<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] $\implies f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{\frac{x+h}{(x+h)^2+1}-\frac{x}{x^2+1}}{h}$
= $ \lim_{h\to 0}\frac{\frac{x+h}{(x^2 + 1 + 2h +h^2}-\frac{x}{x^2+1}}{h}$
$f(x + h) - f(x) = \frac{x+h}{(x+h)^2+1}-\frac{x}{x^2+1} = \frac{(x+h)(x^2 -1)}{((x+h)^2+1)(x^2 + 1)}-\frac{x((x+h)^2+1)}{((x+h)^2+1)(x^2 + 1)}$
$= \frac{h(1-x^2 -hx)}{(x^2 + 1)^2 + (2h + h^2)(x+1) } $
so $f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}= \lim_{h\to 0} \frac{\frac{h(1-x^2 -hx)}{(x^2 + 1)^2 + (2h + h^2)(x+1) }}{h} = \lim_{h\to 0}\frac{(1-x^2 -hx)}{(x^2 + 1)^2 + (2h + h^2)(x+1) } = \frac{1-x^2}{(x^2 + 1)^2}$
I don't know what "set a point" means unless it means try it for x = 0 or something. [In which case we'd get
$f'(0)=\lim_{h\to 0}\frac{f(h)-f(0)}{h}= \lim_{h\to 0} \frac{\frac{h}{h^2 + 1} -0}{h} = \lim_{h\to 0}\frac{1}{h^2 + 1} = 1$
]
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