Math 181 Miniproject 3: Texting Lesson.md
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My lesson Topic
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Hey!! I seriously need your help so bad, I know you took Math 181 last semester and now I'm taking it but I absolutley have no idea what is going on.

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Hi! Yes, I took Math 181 last semester & passed, what can I help you out with?

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My professor kept talking about limits and derivatives and I have no idea how to complete my homework now, do you think you could help? Thank you!

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I have to define something called "the limit definition of the derivitave" and use it to solve for $f'(x)$. Grossssss

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Okay no problem! Well I think the easiest part right now would be to define what you are learning about to understand what "the limit definiton of the derivitave means".
The limit definition of the derivitave (in words) is basically finding the derivitive of a function if it's limit exists.

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We use a function $f(x)$ to find the derivative and plug $x=a$ in. Following?

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Yes! But how do we sove for $f'(x)$?

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Great question, there's numerical definiton that can be used to find the limit definiton of the derivative and it is SUPER important to remember this so you can complete more problems- here ya go!
$f'(x)$ = $\:\lim _{h\to 0}\frac{f\left(x+h\right)-f\left(x\right)}{h}$
$f'(x)$ is the slope of the tangent line to the graph of $f$ at the point x=a.

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Can you break down each component of that formula? I don't understand what all that means. Sorry

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Yeah that's no problem, being thorough can never hurt!

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So the first part you'll write is $\:\lim _{h\to 0}$. YOU WRITE THIS PART EVERY SINGLE TIME YOU CONTINUE SOLVING THE PROBLEM. I CANNOT STRESS THIS ENOUGH. This will ensure you keep your work organized and will come in handy at the end of your computation. I will show you how to utilize this so don't worry!
The second part is $\frac{f\left(x+h\right)-f\left(x\right)}{h}$. This is where you would insert your function and solve. Notice that the $f(x+h)$ is different than $f(x)$. And it's all over $h$. Make sure you solve correctly, and if that means taking it slow and adding a few more computation lines that is totally fine!
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Okay I think I might be starting to understand this! Could we try and example to see everything in action?

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Of course! Here's one:
For the function $f(x)=6x^2-3x$ find the exact formula for $f'(x)$. You can only use the definition of the derivitive.
How would you first approach this problem?

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1) I would start by setting up the limit definiton of the derivitive $\:\lim _{h\to 0}\frac{f\left(x+h\right)-f\left(x\right)}{h}$
2) Then next, I would have to insert the function given to me into the limit definiton of the derivative
$f'(x)=\lim _{h\to 0}\frac{6\left(x\right)^2-3\left(x\right)-\left(6x+h^2-3x+h\right)}{h}$

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Hey, I would take a look at that second part again, notice anything?

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Oh shoot, I got my formula mixed up, it should be like this right?
$\:\lim _{h\to 0}\frac{6\left(x+h\right)^2-3\left(x+h\right)-\left(6x^2-3x\right)}{h}$

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Yes! There you go, what's your next move?

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Well, I would distribute the 6, 3, and negative sign
$\lim _{h\to 0}\frac{6\left(x^2+2xh+h^2\right)-3x-3h-6x^2+3x}{h}$
But hey, I also noticed that I could cancel some like terms, so I simplified my eqaution to
$\lim _{h\to 0}\frac{6\left(x^2+2xh+h^2\right)-3h-6x^2}{h}$
Is this okay?

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Yeah that's totally okay, keep going!

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I'm going to distribute the 6
$\lim _{h\to 0}\frac{6x^2+12xh+6h^2-3h-6x^2}{h}$
Simlifiying I end up with
$\lim _{h\to 0}\frac{12xh+6h^2-3h}{h}$
But now I'm stuck, what do I do now?

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Well I don't want to just give you the answer, but I can definitley push you towards finding the correct answer.

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So, when you have a fraction & there are LIKE TERMS in the top and the bottom, there's a little trick you can do. Can you remember what that might be?

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Let me think for a second...

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LIKE TERMS- I can cancel my like terms in the top and bottom of the equation! In this case that would be $h$.

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Great job! You got my hint, go ahead and and cancel those like terms.

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My equation would be
$\lim _{h\to \:0}\frac{\left(h\right)12xh+6h^2-3h}{h}$=$\lim _{h\to \:0}12x+6h-3$

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Alright, you're almost finished. Now remember earlier when I said I would show you how to utiilize $\lim _{h\to \:0}$ ? Here's how you do it.
When you have a limit, h is approaching zero. After all this math we did, we now have an equation with $h$ So all you have to do is replace $h$ with 0 and you'll have your derivative!

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Also, at this step you can stop writing $\lim _{h\to \:0}$.

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Okay seems simple enough! When I replace $h$ with 0 this is what I get
$12x+6\left(0\right)-3=\:12x-3$

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Perfect, you've solved it! I'm so glad I was able to help you out! But if I'm not available to help you if you're struggling again, here's some tips:

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Always remember $\lim _{h\to \:0}$ in everyline of your computation, expect at the last step.

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Remember it's $f\left(x+h\right)-f\left(x\right)$, and that never changes! And it ALL goes on top of $h$

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And lastly, practice, practice, practice always makes perfect!
Passing Math 181 is possible if you're willing to do all your work and practice!

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I hope I was able to help you out!

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Yes, thank you so much! I will listen to your advice!
Thanks again!

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Good luck!

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