Finding derivative of $\frac{x}{x^2+1}$ using only the definition of derivative

I think the title is quite self-explanatory. I'm only allowed to use the definition of a derivative to differentiate the above function. Sorry for the formatting though.

Let $f(x) = \frac{x}{x^2+1}$

\begin{align} 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+2hx+h^2+1}-\frac{x}{x^2+1}}{h}\\ &= \lim_{h\to 0} \frac{\frac{(x+h)(x^2+1)-x(x^2+2hx+h^2+1)}{(x^2+1)(x^2+2hx+h^2+1)}}{h}\\ &= \lim_{h\to 0} \frac{\frac{hx^2-2hx-h^2+h}{(x^2+1)(x^2+2hx+h^2+1)}}{h} \end{align}

I'm currently stuck with simplify the fraction so that I can finally find the derivative. I'd really appreciate some advice on how to proceed with the problem.

• You can factor an $h$ from the numerator. That cancels with the one in the denominator. – iamwhoiam Sep 12 '17 at 3:37
• Note that you can factor out an h from the numerator and cancel out the h in the denominator. Then you can just evaluate the limit by plugging 0 in for h – themathandlanguagetutor Sep 12 '17 at 3:40
• Alternatively, to see it better, first simplify the compound fraction. Recall $$\frac{\,\frac{a}{b}\,}{c}=\frac{a}{bc}$$ and then factor out $h$ from numerator and denominator. – quasi Sep 12 '17 at 3:41
• I think you should double check the step between your last two lines though – themathandlanguagetutor Sep 12 '17 at 3:43

Your last step is wrong, the actual steps will be,

$$=\lim_{h\to 0} \frac{\frac{-\color{red}{(hx^2-h+h^2x)}}{(1+x^2)(x^2+2hx+h^2+1)}}{\color{red}{h}}$$ $$=\lim_{h\to 0} \frac{\frac{-\color{blue}{(x^2-1+hx)}}{(1+x^2)(x^2+2hx+h^2+1)}}{\color{blue}{1}}$$

$$=\frac{1-x^2}{(1+x^2)^2}$$

\begin{eqnarray} {1 \over h } ({x+h \over 1 + (x+h)^2} - {x \over 1+x^2}) &=& {1 \over h } { (x+h)(1+x^2) - (1+(x+h)^2)x\over (x+x^2) (1 + (x+h)^2} \\ &=& {1 \over h } { h -h x^2 - h^2 x\over (x+x^2) (1 + (x+h)^2} \\ &=& { 1 - x^2 - h x\over (x+x^2) (1 + (x+h)^2} \end{eqnarray} The limit follows by taking $h \to 0$.

\begin{align} 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^2+1)((x+h)^2+1)}\\ &= \lim_{h\to 0} \frac{h - h^2x - hx^2}{h(x^2+1)((x+h)^2+1)} \end{align}

$$= \lim_{h\to 0} \frac{1 - hx - x^2}{(x^2+1)((x+h)^2+1)}$$ $$=\frac{1 - x^2}{(x^2+1)^2}$$

There are two common ways to compute a derivative. This is the other.

\begin{align} \left. f'(x) \right|_{x=x_0} &= \lim_{x\to x_0} \frac {f(x)-f(x)} {x-x_0}\\ &= \lim_{x\to x_0} \frac {\frac{x}{x^2+1}-\frac{x_0}{x_0^2+1}} {x-x_0}\\ &= \lim_{x\to x_0} \frac {x(x_0^2+1)-x_0(x^2+1)} {(x^2+1)(x_0^2+1)(x-x_0)}\\ &= \lim_{x\to x_0} \frac {(x x_0^2 - x_0 x^2) + (x-x_0)} {(x^2+1)(x_0^2+1)(x-x_0)}\\ &= \lim_{x\to x_0} \frac {-x x_0(x-x_0) + (x-x_0)} {(x^2+1)(x_0^2+1)(x-x_0)}\\ &= \lim_{x\to x_0} \frac {-x x_0 + 1} {(x^2+1)(x_0^2+1)}\\ &= \frac{1-x_0^2} {1+(x_0^2)^2}\\ \end{align}