## anonymous 3 years ago using the difintion of derivative only find the derivative of f(x)=x/x^2+1

1. anonymous

-1/x^2

2. anonymous

\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: $\lim_{h\to0}\dfrac{\dfrac{h-h^2x-hx^2}{((x+h)^2+1)(x^2+1)}}{h}\\ \lim_{h\to0}\dfrac{h\left(\dfrac{1-hx-x^2}{((x+h)^2+1)(x^2+1)}\right)}{h}\\ \lim_{h\to0}\dfrac{1-hx-x^2}{((x+h)^2+1)(x^2+1)}\\ \dfrac{1-x^2}{(x^2+1)(x^2+1)}\\ \dfrac{1-x^2}{(x^2+1)^2}$