The limit below represents a derivative \(f'(a)\). Find \(f(x)\) and \(a\).

\(\lim_{h\to 0}\frac{\left(2+h\right)^{2}-4}{h}\)

\(f(x) =\)

\(a = \)

GAMEMASTERX40 Feb 24, 2022

#2**+2 **

Here is a pic for differentiation from forst principals.

The differential of a curve at a particular point is the gradient of the curve at that point.

A secant is a line that joins two poins on the curve. In this case the orange line is a secant.

the gradient of the secant is \(\frac{rise}{run}=\frac{change \;in \;y\; values}{change\; in\; x\; vlaues}=\frac{f(a+h)-f(a)}{(a+h)-a}=\frac{f(a+h)-f(a)}{h} \)

If you make h smaller and smaller then you are letting h tend to 0 and once h is 0 you will have the gradient of the tangent at that x point.

\(gradient \;of \;tangent \;at\;x=a\\ y'(a)= \displaystyle \lim_{h\rightarrow 0} \frac{f(a+h)-f(a)}{h}\)

**Most people just memorize this formula without necessarily understanding where it came from.**

**But for the smarter students it will help a lot of you understand it.**

Lets compare this to what you have in your question

\(y' (a)= \displaystyle \lim_{h\rightarrow 0} \frac{ f(a+h)-f(a)}{h}\\ y' (a)= \displaystyle \lim_{h\rightarrow 0} \frac{(2+h)^2 -4}{h}\\ \)

so

\(f(a)=4\\ f(a+h)=(2+h)^2 \;\;so\\ a=2 \;and\;\\ f(whatever)=(whatever)^2 \;\;so \\ f(x)=x^2 \;\;f(2)=2^2=4 \;\;(\text{which I already knew})\\~\\\)

Melody Feb 25, 2022