How do you use the formal definition of the derivative as a limit to find the derivative of #F(x)= 1/x+1#?

To find the derivative of ( F(x) = \frac{1}{x+1} ) using the formal definition of the derivative as a limit, we start by applying the definition:

[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} ]

Substitute ( f(x) = \frac{1}{x+1} ):

[ f'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h+1} - \frac{1}{x+1}}{h} ]

Combine the fractions:

[ f'(x) = \lim_{h \to 0} \frac{(x+1) - (x+h+1)}{h(x+1)(x+h+1)} ]

Simplify the numerator:

[ f'(x) = \lim_{h \to 0} \frac{x+1 - x - h - 1}{h(x+1)(x+h+1)} ]

[ f'(x) = \lim_{h \to 0} \frac{-h}{h(x+1)(x+h+1)} ]

Cancel out ( h ) in the numerator and denominator:

[ f'(x) = \lim_{h \to 0} \frac{-1}{(x+1)(x+h+1)} ]

Now, substitute ( h = 0 ) into the expression:

[ f'(x) = \frac{-1}{(x+1)(x+1)} ]

[ f'(x) = \frac{-1}{(x+1)^2} ]

So, the derivative of ( F(x) = \frac{1}{x+1} ) is ( F'(x) = \frac{-1}{(x+1)^2} ).