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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 bro! Mind helping me out with this math stuff (assignment)? it looks tricky  

</div></div> 


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

If you want I can help explain it to you. Which problem are you stuck on?  

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


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

It's problem #2. 
Part a) write out the limit definition of the derivative f'(x) of a function f(x)
Part b) for the function f(x)=2x^2+1, find the exact formula for f'(x). Use only the definition for the derivative.

</div></div> 

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

How do I do this problem???

</div></div> 

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

So what this question is asking you to do is to compute the derivative using a single point to find the formula of f'(a). Like this you can find f'(a) at any point of 'a'

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

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

Could you show me the steps pls?

</div></div> 

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

K. The first thing we have to do is take the derivative of f(x) in order to get the formula of f'(x).Do you know how that formula looks?


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

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

Is it this one?$$lim(h->0)f'(x)={\frac{f(x+h)-f(x)}{h}}.$$


</div></div> 

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

Exactly! That's part a) right there.Now for part b) we are trying to find the derivative of f'(x) and we are already given 'X' which we just plug into the derivative formula.

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

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

Ok I think I'm starting to get it but how do we find f'(x)
</div></div> 

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

Mikey, you just plug in your f(x) equation into every x in the derivative formula to get f'(x) 


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

<div><img class="left"/><div class="alert gray">
So like dis??

$$f'(x)={\frac{(2(x+h)^2+1)-(2x^2+1)}{h}}.$$
</div></div>

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

Great, now foil the 2(x+h)^2 by mutipling 2(x+h)(x+h) and the rest is simple.

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

<div><img class="left"/><div class="alert gray">
So like dis??

$$f'(x)={\frac{(2(x+h)(x+h)+1)-(2x^2+1)}{h}}$$
$$={\frac{(2(x^2+2xh+h^2)+1)-(2x^2+1)}{h}}$$
$$={\frac{(2x^2+4xh+2h^2+1)-(2x^2+1)}{h}}$$
$$={\frac{(2x^2+4xh+2h^2+1-2x^2-1)}{h}}$$
then i cancel any opposites to get:
$$={\frac{4xh+2h^2}{h}}$$
lastly I divid it all by h to get:
$$={4x+2h}$$
What do I do after that?? :confused:

</div></div>

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

Now you just find the lim from h->0 by pluggin' in zero to the h  

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

<div><img class="left"/><div class="alert gray">
Is this right bro?

$$im (h->0)f'(x)= 4x+2(0)$$
$$=4x???$$
</div></div>

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

Yup! And that's all you gotta do to solve for problem 2 :thumbsup: 

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

<div><img class="left"/><div class="alert gray">
Thanks Donnie! You know you're my favorite brother right?! I luv u

:kissing_closed_eyes: :rolling_on_the_floor_laughing: 
</div></div>

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

Yeah yeah. Love you too bro :)

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


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