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Algebra and calculus math help from an MIT graduate.
Thank you Nancy for sharing your knowledge with your viewers. I could not understand this for 5 months and now I understood due to watching 1 of your videos! You explained diligently, and you gave your all so thank you and God bless you abundantly for simply giving.
Better yet, set u^2 to (1-x^4) and you'll eliminate the radical altogether. Reference Elements of the Differential and Integral Calculus by Granville, Chapter 26 - Integration By Substitution Of A New Variable. Rationalization. The text (free PDF) is available for download at the Internet Archive.
Almost 9years later since i studied maths in school, this video for the 1st time in my life explained solving equations so lucidly that i can't thank you enough Nancy <3 and you proved that literally Beauty with Brains is a thing <3 thank you so much
Thanks NancyPi for your lecture on a fundamental understanding about the meaning of limits and therefore derivatives. Gracias desde Uhland, Texas. Una bendicion y abrazos donde sea que se encuentre. Hasta pronto!!
i may be sounding weird but for a milisecond i literally thought "OLIVIA RODRIGO IS TEACHING MATH??! THE HECK?" and then realised its not liv.. sorry 😂
For curious viewers, If you study the video carefully, she writes with the left hand. From her angle, the writings would be reversed which doesn't make sense unless she's an expert in writing mirror inverted images. So in reality, she actually writes on the glass with the right hand, but for us to see the writings in the right order, the whole image is flipped making it look like she writes with her left hand.
This is COMPLETELY FALSE. You can factor any quadratic. She's trying to FOOL you all. This method works with any quadratic : Set the quadratic equal to zero, find its two roots ( either by the *quadratic formula* or *completing the square* ), say *α & β* , then write the *Quadratic = (x-α)(x-β)* . There you go, it is factored! This works for the *factor theorem* : If *d* is a root of f(x), then *x-d* is a *factor* of f(x). I'm greatly surprised seeing none caught hold of Nancy's deception. Your looks and charisma can't fool me, Nancy 😆
We know by the derivative rule that, d/dx [ f(x) * g(x)] = d/dx (f(x)) * g(x) + d/dx(g(x)) * f(x) let u=f(x) du = f ' (x) let v=g(x) dv = g'(x). dx d/dx [ u * v ] = du * v + dv * u Take integral of both sides: u * v = intSign(du * v ) + intSign(dv*u)