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>
</div>
<div>
<i class="fa fa-user-plus fa-2x" aria-hidden="true"></i>
</div>
</div>

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

Hi, I was wondering if you could explain what this problem is asking and why it is important? It looks tricky.

</div></div>

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

Sure, which problem did you need help with?

</div><img class="right"/></div>

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

"For the function $f(x)=8x^2+4x+3$ find the exact formula for $f'(x)$ using only the definition of the derivative."

</div></div>

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

I don't even know where to start...

</div></div>

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

Okay, first lets look at what they are asking. What exactly is $f'(x)$?

</div><img class="right"/></div>

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

It is the derivative of $f(x)$, right?

</div></div>

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

Yes, thats right. In other words, it is the slope of the function $f(x)$. The problem states that we must use the definition of the derivative. Do you remember what that formula looks like?

</div><img class="right"/></div>

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

$f'(x)=\frac{f(x+h)-f(x)}{h}$

</div></div>

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

That is super close! You are just missing the part that tells us we are looking for a LIMIT definition. This one is just the general one where the limit is for $h$ as it goes toward $0$. It should look like this:

$f'(x)=$$\lim_{h \to 0}$$\frac{f(x+h)-f(x)}{h}$

</div><img class="right"/></div>

<div><img class="left"/><div class="alert gray">
Oh yeah! That looks right!
</div></div>

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

The limit definition is important because once we find it ONCE for a function, we can find all the derivatives for any value of $x$ of the function using the new formula- which is usually way easier!

</div><img class="right"/></div>

<div><img class="left"/><div class="alert gray">
I sure hope so!
</div></div>

<div><div class="alert blue">
Okay. So now lets put our function into this formula. It should look like this:

$f'(x)=$$\lim_{h \to 0}$$\frac{[8(x+h)^2+4(x+h)+3]-(8x^2+4x+3)}{h}$
Now show me how you would simplify it and combine like terms.
</div><img class="right"/></div>

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

$f'(x)=$$\lim_{h \to 0}$$\frac{[8(x+h)^2+4(x+h)+3]-(8x^2+4x+3)}{h}$
$f'(x)=$$\lim_{h \to 0}$$\frac{8(x^2+2hx+h^2)+4x+4h+3-8x^2-4x-3}{h}$
$f'(x)=$$\lim_{h \to 0}$$\frac{8x^2+16hx+8h^2+4x+4h+3-8x^2-4x-3}{h}$
$f'(x)=$$\lim_{h \to 0}$$\frac{16hx+8h^2+4h}{h}$

</div></div>

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

Great! Now lets pull the h's out of the numerator so we can
get rid of the h in the denominator.

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

$f'(x)=$$\lim_{h \to 0}$$\frac{h(16x+8h+4)}{h}$
$f'(x)=$$\lim_{h \to 0}$$(16x+8h+4)$

</div></div>

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

Awesome! Now we can plug $0$ in for $h$. We know that $8*0=0$. What is our formula now for the derivative of $f(x)$ ?
</div><img class="right"/></div>

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

$f'(x)= 16x+4$

That was way easier when we broke it down into parts and now I know why we do it! Thanks!

</div></div>

<div><div class="alert blue">
You are welcome! See you in class!
</div><img class="right"/></div>
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