives will only get harder from here. Does anyone have any helpful tips or tricks that helped them pass their calculus class? Any help would be greatly appreciated

[@ForsakenedMilk]

So hear me out; What if they just had one bad precalc teacher and then an amazing calculus teacher. From school to school, it’s different.

[@SarahTheAmbiguous]

If you pretend like integrals don't exist, for me precalc was harder than calculus. However, and I have no idea why, I just absolutely suck at integrals, and no matter how much practice I get with them I still need help choosing which strategy to use. The strategies themselves are easy for the most part (integration by parts 3 times in a row is everyone's favorite thing) but I never know how to set them up or which ones to use

[@fgvcosmic6752]

I am much beyond both precalc and calculus. I do still believe Calculus is easier, much easier in fact. Calculus 1 concepts _are_ very easy, and honestly more digestible than precalc. A lot of precalc work implicitly assumes calculus in a weird way, in that calculus is later used to post-justify what you were taught in pre calc. As well as that, the step up is definitely smaller. Transitioning from pre calc to calculus is a much smaller step than from basic algebra to pre calculus.

[@piggyjain8720]

Thanks for the deep analysis 🤣

[@TreyWilliamson-ku7bi]

The other consideration here is the difficulty jumps between levels of math. For many people, going from high school algebra/geometry to the trigonometry, college algebra, complex numbers, polar coordinates, etc. found in precalculus is a greater increase in difficulty than the jump from precalculus to the derivative rules and simple integration in calculus I. The material is still harder, and the concepts from precalc were necessary to understand the new material, but you might find yourself struggling less with Calc I, for this reason.

[@cyborgvison2024]

So now, correct me if im wrong, havent done any sort of calculus in a year or so, but cant you do f prime, get rid of the integral so f'(y)=g(x) plug in g(x) then take the derivative 9 more times? I know the answer is 0, but cantyou do that?

[@lamalamalex]

As one of those students. Nothing before calculus made sense to me. Yet calculus solved everything for me.

[@its_puggy_pugster8469]

Yeehah!! Algebra and calculus in one video!? What is it, my birthday?? 😜

[@CalistoArc]

Okay Im biased because I am a decent calc student but like I still would argue that problem 2 is easier. I think tedium and difficult are two different things but I also think that when people go into calculus they are expecting it to be plug and chug or which formula do I use. Which I would argue is more so an issue with the way math is taught. (This opinion will definitely be unpopular)

[@onurruzgar4635]

Yes!

[@sharingheart13]

I disagree. Calculus 1 is far easier than pre-calculus. I mean, the rules for basic derivatives are a lot easier to remember than the quadratic formula as is. You could easily teach Calculus 1 before college algebra. It won't make as much sense, but people would still ace it far more often than college algebra. I still remember the first thing I was taught in calculus 1: "A line without a break in it is called a continuous line. If it has a break in it, it's called a non-continuous line." The professor even showed drawings and went into detail. Calculus 2, however, is a very cruel reminder for why some people have to choose between STEM subjects and good mental health.

[@KotikKelTuzada]

It took me 11 seconds to solve bottom one... Ig I don't good enough(

[@user-vq5eb2xr8c]

Why does he sound like summoningsalt

[@mathgenius199-n8e]

x = 12/5

[@zqwas6306]

W post

[@lucalal7093]

Solution of f^{(10)}(y) : f^{(10)}(y)=\int_{0}^{y} x^{8}-126dx By integrating, we get the antiderivative of g expressed in y, so f^{(1)}(y) is simply y^{8}-126 By derivating again, we get 8y^{7} Since f^{(2)}(y)=8y^{7}, we see that x is at the power of 7, so 7 derivatives later we will get a constant that i'll name "a", so we get: f^{2+7}(y)=a f^{9}(y)=a f^{10}(y)=0

[@NotGleSki]

Bro asked ChatGPT 😭🙏

[@angeld23]

@@NotGleSki You need ChatGPT to take the derivative?

[@ensomosa5137]

​​​​​@@NotGleSki When you take a derivative of a function like x^8-126, you remove the constant, drop the exponent by 1, and multiply the variable by the former exponent, which at f(2) would give you 8x^7 Then, once you reach f(9), you'll get x*(8^7), which is a constant. Doing the derivative one more time would just get you 0, since the derivative of any constant is 0.

[@NotGleSki]

@@angeld23 no it cause the way he answered was so ai and ontop of that it’s in latex

[@NotGleSki]

@@ensomosa5137 ik how to differentiate it bro dw i said that cause he wrote it in latex 😭🙏

[@RandolphPereira-f1y]

I can solve both in 5 seconds😢

[@DaFeMaiden]

I’m sorry hands down calc 3 is way easier than calc 2. Maybe you could argue the jump in difficulty is way higher for calc 2. Everyone I’ve asked agrees

[@scar6073]

I disagree. Precalc has a lot of topics that are way harder to warp you head around than Calculus. The kids who say Calculus is hard are usually the ones with bad precalc foundation.

[@alithemathguy]

No Calculus 3 was easier than Calculus 2 and Calculus 1 was easier than PreCalc My grades in Calc 3 were so high, and the stuff was actually so easy to understand In Calc 2 there was stuff coming out of nowhere and I couldn’t handle Calc 1 is a lot simpler concepts like using limits to introduce the derivative or implicit compared to exact values of trig

[@nebula3415]

Na this is dumb anything involving proofs is automatically a lot easier than tedious stuff

[@anujparekh752]

Hmm

[@shachirai6873]

That is a college level problem? Man that looks easy

[@straightupsauce9210]

so?

[@arkdoesstuff7362]

Could be calc one, or calc ab in highschool. Stop tryna be different

[@midmeh]

stop commenting on the problem and focus on the message.

[@joaomiguel2100]

X = 12/5 and x = 0

[@ahmadsyauqi765]

X=12/5

[@darkzzgaming6153]

Could you explain it please

[@gustavoshigueo]

@@darkzzgaming6153 (10 + 3srqt(5x - 2))/sqrt(5x - 2) = 3 + sqrt(10) Multiply both sides by sqrt(5x -2) and you get 10 + 3sqrt(5x - 2) = sqrt(5x - 2) * (3 + sqrt(10)) 10 + 3sqrt(5x - 2) = 3sqrt(5x - 2) + sqrt(10) * sqrt(5x - 2) 10 + 3sqrt(5x - 2) = 3sqrt(5x - 2) + sqrt(10 * (5x - 2)) 10 + 3sqrt(5x - 2) = 3sqrt(5x - 2) + sqrt(50x - 20) Now subtract 3sqrt(5x - 2) from both sides: 10 = sqrt(50x - 20) Square both sides: 100 = 50x - 20 Divide both sides by 10: 10 = 5x - 2 Put 10 on the right side (not necessary, I just think it looks nicer): 5x - 2 = 10 Add 2 to both sides: 5x = 12 x = 12 / 5

[@N3RO_]

Let √5x-2 = y So the problem becomes 10 + 3y /y =3 + √10 Multiply both sides by y to get 10 + 3y = 3y + y√10 Remove the 3y from both sides to get 10 = y√10 Square both sides so 100 = 10y^2 divide by 10 and take the square root of both sides to get y = √10 And y is √5x-2 so √5x-2 = √10 Square both sides to get 10 =5x - 2 so 12 =5x divide by 5 to get x =12/5

[@N3RO_]

​@@darkzzgaming6153 there you go bud

[@ensomosa5137]

​@darkzzgaming6153 10+3(sqrt[5x-2]/sqrt(5x-2)=3+sqrt(10) Multiply both sides by the denominator 10+3(sqrt[5x-2)]=3(sqrt[5x-2])+sqrt(50x-20) Get rid of the 3* by multiplying the numbers inside the sqrt by 9 (you don't have to but i will) 10+sqrt(45x-18)=sqrt(45x-18)+sqrt(50x-20 Drop sqrt(45x-18) 10=sqrt(50x-20) Square both sides and solve 100=50x-20 120=50x 120/50=x 12/5=x