+73 The limit below represents a derivative \(f'(a)\). Find \(f(x)\) and \(a\).
\(\lim_{h\to 0}\frac{\left(2+h\right)^{2}-4}{h}\)
\(f(x) =\)
\(a = \)
+118608 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
\(f(a)=4\\ f(a+h)=(2+h)^2 \;\;so\\ a=2 \;and\;\\ f(whatever)=(whatever)^2 \;\;so \\ f(x)=x^2 \;\;f(2)=2^2=4 \;\;(\text{which I already knew})\\~\\\)
