Question

Using the limit definition, how do you find the derivative of `y =3/(x+1)`?

Answer 1

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

Explanation:

The limit definition of a derivative states that

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

Substituting `f(x)=3/(x+1)` into `f'(x)`,

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

From this point on, you want to expand and simplify.

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

`f'(x)=lim_(hrarr0)((3(x+1))/(x^2+xh+x+x+h+1)-(3(x+h+1))/(x^2+xh+x+x+h+1))/h`

`f'(x)=lim_(hrarr0)((3x+3)/(x^2+2x+xh+h+1)-(3x+3h+3)/(x^2+2x+xh+h+1))/h`

`f'(x)=lim_(hrarr0)((-3h)/(x^2+2x+xh+h+1))/h`

`f'(x)=lim_(hrarr0)(-3h)/(x^2+2x+xh+h+1)*1/h`

`f'(x)=lim_(hrarr0)-3/(x^2+2x+xh+h+1)`

Plugging in `h=0`,

`f'(x)=-3/(x^2+2x+x(0)+(0)+1)`

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

`color(green)(|bar(ul(color(white)(a/a)color(black)(f'(x)=-3/((x+1)^2)color(white)(a/a)|)))`