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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">

Hey! So I'm trying to do this homework assignment and I'm a little confused about derivatives and how to use the limit definition of a derivative. Could you explain it to me?

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

Sure, no problem! Let's start with what we're saying when we say find the derivative. Really, it's all about slope! The derivative is the slope of a tangent line. In algebra, we're used to seeing it as rise over run or the change in y over the change in x. That is to say, we were finding the slope of a secant line. And up until now we could find an average slope between two points. But that raises the question how do we find the slope at a specific point when there's nothing to measure? That's where derivatives come in! It's very useful when trying to find the slope in a specific instant. We use a small difference and then have that difference shink towards zero. In the equation, that's where h comes in. We want h, or that small difference, to approach zero.

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

Okay! So, the derivative is the slope of a tangent line and we get a single point by having h approach zero. So how do I use the formula?

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

Can you walk me through a problem?

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

Sure! Let's say that we have the function $f(x)=-5x^2+10x+5$. Calculate the exact value of $f'(2)$. To do that, we use this formula: $f'(x)=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}$
As you can see, h is approaching zero, meaning we're closing in on a single point. What modifications will you make for this specific problem? Where do you plug in 2?

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

Since we're looking for $f(2)$, you plug in 2 where ever there is an x. So you'd end up with $f'(2)=\lim_{h \rightarrow 0}\frac{f(2+h)-f(2)}{h}$

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

Great! Now, we get to do some algebra. If you plug $f(2+h)$ and $f(2)$ into our original formula for $f(x)$, and then put those into our formula for $f'(x)$, what would that look like?

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


So I got: $f'(2)=\lim_{h \rightarrow 0}\frac{[-5(2+h)^2+10(2+h)+5]-[-5(2)^2+10(2)+5]}{h}$

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

Looks good to me! Now you just need to distribute everything in the numerator and cancel out like terms like you would in an algebra class, and we can go on to the next step.

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

Once I did all of the distributing and cancelling of like terms, my equation looked like: $f'(2)=\lim_{h \rightarrow 0}\frac{-5h^2-20h}{h}$

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


Awesome. The next thing we need to do is factor out an h in the numerator. Once we do this, that factored out h will cancel out with the h in the denominator. That would look like this:$f'(2)=\lim_{h \rightarrow 0}\frac{h(-5h-20)}{h}$ 

$f'(2)=\lim_{h \rightarrow 0}-5h-20$

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

And this is when we'd take the limit right?

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

Yep! Now we plug in 0 for our h. What do you get as answer when you do that?

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

I got $f'(2)=-20$

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


Good job! You did it! Do you have any more questions for me, or does it make sense now?

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

No, I think I got it. Thank you so much!

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