Formula Derivation for First Principle of Derivatives

In this video, we will discover how the formula for first principle of derivatives is derived, which is the formula derivation for the first principle. This is also known as the limit definition, or the first definition. So, how to derive the formula for the first principle of differentiation?
For a given function represented by a curve, we plot two different points lying on it. We first find the coordinates of both the points in terms of x and/or h. Then, we find the gradient of the secant line that passes though the points. However, in finding the derivative of a function, we actually want to find gradient of the tangent line at a specific point on the curve. Therefore, we try to move the second point closer to the first point, so that the gradient of the secant line that passes through both the points can be used to approximate the gradient of the tangent line. This also means that the distance between both points is close to 0, and can be represented by the limit of h approaches 0. Thus, we will obtain a formula for calculating the gradient of a secant line, where the distance between the 2 points is very close to each other, and this is what we are actually calculating for when we wanted to find the derivative of a function.
#differentiation #derivatives #calculus #firstprincipal #firstdefinition
TIMECODES:
0:00 Intro
0:10 Finding Coordinates of Points
1:23 Drawing a Secant Line
1:40 Calculating Gradient of Secant Line
2:51 Moving the 2nd point closer to the 1st point
3:12 Obtaining Tangent Line
3:52 Formula Is Derived

[@KUDIYARASAN-]

to the point explanation very great.

[@YeahMathIsBoring]

Glad to hear that! I've put everything I have into this explanation video.

[@bruhifysbackup]

but if we look at x^3, wouldn't y= x+1 be a tangent to x^3 at x = 3 be x+24? The slope of x + 24 is 1 but with differentiation, slope of tangent = 2*3^2 = 2*9 = 18?

[@YeahMathIsBoring]

If a line intersects the curve at only one point but does not have the same slope as the curve at that point, it is not a tangent line. It would simply be a line intersecting the curve at that point. The key aspect of a tangent line is that it is aligned with the curve's slope at the point of contact.
I totally agree with you where by using differentiation, even with the first principle, we get our gradient function as 2x^2. If x=3, then 2(3)^2 would be 18, meaning that the gradient of the tangent line is 18 when x equals 3.