Math 181 Miniproject 3: Texting Lesson.md
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My lesson Topic
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<i class="fa fa-envelope fa-2x"></i>
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<i class="fa fa-camera fa-2x"></i>
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Problem 2 on the exam wanted us to write out the limit definiton of the derivative function f'(x) of the function f(x). Finding the derivative of functions is finding the instantanous slope of that function.

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This is helpful information because in the work place; this type of formula can be used to help a company maximize returns. This will deliver the rate of return at any given moment based on the cost and or the demand. This can help a business owner understand at what point during prodution he or she begins to see a profit. That will also help that owner undertand how much product they must produce or sell in order to not be in the negative at the end of every day.
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The equation for calulating the Limit Definition of the derivative is
$f'(x)$ $\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$
Next we would be provided with a function that we would use in this equation. In this example I will use

$f(x)=4x-x^2$

which would then be subsituted into the limit definition equation
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$f(x)=4x-x^2$

$f'(x)\lim_{h \to 0}\frac{[4(x+h)-(x+h)^2]-[4x-x^2]}{h}$

The next set of steps will test our algebra skills as we will need to simplify and reduce the equation.
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$f'(x)\lim_{h \to 0}\frac{[4x+4h-(x^2+2xh+h^2)]-[4x-x^2]}{h}$

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The next step combine like terms and simplify

$f'(x)\lim_{h \to 0}\frac{4x+4h-x^2-2xh-h^2-4x+x^2}{h}$

The $x^2$ and $4x$ will cancel out and $h$ will factor out leaving the simplified equation to be

$4-2x-h$

because we are trying to find the $f'(x)\lim_{h \to 0}$ the $h$ in the equations becomes zero leaving the answer for $f'(x)\lim_{h \to 0}= 4-2x$

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The best study tips I can provide for you is try and keep your work as neat as possble, be careful with your algebra. Also, be mindful of your signs. If at the end if the problem you notice most things didnt cancel out go back and take a look at your signs during the start of your simplification.
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