I am attempting to find the derivative of $(x-1)^2$ using the definition method for solving derivatives. $$F'(a) = \lim_{h \to 0} \frac{F(a+h)-F(a)}h $$
Would the RHS become $\lim_{h\to0} \dfrac{(a+h-1)^2-(a-1)^2}h$?
I have attempted this previously turning the formula into sqrt rather than exponent 2, but failed miserable. Any advice?
On
expanding your right Hand side we arrive at $$\frac{x^2+h^2+1+2xh-2x-2h-(x^2-2x+1)}{h}$$ with $$h\ne 0$$ can you finish this?
On
Well its simple , Just expand
$$\frac{(a+h-1)^2-(a-1)^2}{h}$$ $$=\frac{a^2+h^2+2ah-2a-2h-a^2-1+2a}{h}$$ $$=\frac{h^2-2h+2ah}{h}=h-2+2a=\lim_{h \to 0}h+2(a-1) \to 2(a-1)$$
Note: You don't need square roots to evaluate this. Just focus on manipulating to remove the $h$ from denominator to avoid division by $0$.
On
Yes, that works. I personally prefer to use: $$f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$ Then your result is in terms of $x$ instead of $a$. Therefore, your result, as you've mentioned well becomes: $$f'(x)=\lim_{h\to 0} \frac{(x+h-1)^2-(x-1)^2}{h} \tag{1}$$ Now, the next thing to do would be to expand the terms on the numerator. Note that: $$(a+b+c)^2=a^2+2ab+2ac+b^2+2bc+c^2$$ And, obviously: $$(a+b)^2=a^2+2ab+b^2$$ Using these expansions and substituting into $(1)$ gives you an easy limit to solve.
On
Derivative = $f'(x)$ = $\lim_{h\to 0}$$\frac{f(x + h) - f(x)}{h}$ = $\lim_{h\to 0}$$\frac{(x + h - 1)^2 - (x -1)^2}{h}$
Both the numerator and denominator tend to $0$, so it's safe to use L'Hopital's Rule, which gives us the following after deriving both the numerator and denominator with respect to h then replacing h by 0 in the final solution:
$f'(x) = lim_{h\to 0}$$\frac{2(x + h -1)}{1}$ = $2(x - 1)$
which is the correct answer.
An easier way without using L'Hopital's would be to factorize the numerator and find the solution directly as follows:
Derivative = $f'(x)$ = $\lim_{h\to 0}$$\frac{f(x + h) - f(x)}{h}$ = $\lim_{h\to 0}$$\frac{(x + h - 1)^2 - (x -1)^2}{h}$ = $\lim_{h\to 0}$$\frac{h(h + 2x - 2)}{h}$ = $\lim_{h\to 0}$$(h + 2x - 2)$ = $2x -2$
(Or just expand everything, but I think it wouldn't be as elegant :P)
Yes, that's exactly what it would become.
Square roots have nothing to do with this problem. Functions involving square roots are still possible to do with this process but are more difficult to work with than functions involving squares. Don't think about square roots for this particular problem at all.
For this problem, just expand $(a+h-1)^2$ and $(a-1)^2$, simplify while being very careful with your minus signs, and if done correctly then eventually you'll be able to factor an $h$ out of the numerator, cancel it with the $h$ in the denominator, and get your final answer of $2(a-1)$.