Question 1

At 6:35, how is Sal using the Power Rule to find the derivative? I don't understand how he got 1/2x^-1/2.

Accepted Answer

I think this video is a little out of order in the current calculus playlist. We are in "Derivatives Introduction," but the power rule is not introduced until the next section: "Derivative Rules."  See https://www.khanacademy.org/math/ap-calculus-ab/ab-derivative-rules/ab-power-rule/v/power-rule for details.

Question 2

I'm in the same situation as "willwassell".  I thank you for the explanation of how to plug values into L(x), but the important thing is that we're not told where the evidently general form L(x) =f(a)+f'(a)(x-a) comes from. First, isn't willwassell is correct in asking whether the f'(a)(x-a) isn't simply a manipulation of the slope formula?  I.e., aren't  we saying  " m=dy/dx and therefore dy=mdx " ?   Secondly, even if this is correct, there's STILL the problem of why L(x) = f(a) PLUS this value f'(a)(x-a). The explanation just before 6:00 is too circular. It assumes we know why L(x) equals this.
Finally, the glitch in the audio at 5:59, about "4.36" vs ".36" comes at too critical a point not to cause real confusion.    I'm sorry to be such a dweeb about this, but I really feel the need to know where this formula comes from.  Thank you for your patience, and I still really love your program.  I really do :)

Accepted Answer

I was confused about this too; however, it's simply another way to write point-slope form: y - y1 = m (x - x1), where (x1, y1) is a point on the line.
In L(x) = f(a) + f'(a) (x - a), the point is (a, f(a)) instead of (x1, y1). As you already know, f'(a) represents the slope (m) of the equation at the point where x = a. Therefore:
y - y1 = m (x - x1)
is the same as
f(x) - f(a) = f'(a) (x - a)
Add f(a) to both sides:
f(x) = f(a) + f'(a) (x - a)
And we call f(x) L(x) just to be fancy, I suppose. I hope this is accurate and helps!

Question 3

How does a calculator compute square roots? does it employ the same steps/reasoning as in the video?

Accepted Answer

There are a lot of algorithms to approximate these things. A calculator may use any one of them, after looking at the trade offs. Another, very much related way is Newton's method.

If you look at your calculator's user manual, they might provide the actual functions used. My Casio calculator for example, uses newton's method to solve simple 1=variable equations

Question 4

Can local linearization be used to calculate the value of `f(x₁)` which is far away from `f(x₀)`? Say, in this scenario, can I calculate `√16` by using L(x)?

Accepted Answer

The answer to your question is yes/no. The farther away you move from your original point, the more precision you lose. It will all depend on how good of an approximation you want. Majority of the time local linearization is only  used on points that are close to your original point. I will show you why....

Let us say f(x)=x^(1/2) and we wanted an estimation of (25.1)^(1/2). Therefore my a=25
when you work everything out your L(x)= 5 + (1/10)(x-25)
now we want L(25.1)=5+ (1/10)(25.1-25)=5.01
The actual answer to five decimal places for (25.1)^(1/2)=5.00999...
You see how close of an answer we got!!
Now if we do the same procedure and if we wanted to figure out (37)^(1/2). a=25
L(x) = 5+(1/10)(x-25)
L(37)=5+(1/10)(37-25)=6.2
The actual answer to five decimal places for (37)^(1/2)=6.08276...
You see that our estimation is not as precise as the last. The farther away you go from "a" the more precision you lose. To fix that choose an "a" that is close to the number you want an estimation. In our case if you let a = 36, L(37)=73/12 which is 6.08333... You see how better our approximation became by choosing a different "a"! Hope this helps

Question 5

at 3:40, why the formula of the line is f(4)+f'(4)(x-4)?

Accepted Answer

Remember point-slope form from Algebra? It went:
y-y1 = m(x2-x1)

(This comes from rearranging the slope formula:
m = (y2-y1)/(x2-x1) )

Anyway, if you take the point-slope formula and add y1 to both sides you get:
y= m(x2-x1) +y1
Now, the slope of a tangent line at a certain point is the derivative of the function at that point. The point we were looking at was 4, and the function was named f(x), so m was replaced by f'(4). x1 or the x value that we chose was 4. So we replace (x-x1) with (x-4). Finally, y1 is the output of the function at the point x1. x1 was 4, so f(4) would be the output of the function at the x-value 4. Now we have derived the formula used.

(The slope formula that was shown in parenthesis is derived from rise over run, where the rise is the difference between the y outputs and the run is the difference between the x inputs.

Question 6

Around 2:56, Sal calls this technique local linearization. Just to clarify, is local linearization equivalent to the linear approximation? And is process creating a tangent line to a curve and looking at a specific point? I'm unsure about the definition.

Accepted Answer

Yes, my AP Calculus teacher also introduced this topic as "Tangent Line Approximation." They all mean the same thing: using the tangent line at a point on the graph to approximate the value of another point very close to the original point.

Question 7

Starting at about 3:35, Sal write the equation for L(x). Where does he get f(4) from? Also, is L(x) just the tangent line to f(x) at x = 4?

Accepted Answer

Yes, L(x) is the tangent line to f(x) at x=4.  Sal uses f(4) because he is writing the equation of the line tangent to f(x) at the point (4, f(4)) (if the x-value is 4, then the y-value is f(4), right?).  You might be more used to always writing y = mx +b, where you put in the slope (which in this case would be f'(4)), and then plug in a point to find b, but Sal's method is faster and easier.  Basically, he is using the "point-slope" form instead of the "slope-intercept" form.

Question 8

Would it be possible to use the second derivative to do "local linearization" on the first derivative, THEN use the first derivative to do "local linearization" on the function?  Would this make it more accurate?  What if you used even higher-order derivatives?  Could you ever reach the actual value of the function or would you just get better approximations?

Accepted Answer

You've hit on a very powerful concept in calculus and analysis.

Yes, you can use higher-order derivatives to approximate a function as closely as you like. When you construct the approximation, you end up with a polynomial called a Taylor polynomial. It looks like this:

Say you want to approximate a function f(x) near a point c. The zeroth term of the polynomial is f(c).

The first term is f'(c)(x-c).

Second term is (f''(c)*(x-c)^2)/2!.

Third is (f'''(c)*(x-c)^3)/3!

And in general, the nth term is (f^n(c)*(x-c)^n)/n! , where f^n(c) means the nth derivative of f(x), evaluated at c.

Taylor polynomials are covered in the integral calculus playlist on Khan Academy.

Question 9

Could I also do this by finding the slope of the tangent line at x = 4, which is 1/4, then make an equation with the slope (y = 1/4x + 1), and then just plug in 4.39 as the x value to get 2.09?

Accepted Answer

Yes, that works fine!
It's basically what Sal did, only he wrote the line in point-slope form instead of slope-intercept form.

AP®︎/College Calculus BC

Unit 4: Lesson 6

Local linearity

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