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">

Hey math partner! So I am having a pretty hard time with one of the concepts we've been learning about in Calculus. Do you think you could help me out?

</div></div>

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

Hey! I'm no math guru like Dr. Ballif, but I can certainly give it a shot! What's up?

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

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

Awesome! So basically, I am stuck on this problem that is asking to find the equation of a tangent line using only the definition of the derivative. The problem is, I don't know where to start! What the heck is the definition of a derivative anyway?

</div></div>

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

And the second part of the question wants me to Find the equation of the line tangent to f(x)=x^2 at x=2! Yikes!

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

Okay, I can definitely help you with that. First, take a deep breath. This concept seems difficult at first, but it's actually pretty manageable once you get a hang of it. So first, let's talk about what the definition of a derivative is. I don't want you to overthink this, because this is literally just a formula that you will use in all problems such as these! It is no different than any of the other cool formulas we learn about in Professor Martinelli's chemistry class, like PV=nRT!

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

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

What, really? It's that easy? I was totally overcomplicating it. What is the formula?

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

Yep, it really is that easy. Obviously we must first understand some basics though, so before I introduce you to the definition, let's go over some key concepts. First, A tangent line is a line which touches a curve at one and only one point.

Also, the slope-intercept formula for a line is y = mx + b, where m is the slope of the line and b is the y-intercept.

Furthermore, the point-slope formula for a line is y – y1 = m (x – x1). This formula uses one point on the line, represented by (x1, y1), and the slope of the line, represented by m, to calculate the slope-intercept formula for the line.

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

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

Now here is where things get more fun because we are in Calculus and not Algebra, which means we get to do more complicated stuff :)

The first derivative is an equation for the slope of a tangent line to a curve at an indicated point. And now we can define it. The "formula" or limit definition of the derivative f'(x) of a function f(x) is as follows:

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

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

Ooooh I am so excited!

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

f'(x)= The limit as h approaches 0 * (f(x+h)-f(x))/(h). Now, we all these definitions and formulas, you can find the equation of a tangent line!

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

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

Ok, that doesn't seem so terrible, but I am still pretty confused on how to find the tangent line with all that new information?

</div></div>

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

The equation for the slope of the tangent line to f(x) = x2 is f '(x), the derivative of f(x).
If we use the power rule we will get the following:
f(x) = x2
f '(x) = 2x

So, at x = 2, the slope of the tangent line is f '(2).
f '(2) = 2(2)= 4

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

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

f(x)=(2)^2=4, so y=4
Therefore, you have found the coordinates, (2, 4), for the point shared by f(x) as well as the line tangent to f(x) at x = 2. We now have a point on the tangent line and the slope of the tangent line.

The only step left is to use the point (2, 4) and slope, 4, in the point-slope formula for a line. Thus:

y-y1=m(x-x1)
y-4=4(x-2)
y-4=4x-8
y=4x-4

This is the equation of the tangent line! That's all there is to it!

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

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

Wow, that's not so bad. Thank you so much for your help!

</div></div>

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