Math 181 Miniproject 3: Texting Lesson.md
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My lesson Topic
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<style>
body {
background-color: #eeeeee;
}

h1 {
color: maroon;
margin-left: 40px;
}

.gray {
margin-left: 50px ;
margin-right: 29%;
font-weight: 500;
color: #000000;
background-color: #cccccc;
border-color: #aaaaaa;
}

.blue {
display: inline-block;
margin-left: 29% ;
margin-right: 0%;
width: -webkit-calc(70% - 50px);
width: -moz-calc(70% - 50px);
width: calc(70% - 50px);
font-weight: 500;
color: #fff;
border-color: #336699;
background-color: #337799;
}

.left {
content:url("https://i.imgur.com/rUsxo7j.png");
width:50px;
border-radius: 50%;
float:left;
}
.right{
content:url("https://i.imgur.com/5ALcyl3.png"); width:50px;
border-radius: 50%;
display: inline-block;
vertical-align:top;
}
</style>

<div id="container" style=" padding: 6px;
color: #fff;
border-color: #336699;
background-color: #337799;
display: flex;
justify-content: space-between;
margin-bottom:3px;">
<div>
<i class="fa fa-envelope fa-2x"></i>
</div>
<div>
<i class="fa fa-camera fa-2x"></i>
</div>

<div>
<i class="fa fa-comments fa-2x"></i>
</div>
<div>
<i class="fa fa-address-card fa-2x" aria-hidden="true"></i>
</div>
<div>
<i class="fa fa-phone fa-2x" aria-hidden="true"></i>
</div>
<div>
<i class="fa fa-list-ul fa-2x" aria-hidden="true"></i>
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<div>
<i class="fa fa-user-plus fa-2x" aria-hidden="true"></i>
</div>
</div>

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

How can I find the derivative of $f(x)= x^3-2x+x-1$?

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

Before you start, you have to understand what a derivative is. A derivative function $f'(x)$ is the slope of the function $f(x)$ at a point. The derivative is also classified as the instantaenous rate of change at $x=a$, where $a$ is a real number. The derivative $f'(x)$ of a function helps us see how the function $f(x)$ is changing across a time interval.

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

Ok, I see. How would I start to solve for $f'(x)$?

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Would I need to use a formula? Can I use the limit defintion of a derivative?

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Yes, you can use the limit definition of a derivative. The formula would be $f'\left(x\right)=lim_{h->0}\ \frac{\left(f\left(x+h\right)-f\left(x\right)\right)}{h}$
Here $h$ represents the change in $x$.

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How would I use it? Do I just plug in $(x+h)$ for every $x$ in the equation $x^3-2x^2+x-1$ and then subtract by $x^3-2x^2+x-1$ ? And it will all be divided by $h$? Will it look like $f'\left(x\right)=lim_{h->0}\ \frac{\left(\left(\left(x+h\right)^{3}-2\left(x+h\right)^{2}+\left(x+h\right)-1\right)-\left(x^{3}-2x^{2}+x-1\right)\right)}{h}$

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

Yes! Now you will have to foil parts of the equation that can be foiled and when done combine/cancel like terms. That would be numbers or variables opposite or alike to each other. Be careful! You can easily lose track of your numbers so it's important that you work slowly. Do not try to do the calculations in your head. Trust me on this. Write out all your work and steps so that you can easily go back and review it. That way you can spot your mistakes, or you can use it later to study for the quiz.

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

Ok, so after I foiled my equation looked like this $f'\left(x\right)=lim_{h->0}\ \frac{\left(x^{3}+3x^{2}h+3xh^{2}+h^{3}-2x^{2}-4xh-2h^{2}+x+h-1-x^{3}+2x^{2}-x+1\right)}{h}$
After the cancelling like terms it was
$f'\left(x\right)=lim_{h->0}\ \frac{\left(3x^{2}h+3xh^{2}+h^{3}-4xh-2h^{2}+h\right)}{h}$
But now I don't know how to get rid of the $h$ in the denominator...

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

To simplify the equation further you can factor an $h$ from the numerator, and that way the $h$ will cancel in the denominator. Also, since we are figuring out the limit of $h$ as it approaches $0$ we plug in $0$ for any $h$'s that are left in the equation. This will leave us with the instantaneous rate of change or the derivative of f(x).

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

Factored out an $h$.
$f'\left(x\right)=lim_{h->0}\ \frac{h\left(3x^{2}+3xh+h^{2}-4x-2h+1\right)}{h}$

Plugged in $0$ for all the left over $h$'s.
$f'\left(x\right)=lim_{h->0}\ \left(3x^{2}+3x\left(0\right)+\left(0\right)^{2}-4x-2\left(0\right)+1\right)$

This leaves me with the expression $3x^{2}-4x+1$ so therefore $f'\left(x\right)=3x^{2}-4x+1$.

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<div><div class="alert blue">
Yes! Good job. Also, when showing your work on a test or quiz do not stop writing the limit until you actually take the limit.

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