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?
 A: 
question is,, find functions at an arbitrary point.. i know i can just use that rule,, but the thing is i have to use an arbitrary point

$x$ is already arbitrary. It doesn't get any more arbitrary by using a different letter as a variable name. The whole point of a function is to associate one element of a set that you provide with another element (of a possibly different set).
And $x$ is any number of the real numbers (probably at least, you didn't specify what set $x$ comes from) without specifying which one. It's not $12$, not $-245234234.4564576$, it's any element of the set that the function "accepts". After all, this is why we use letters to denote some variable, because it's not one distinct number.
A: 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 $x=a$ 
$$f'(a)=\frac{1-a^2}{(a^2+1)^2}$$
A: $$
\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}  = \lim_{h \to 0} \frac{\frac{a+h}{(a+h)^2 + 1} - \frac{a}{a^2+1}}{h} = \underbrace{\cdots \cdots \cdots \cdots \cdots \cdots \cdots}_{\text{and ?}} = \frac{ 1-a^2}{(a^2+1)^2} 
$$
In some contexts, jumping from the limit to its value without saying how it is derived is appropriate.  But here since you say the problem is to find the derivative, and normally one should include those specifics.  First some routine algebraic simplifications, then the crucial cancellation of the $h$, then substituting $0$ for $h$ in an expression that defines a continuous function.
Whether you should do this by finding a limit or by some other method (in this case, probably the quotient rule) also depends on the context, and you haven't told us about that.
A: $$\ f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{\frac{x+h}{(x+h)^2+1}-\frac{x}{x^2+1}}{h}$$
$$=\lim_{h\to 0}\frac{1-x^2}{((x+h)^2+1)(x^2+1)}$$$$=\frac{1-x^2}{((x+0)^2+1)(x^2+1)}=\frac{1-x^2}{(x^2+1)^2}$$
As far as I know, my answer also the same with the others. I guess you can substitute x=a. (not that important)
A: $\implies f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{\frac{x+h}{(x+h)^2+1}-\frac{x}{x^2+1}}{h}$
= $ \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
$f'(0)=\lim_{h\to 0}\frac{f(h)-f(0)}{h}= \lim_{h\to 0} \frac{\frac{h}{h^2 + 1} -0}{h} = \lim_{h\to 0}\frac{1}{h^2 + 1} = 1$
]
