I am extremely grateful to you !!!! These videos are so comprehensive so good the explanation are intuitive and easy to follow ❤️ You give attention and time to all students. you are perhaps the best teacher I have every came across. Love and best wishes from India 👋
Respected Sir, am Ganesh from India. Sir, you are by the far the best Mathematician i have ever come across. The simplest way of ur explaining topics to the pupil it's just awesome. Thank you
you literally explained everything from a to z. even the things that i thought didn need an explanation, and you really just enlighten me. your lectures are edifying
You are calculating the gradient of a line going through the origin and point (2,12), instead of the tangent of the function at that location. (He uses a derivative function to find this tangent in the video, which is how he reached a value of 7). These two functions will only give the same result for a linear function, (or at very specific value(s) on a non-linear function).
y - y1 = m(x - x1) x1/y1 -> known coordinate points of the tangent in question m -> gradiant of the tangent in question clean up for the actual equation.
My school's math curriculum has been pretty dry these past few years, so I decided to learn this ahead of time! Thank you for rejuvenating my interest in math
Slope of line is m=(y2-y1)/(x2-x1)...cross multiply to get y2-y1=m(x2-x1)...if you have slope, m, and a point (x1, y1) you can plug in all three values into that equation and get an equation for a line in the form y=mx+b, with b being y-intercept.
