MASTER Derivatives In Less Than A Minute!! 

Yeah me too, i am learning vector calculus rn and getting into Laplace transformation and Fourier series. But anyways i still watch these videos for aesthetic purposes lmao.

@@saudyassin5352 what the fuck does that mean😭

Calc 4 bruh? Calc 1 is already a nightmare

​@@greygousse9359 tbh it gets easier after calc 1 imo becuase with 1 you are introduced to these new concepts and going past that its mostly building off of the concepts you learned in 1 and 2 same goes for algebra.

@@greygousse9359math shouldn’t keep getting harder if you know what your doing

Yes but that's not always the case

U Weren’t paying attention in class but now you really needed to learn that’s why

@@TK-xi1jbyup

@@TK-xi1jbidk man, good teachers are getting rarer by each day.

@@TK-xi1jb not when your teacher in COLLEGE doesnt even speak english...to the point he doesnt even understand simple questions

Slope of tangent at x value if function

Bro it was 3 raise to 3 then he changed to 3 × 3^2 both are 27

Idk, but it works 😅

That is given in the question.

Yup and f'''(x)= 6

The answer would be 0 only when we differentiating a constant.The derivative of 5x is 5. You can think of it as x is raised to the power of 1. By using power rule, it would be 1(5)x^(1-1), which results in 5(x^0). x^0 is equal to 1. Therefore, the answer is 5. Hope it helped.

@@YeahMathIsBoring It did help, I appreciate it!

@@andyjustandy8721 You're welcome

(low*d-high minus high*d-low) all over low squared, down below.

seriously

Bro this is taught in Calc BC, like what do you expect 💀

It's the way the power rule works. Given a term in the form of k*x^n, its derivative is k*n*x^(n - 1). Drop the power by 1, and have the original power join the coefficient. Do this with x^3, and we get 3*x^2.

It is supposed to be given by the question. Depends on the question, and for this it is given x=3.

#flower bd Depends on the question u should only do if they ask for that

it's just a random value he chose

First principles is the 'correct' way. The reason why this shortcut is valid is due to first principles bring applied to variable with a power.

@@FootballPsychoPS3T oh, the original method is the first principle, I learnt it a few months ago in math class 🤦🏼

There is an actual way to do it

@@FootballPsychoPS3T First principles is reinventing the wheel. Anything you need to do in practice will use one of the standard rules derived from first principles. Unless you have a very exotic function where standard rules don't work.

There are limit proofs of the power rule that you can look up, if you care to see why it works. A basic summary of where it comes from, is expanding the general binomial of (x + h)^n and x^n. Then take the difference and divide by h. Since h is a really small number, all the h^2, h^3, etc terms will diminish to zero, so we just make them equal to zero. The term with h^1 will remain, and has a coefficient equal to 1 (see Pascal's triangle, and the binomial theorem). This leaves us with: (x^n + n*h*x^(n-1) - x^n)/h Cancel the two x^n terms: (n*h*x^(n - 1))/h = n*x^(n-1)