# If f(x)= \frac{x^2 - 1}{x^2 + 1}, find f '(x) and f "(x).

## Question:

If {eq}f(x)= \frac{x^2 - 1}{x^2 + 1} {/eq}, find f '(x) and f "(x).

## Quotient Rule:

For this problem, we will be using the quotient rule twice to get the first derivative and the second derivative. It is the differentiation rule that we will be using as the indicated function is a quotient of two functions and such a form of functions is differentiated using this specific rule.

As mentioned, we must differentiate {eq}\displaystyle f(x)= \frac{x^2 - 1}{x^2 + 1} {/eq} via the quotient rule, whose formula is shown below:

{eq}\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left(\frac{f(x)}{g(x)} \right) = \frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2} {/eq}

Finding the first derivative:

{eq}\begin{align*} \displaystyle f(x)&= \frac{x^2 - 1}{x^2 + 1}\\ \displaystyle f'(x)&= \frac{(x^2+1)\frac{\mathrm{d}}{\mathrm{d}x}(x^2-1) - (x^2-1)\frac{\mathrm{d}}{\mathrm{d}x}(x^2 + 1)}{(x^2 + 1)^2}\\ \displaystyle f'(x)&= \frac{(x^2+1)(2x) - (x^2-1)(2x)}{(x^2 + 1)^2}\\ \displaystyle f'(x)&= \frac{2x^3+2x-2x^3+2x}{(x^2 + 1)^2}\\ \implies \displaystyle f'(x)&= \frac{4x}{(x^2 + 1)^2}\\ \end{align*} {/eq}

As the first derivative is also a quotient of two functions, the second derivative can be attained using the same derivative rule:

{eq}\begin{align*} \displaystyle f''(x)& =\frac{\mathrm{d}}{\mathrm{d}x}\left( \frac{4x}{(x^2 + 1)^2}\right) \\ \displaystyle f''(x)& =\frac{(x^2 + 1)^2(4) - (4x) (2(x^2 + 1))(2x)}{((x^2 + 1)^2)^2}\\ \displaystyle f''(x)& =\frac{(x^2 + 1) (4x^2+4-16x^2)}{(x^2 + 1)^4}\\ \implies \displaystyle f''(x)& =\frac{ -12x^2+4 }{(x^2 + 1)^3}\\ \end{align*} {/eq}