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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">

Hi, I have a question about one of our exam problems. How do I find the derivative of 
$$f\left(x\right)=5x^{2}+11x-6?$$

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

To begin, you must understand what a derivative function is. The derivative f'(x) is the slope of a function f(x) at a point. We can also classify the derivative as the instantaneous rate of change at x=a where a is a real number. The derivative of a function f'(x) allows us to see how he function f(x) changes across a time interval.  

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

I see. How do I begin?

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Is there any equation I need to use? Maybe the limit definition of a derivative?

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

Yes, to find the derivative we use the equation $$f'(x)=\lim _{h\to \:0}\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}\right)$$
where h represents the change in x

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

How do we use $$f\left(x\right)=5x^{2}+11x-6$$
and $$f'\left(x\right)=\lim _{h\to \:0}\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}\right)$$ to find the derivative? Do we simply plug in (x+h) for every x in $$5x^{2}+11x-6$$ subtracted by $$5x^{2}+11x-6?$$

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

Exactly, just don't forget that it is all divided by h.
$$f'\left(x\right)=\lim _{h\to \:0}\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}\right)$$



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

So our equation should look like:$$f'\left(x\right)=\lim _{h\to \:0}\frac{\left[5\left(x+h\right)^2+11\left(x+h\right)-6\right]-\left[5x^2+11x-6\right]}{h}$$
and after a little algebra and foiling, I came up with this:$$f'\left(x\right)=\lim _{h\to \:0}\frac{\left[5\left(x^2+2hx+h^2\right)+11x+11h-6\right]-\left[5x^2+11x-6\right]}{h}$$
Now all I have to do is multiply $$\left(x^{2}+2hx+h^{2}\right)$$ by 5 and $$\left[5x^{2}+11x-6\right]$$ by -1. Also combine like terms. So now I have:
$$f'\left(x\right)=\lim _{h\to 0}\left(\frac{10hx+5h^2+11h}{h}\right)$$ 
Then we can eliminate the two h leaving us with $$f'\left(x\right)=\lim _{h\to 0}10x+5h+11$$

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

Since we are finding the limit of h as it approaches 0, we plug in 0 for any h's left in the equation. This leaves us with our instantaneous rate of change or derivative of f(x). 

So if you plug in 0 for any h, you are left with:
$$f'\left(x\right)=10x+5\left(0\right)+11$$
$$f'\left(x\right)=10x+11$$
The derivative of 
$$f\left(x\right)=5x^2+11x-6$$
is:
$$f'\left(x\right)=10x+11$$




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