Stuck simplifying a fractional expression $$ \frac { \frac { 1 }{ 1+x+h } -\frac { 1 }{ 1+x }  }{ h }  $$
$$ \frac { 1(1+x) }{ 1+x+h(1+x) } -\frac { 1(1+x+h) }{ 1+x(1+x+h) } $$
$$ \frac { -h }{ (1+x+h)(1+x) } \quad *\quad \frac { 1 }{ h } $$
This is what I have so far. I have no idea what my next step is. I get to:
$$\frac { -h }{ (1+x+h)(1+x)(h) } $$
and don't know where to go from here..

$$\frac { \frac { 1 }{ 1+x+h } -\frac { 1 }{ 1+x }  }{ h }\cdot \frac{1/h}{1/h} = \frac{1}{h(1+x+h)} - \frac{1}{h(1+x)} = \frac{(1+x) - (1+x+h)}{h(1+x+h)(1+x)}=\frac{-h}{h(1+x+h)(1+x)}$$
The $h$ cancels, leaving us with $$\frac{-\require{cancel}\cancel h}{\cancel{h}(1+x+h)(1+x)} = \left(-\frac {1}{(1+x+h)(1+x)}\right)$$
