Question

using the difintion of derivative only find the derivative of
f(x)=x/x^2+1

Answer 1

-1/x^2

Answer 2

\[\begin{align*}f(x)&=\frac{x}{x^2+1}\\ \\
f'(x)&=\lim_{h\to0}\dfrac{\dfrac{x+h}{(x+h)^2+1}-\dfrac{x}{x^2+1}}{h}
\end{align*}\]
Focusing on the numerator:
\[\left(\dfrac{x+h}{(x+h)^2+1}\cdot\dfrac{x^2+1}{x^2+1}\right)-\left(\dfrac{x}{x^2+1}\cdot\dfrac{(x+h)^2+1}{(x+h)^2+1}\right)\\ \\
\dfrac{(x+h)(x^2+1) - x((x+h)^2+1)}{((x+h)^2+1)(x^2+1)}\\
\dfrac{x^3 + hx^2+x+h - x^3-2x^2h-xh^2-x}{((x+h)^2+1)(x^2+1)}\\
\dfrac{h-h^2x-hx^2}{((x+h)^2+1)(x^2+1)}\]
Now back to the limit:
\[\lim_{h\to0}\dfrac{\dfrac{h-h^2x-hx^2}{((x+h)^2+1)(x^2+1)}}{h}\\
\lim_{h\to0}\dfrac{h\left(\dfrac{1-hx-x^2}{((x+h)^2+1)(x^2+1)}\right)}{h}\\
\lim_{h\to0}\dfrac{1-hx-x^2}{((x+h)^2+1)(x^2+1)}\\
\dfrac{1-x^2}{(x^2+1)(x^2+1)}\\
\dfrac{1-x^2}{(x^2+1)^2}\]