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Math 181 Miniproject 3: Texting Lesson.md
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My lesson Topic
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h1 {
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<div><img class="left"/><div class="alert gray">

Hey I heard you were stuck on a math problem in class today, which one was it?

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

It was $f'\left(x\right)=5x-x^{2}$ 

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

I see. 

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

What part are you stuck on? the whole thing or completely lost?

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

I'm pretty much entirely lost...

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

It's ok. I'll explain it to you.

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

Thank you!


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<div><img class="left"/><div class="alert gray">
First you take whats given,        

$f'\left(x\right)=5x-x^{2}$
Then we We define the derivative of f with respect to x
$\lim _{h\to 0}\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}\right)$
produces a value for each x
 at which the derivative is defined, and this leads to a new function $y=f'(x)$
then plug in your given info
$\lim _{h\to 0}\left(\frac{\left[5\left(x+h\right)-\left(x+h\right)^2\right]-\left[5x-x^2\right]}{h}\right)$ this simplifies to 
$\lim _{h\to 0}\left(\frac{\left[5x+5h-x^2-h^2-2xh-5x+x^2\right]}{h}\right)$
this gets worked out to $\lim _{h\to 0}\left(\frac{5h-h^2-2xh}{h}\right)=\lim _{h\to 0}\left(\frac{h\left(5-h-2x\right)}{h}\right)$
and finally gives you $\lim _{h\to 0}\left(5-h-2x\right)=5-0-2x$

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

Thank you! I understand now! :)

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


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