Math 181 Miniproject 3: Texting Lesson.md
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My lesson Topic
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<div><img class="left"/><div class="alert gray">

So... what exactly is this assignment? it looks tricky.

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Hey! I could help explain it to you, if you want. Which question is it?

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It's question 2, it states:
"a.)
Write out the limit definition of the derivative $f'(x)$ of a function f(x)
b.)
For the function $f(x)= 4x^2 - 8$, find the exaxt ormula for $f'(x)$. Use only the definition for the derivative."

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Is this question about computing the derivative at a single point to comput a formula for f′(a) at any point a?

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Yes!!! Can you help me solve this?
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<div><div class="alert blue">
Yeah of course. So what we're doing is “taking the derivative” generates a new function, denoted by $f'(x)$, derived from the original function f(x).

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For the first part of this question all we need to do is write the function. Which is $f'(x)= \ lim_{h->0}\frac{f\left(x+h\right)\ -\ f\left(x\right)}{h}$.
f'(x) is simply the derivative we are trying to find, while x is the vaule we are given, that we are plugging in.
Now $f(x+h)$ and $f(x)$ is where we are going to plug x+h and x into the equation. However, if we are given a value for x it would look like $f(a+h)$ and $f(a)$. A is the vaule given to us.

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<div><img class="left"/><div class="alert gray">

Ok it's starting to make sense. How do we find the formula for $f'(x)$.

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Ok, since we are given the derivative which is $f'(x)$, our value we are solving for is x. So were going to use the eqaution from part a and leave it how is is. This is because the value given to us is x. Then what we're going to do is replace the f from $f(a+h)$ and $f(a)$ with the eqaution $f(x)= 4x^2 - 8$. Also, when you do the subitution for $f(x)$ put parentheses around it because you're going to distribute the negative. One more thing do not forget to write $=\ lim_{h->0}$ everytime you "change" the problem, while sovling it.

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So like this? $=\ lim_{h->0}\frac{4\left(x+h\right)^{2}\ -\ 8\ -\left(4\left(x\right)^{2}-8\right)}{h}$.

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<div><div class="alert blue">

Yup, looking good. Next you're going to mutiply $(x+h)$ twice, which would look like this $(x+h)(x+h)$. Then sqaure the x in $4(x)^2$, and then distribute that negative sign. And don't forget $=\ lim_{h->0}$

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Yeah, yeah I won't forget. This is what I got.
$=\ lim_{h->0}\frac{4\left(x^{2}+2xh+h^{2}\right)\ -\ 8\ -4x^{2}+8}{h}$.

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<div><div class="alert blue">

That's right. You're almost done, just few more steps. First you're going to distribute the 4. After that you're going to cancel out the positive and negative like terms. When you're done canceling, factor out an h. This is because we need to get rid of the h in the denominator. Lastly you're going to take the $lim_{h->0}$ and plug zero into any remaining h or h's. After that you add, subtract, mutiply, and/or divding normaly to get your formula. Now when you plug zero where h is you no longer need to write $=lim_{h->0}$ because you're taking that limit and plugging it into the h or h's.

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<div><img class="left"/><div class="alert gray">
Ok, I'm going to break my solvings into parts, ok.

DISTRIBUTING 4:
$=lim_{h->0}\frac{4x^{2}+8xh+4h^{2}\ -\ 8\ -4x^{2}+8}{h}$

CANCELING LIKE TREMS:
$=lim_{h->0}\frac{8xh+4h^{2}\ }{h}$
I canceled out $4x^2$ and $8$.

FACTORING h:
$=lim_{h->0}\frac{h\left(8x+4h\ \right)}{h}$
The h's in the denominator and outside the parentheses will cancel out.

PLUGGING 0 INTO THE h's:
$=8x+4(0)$
So zero times 4 eqauls zero, so the new eqaution looks like:
$=8x+0$

FORMUAL/ANSWER:
Now because anything plus zero eqauls itself, our formual is:
$f'(x)=8x$

Is that right?

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<div><div class="alert blue">

Yup!

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Really? Omg, thank you so much for your help!

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Anytime. See you Monday.