<s>[INST] Find the derivative of the function $f(x)=x/(x^2+1)$ at a point a $$\frac{x}{x^2+1}$$
I have to find the derivative and set a point by myself.
So I just set $x=a$ ( is it correct?)
and using the limit,
$$ \lim_{h\to 0}\frac{f(a+h)-f(a)}{h}  = \lim_{h \to 0} \frac{\frac{a+h}{(a+h)^2 + 1} - \frac{a}{a^2+1}}{h}$$
therefore, the answer is $$\frac{ 1-a^2}{(a^2+1)^2}$$
is it correct?
 [/INST] Your answer is correct. 
Notice, $$f(x)=\frac{x}{x^2+1}$$
$$\implies f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{\frac{x+h}{(x+h)^2+1}-\frac{x}{x^2+1}}{h}$$
$$=\lim_{h\to 0}\frac{(x+h)(x^2+1)-x((x+h)^2+1)}{h((x+h)^2+1)(x^2+1)}$$
$$=\lim_{h\to 0}\frac{x^3+hx^2+x+h-x^3-2hx^2-h^2x-x}{h((x+h)^2+1)(x^2+1)}$$
$$=\lim_{h\to 0}\frac{h-hx^2}{h((x+h)^2+1)(x^2+1)}$$
$$=\lim_{h\to 0}\frac{1-x^2}{((x+h)^2+1)(x^2+1)}$$$$=\frac{1-x^2}{((x+0)^2+1)(x^2+1)}=\frac{1-x^2}{(x^2+1)^2}$$
Now, substituting arbitrary value $x=a$ 
$$f'(a)=\frac{1-a^2}{(a^2+1)^2}$$
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