Given the function `f(x)=1/(x+4)`

We have to use the alternative form of derivative to find the derivative at x=c.

Here given that c=3.

So by definition of alternative derivative we have,

`f'(c)=lim_(x->c)(f(x)-f(c))/(x-c)`

`f'(3)=lim_(x->3)(f(x)-f(3))/(x-3)` ``

`=lim_(x->3)(1/(x+4)-1/7)/(x-3)`

`=lim_(x->3)(7-x-4)/(7(x+4)(x-3))`

`=lim_(x->3)(3-x)/(7(x+4)(x-3))`

`=lim_(x->3)-1/(7(x+4))`

`=(-1)/49`

``

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

Given the function `f(x)=1/(x+4)`

We have to use the alternative form of derivative to find the derivative at x=c.

Here given that c=3.

So by definition of alternative derivative we have,

`f'(c)=lim_(x->c)(f(x)-f(c))/(x-c)`

`f'(3)=lim_(x->3)(f(x)-f(3))/(x-3)` ``

`=lim_(x->3)(1/(x+4)-1/7)/(x-3)`

`=lim_(x->3)(7-x-4)/(7(x+4)(x-3))`

`=lim_(x->3)(3-x)/(7(x+4)(x-3))`

`=lim_(x->3)-1/(7(x+4))`

`=(-1)/49`

``