I need to use the definition of $f'(1)$ to calculate $f(x) = \frac{1}{1+2x^2}$
I can give the definition of Differentiate as limit:
$$\lim_{h\to0}\frac{f(x_0+h) - f(x_0)}{h}$$
but I'm not sure that this is the right answer for $f'(1)$, is it?
If it is the correct answer how should I prcede? maybe in this way?
$$\lim_{h\to0}\frac{\bigg(\frac{1}{1+2(1+h)^2}\bigg) - \frac{1}{1+2(1)^2}}{h}$$
Edit:
here how I proceded
$$\lim_{h\to0}\frac{\bigg(\frac{1}{1+2(1+h^2+2h)}\bigg) - \frac{1}{3}}{h}$$ $$\lim_{h\to0}\frac{\bigg(\frac{1}{1+2+2h^2+4h}\bigg) - \frac{1}{3}}{h}$$ $$\lim_{h\to0}\frac{\bigg(\frac{3-(3+2h^2+4h)}{3(3+2h^2+4h)}\bigg)}{h}$$ $$\lim_{h\to0}\frac{3-3-2h^2-4h}{3h(3+2h^2+4h)}$$ $$\lim_{h\to0}\frac{h(-2h-4)}{3h(3+2h^2+4h)}$$ $$\lim_{h\to0}\frac{-2h-4}{3(3+2h^2+4h)} = -\frac{4}{9}$$
is this correct?