Derivative of Sqrt (x) from first principles

Рет қаралды 6,809

Prime Newtons

5 ай бұрын

In this video , I showed how to differentiate square-root of x from first principles

Пікірлер: 39

@punditgi 5 ай бұрын

The first principle is to watch all the videos by Prime Newtons! ❤🎉😊

@CharlesShorts 5 ай бұрын

Never stop learning

@AliM-gf7tq 4 ай бұрын

Wow, No one can explain it better than this, this guy does not skip even the obvious step….impressive, clear.

@mustaphaballaji4628 5 ай бұрын

You reminded me of this lesson when I was 17 years old. Now I am 60 years old. Thank you.

@RodMartinJr 5 ай бұрын

*_Never stop learning._* Advice I liked to give my own students. And when you master the understanding, you can more easily do word problems. 😎♥✝🇺🇸💯

@richardbraakman7469 5 ай бұрын

Along with the math, dispensing valuable life advice like "never touch the bottom"

@brrrayday 5 ай бұрын

Once i learned the power rule, I never looked back. But this is so important to know...its how you solve complex stuff

@kingbeauregard 5 ай бұрын

What I like about this is, the same skills by which you find the derivative will come into play again when you do epsilon-delta. For a more general proof of the derivate of x^a, I had to resort to logarithmic differentiation.

@PAGALM-jy9qq 2 ай бұрын

Finally i got what i was Searching for ☺️☺️♥️♥️ love your sir from 🇳🇵♥️

@johnnolen8338 2 ай бұрын

Very clean proof, my friend. 👍

@edgardojaviercanu4740 5 ай бұрын

It is just beautiful! Good job.

@user-il8mt2wz9t 5 ай бұрын

How fabulous the presentation on the math question😊

@Ahmad-yi6d 5 ай бұрын

👍

@stanwellmtonga5103 Ай бұрын

Awesome 🙌

@saarike 3 ай бұрын

Just wonderful!!!

@anirudhk.t7500 5 ай бұрын

Nice broh 🙌🙌

@ReyazulislamReayal 5 ай бұрын

You are a good teacher ❤

@surendrakverma555 5 ай бұрын

Very good. Thanks 🙏

@Samir-zb3xk 5 ай бұрын

i did this by using newton's binomial theorem to expand (x+h)^(1/2) but then i realized you can just multiply by the conjugate 😅😅

@JourneyThroughMath 5 ай бұрын

All first principle videos: Square root, all arctrig except arccot, plus some other various functions. Gotta say, I feel like a bit of a teachers pet doing this, he has a playlist. But its here for the record too

@PrimeNewtons 5 ай бұрын

Thank you. Now I know I have to do arccot x

@Th3OneWhoWaits 5 ай бұрын

Thank you very much sir!

@EnzoMariano 5 ай бұрын

Is it possible to use the same principle to prove the derivation rule for generic radials (derived from the nth root)?

@theshift2004 5 ай бұрын

You are the best. Many Thanks

@prakrit1280 5 ай бұрын

The coolest maths teacher I've ever seen 😎👍Your explanations are very clear and easy to understand 😊 Thank you for such lovely videos 😇

@blackovich 5 ай бұрын

I love you ❤

@DefenderTerrarian 5 ай бұрын

I've always wondered: why does the limit definition give the slope? Perhaps I never fully understood it, l'm in Calc3 now, but it seems like you'd get 0 in the denominator, of course that's undefined. Logically, it makes sense, but it's just never worked conceptually in my head.

@Nameless-qe9hu 4 ай бұрын

It’s the (y2-y1)(x2-x1) definition of the slope of a line at two points very close together. The goal is to get the “h,” which is technically nonzero, to cancel so that there is no division by 0 when you approximate “h” to be 0

@doctorno1626 5 ай бұрын

Please explain the derivative of a factorial function

@Th3OneWhoWaits 5 ай бұрын

Yes! It cant be calculated via first principles but interesting nonetheless.

@Th3OneWhoWaits 5 ай бұрын

Where is the 1/root x derivative video? I can't find it on your channel sir.

@PrimeNewtons 5 ай бұрын

kzfaq.info/get/bejne/eL5hYMWqrq6uZGg.html

@Th3OneWhoWaits 5 ай бұрын

@@PrimeNewtons thank you very much sir.

@jan-willemreens9010 5 ай бұрын

... f'(X) = LIM(h -> 0) [ SQRT(X + h) - SQRT(X) / h ] ... [ rewriting the denominator in its original form ... h = (X + h) - X ] .... LIM(h -> 0) [ SQRT(X + h) - SQRT(X) / ( (X + h) - X ) ] .... now treating the denominator as a Difference of 2 Squares as follows ... h = (X + h) - X = ( SQRT(X + h) - SQRT(X) )( SQRT(X + h) + SQRT(X) ) ... observing a common factor ( SQRT(X + h) - SQRT(X) ) between numerator and denominator, so after cancelling this factor we obtain the solvable LIMIT form ... f'(X) = LIM(h -> 0) [ 1 / ( SQRT(X + h) + SQRT(X) ) ] = 1 / ( SQRT(X) + SQRT(X) ) = 1 / 2*SQRT(X) ... the derivative of f(X) = SQRT(X) ...