Differentiation of 1/(1+x): Find the Derivative and Tangent Line at x = 2

Answer

Using the limit definition,

$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h} =\lim_{h\to0}\frac{\frac1{x+h+1}-\frac1{x+1}}{h}.$$

Combine the fractions in the numerator:

$$\frac1{x+h+1}-\frac1{x+1} =\frac{(x+1)-(x+h+1)}{(x+h+1)(x+1)} =\frac{-h}{(x+h+1)(x+1)}.$$

Therefore,

$$f'(x)=\lim_{h\to0}\frac{-1}{(x+h+1)(x+1)} =-\frac{1}{(x+1)^2}.$$

So

$$\boxed{f'(x)=-\frac{1}{(x+1)^2}}.$$

At $x=2$,

$$f(2)=\frac13,\qquad f'(2)=-\frac19.$$

Using point-slope form, the tangent line is

$$\boxed{y-\frac13=-\frac19(x-2)}.$$

Equivalently,

$$\boxed{y=-\frac19x+\frac59}.$$

The original function is undefined at $x=-1$, so neither the function value nor its derivative exists there.