Question

Using the limit definition, how do you differentiate `f(x)= 3/(x+1)`?

Answer 1

`f'(x)=-3/(x+1)^2`

Explanation:

The limit definition of the derivative:

`f'(x)=lim_(hrarr0)(f(x+h)-f(x))/h`

We know that `f(x)=3/(x+1)`, so `f(x+h)=3/(x+h+1)`.

We can plug this in to find that:

`f'(x)=lim_(hrarr0)(3/(x+h+1)-3/(x+h))/h`

Find a common denominator.

`f'(x)=lim_(hrarr0)((3(x+1))/((x+h+1)(x+1))-(3(x+h+1))/((x+h+1)(x+1)))/h`

Simplify.

`f'(x)=lim_(hrarr0)(cancel(3x)cancel(+3)cancel(-3x)-3hcancel(-3))/(h(x+h+1)(x+1))`

`f'(x)=lim_(hrarr0)(-3cancel(h))/(cancel(h)(x+h+1)(x+1))`

`f'(x)=lim_(hrarr0)(-3)/((x+h+1)(x+1))`

Now, calculate the limit and plug in `0` for `h`.

`f'(x)=-3/(x+1)^2`