Question

How do you use the limit definition to find the derivative of `f(x)=x/(x^2-1)`?

Answer 1

`f'(x)=-(x^2+1)/(x^2-1)^2`

Explanation:

The limit definition of the derivative is as follows:

`color(red)(f'(x)=lim_(h->0)(f(x+h)-f(x))/h)`

`f'(x)=lim_(h->0)((x+h)/((x+h)^2-1)-x/(x^2-1))/h`

`f'(x)=lim_(h->0)((x+h)(x^2-1)-x((x+h)^2-1))/(h(x^2-1)((x+h)^2-1))`

`f'(x)=lim_(h->0)((x^3-x+x^2h-h)-x(x^2+2xh+h^2-1))/(h(x^2-1)((x+h)^2-1))`

`f'(x)=lim_(h->0)((x^3-x+x^2h-h)-(x^3+2x^2h+xh^2-x))/(h(x^2-1)((x+h)^2-1))`

`f'(x)=lim_(h->0)(x^3-x+x^2h-h-x^3-2x^2h-xh^2+x)/(h(x^2-1)((x+h)^2-1))`

`f'(x)=lim_(h->0)(color(blue)cancel(x^3)color(purple)cancel(-x)color(red)(+x^2h)-h-color(blue)cancel(x^3)color(red)(-2x^2h)-xh^2color(purple)cancel(+x))/(h(x^2-1)((x+h)^2-1))`

`f'(x)=lim_(h->0)(color(red)(-x^2h)-h-xh^2)/(h(x^2-1)((x+h)^2-1))`

`f'(x)=lim_(h->0)(h(-x^2h-h-xh^2))/(h(x^2-1)((x+h)^2-1))`

`f'(x)=lim_(h->0)(-x^2-1-xh)/((x^2-1)((x+h)^2-1))`

`f'(x)=(-x^2-1-0)/((x^2-1)((x+0)^2-1))`

`f'(x)=(-x^2-1)/((x^2-1)(x^2-1))`

`f'(x)=-(x^2+1)/(x^2-1)^2`