Blog #2 - Stephen Ledbetter

This week was pretty good actually. I'm so glad we have a 3-day weekend haha. I found out in the lesson this week that derivatives and slope stuff is really easy. There are two formulas that we learned this week. Slope of secant line and slope of tangent line formulas (also known as the derivative formula). And the two formulas are:

(D = delta)

slope of secant line formula: ( f(x + Dx) - f(x) ) / (Dx)

slope of tangent line formula (derivative): lim as Dx approaches 0: ( f(x + Dx) - f(x) ) / (Dx)

The only difference is that for the derivative formula you plug in 0 for Dx towards the end of the "solving" process.

Example:

Find the slope of the graph of f(x) = x^2 + 3x + 1 at (1, 3)

lim as Dx approaches 0: ( f(x + Dx) - f(x) ) / (Dx)

What you have to do is plug in (x+ Dx) for every x in the formula.

So...

(x + Dx)^2 + 3(x + Dx) + 1 - (x^2 + 3x + 1) / (Dx)

That simplifies to...

x^2 + 2xDx + Dx^2 + 3x + 3Dx + 1 - x^2 - 3x -1 / Dx

2xDx + Dx^2 + 3Dx / Dx

2x + Dx + 3

plug in 0 for Dx

2x + 3 is your equation

Now just plug in the x coordinate to find the slope.

2(1) + 3 = 5

The slope is 5.

There's not much I don't really understand except that I'm having a little trouble with the derivatives with the square roots in them. I'm not good with conjugates.