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## Differential Calculus

### Course: Differential Calculus>Unit 4

Lesson 6: Approximation with local linearity

# Local linearity and differentiability

Intuition for how local linearity relates to differentiability using the Desmos graphing calculator.

## Want to join the conversation?

• So basically, "looks like a line" means "differentiable"? Is there a more precise way to define local linearity? • Not quite; that is just an intuitive introduction Sal uses in the video. A function is differentiable over an interval iff it is continuous over that interval and there are not any bends, cusps, or vertical asymptotes. Local linearity means just what it says. A function is locally linear over an interval iff that interval is sufficiently small for a tangent line to closely approximate the function over the interval.
• Let's take an example like the x^1000. You really can't tell if it's a hard corner or not, especially if you're only given the zoomed-out graph. Is there a way to determine local linearity/differentiability if you don't have the equation of the graph? • If there is a cusp, hard corner, vertical asymptote, or any sort of discontinuity then the function is not differentiable. Typically if you are asked to look at a graph and say if the graphed function is differentiable or not, the graph in the question will make it pretty clear that one of the conditions is violated (you probably won't have to strain your eyes to see if there is a hard corner or not etc.). As for 𝑥¹⁰⁰⁰, it is differentiable because all polynomials are differentiable. Though trying to tell from the graph may be difficult for the function because it may seen like a hard corner but if you zoom in sufficiently you'll see that it is smooth. I wouldn't worry about these extreme examples of graphs because I doubt anyone testing you in math would try to trick you based on examples that are difficult to physically see to gauge your mathematical ability.
• Oh! Can I ask about the name of the website you did the functions graphs? Thanks so much!
(1 vote) • I wonder how many flat earthers have taken calculus? • At , we are unable to differentiate the function around (2,0) because it's basically a vertical line. At , most of the curve looks a lot like a vertical line, yet the function is differentiable for all x values. Can anyone please tell me what am I missing? What's the difference between these two cases? • Do we need this for AP Calculus? • I don't think until now, the "power rule" has been discussed. Am I missing a lecture? • The power rule is a straightforward rule to learn.
If n is a constant, then the derivative of x^n, with respect to x, is nx^(n-1).
This is true for all real values of n, including whole numbers, zero, positive and negative integers, fractions, and even irrational numbers.

Example: The derivative of x^4 is 4x^3.
Harder example: Suppose we need to find the derivative of 5/cuberoot(x).
First rewrite 5/cuberoot(x) as 5x^(-1/3).
The derivative is then (-1/3)[5x^(-4/3)] = -5/[3 cuberoot(x^4)] = -5/[3x cuberoot(x)].

Have a blessed, wonderful day!
• So I just did the Practice: Approximation with local linearity. About half of the questions were of the sort "What is f(x) + f'(x) ?" I don't understand the purpose of this question. What is the value of the function plus the slope of the of local linearization? Why would you want to add the slope to the value? Usually these practice questions have an apparent purpose, so I feel like I'm missing something. • f(x) + f'(x) this can be used to approximate value for points which are very close to x. The function actually should look like this

f(x) + f'(x) * C

Here C is some scalar. If you zoom in to the point x sufficiently then x and its close points reside in a straight line. So if you know the slope at the point x which is f'(x) then f(x)+f'(x) will give you the value of the point closest to x. If you do

f(x) + f'(x) * C

Then this will give you value to the point that are C th closest to x.

Hope this helps.
• How would you solve for the linear approximation on arctan(1/2)? • Using a linear approximation, f(x) is approximately equal to f(a)+f'(a)delta x. The derivative of arctan(x) is 1/(x^2+1) and we know that arctan(1/sqrt(3)) = pi/6. So, arctan(1/2) is approximately pi/6 + 1/((1/sqrt(3))^2+1)*(1/2-1/sqrt(3)) = pi/6 + 3/4(1/2-1/sqrt(3)) which is about 0.4656. (The actual answer is 0.4636 so the approximation is fairly close.)

(Sorry for the random bold portion of text, but I cannot change it for some reason.) 