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$$ \frac { \frac { 1 }{ 1+x+h } -\frac { 1 }{ 1+x } }{ h } $$ $$ \frac { 1(1+x) }{ 1+x+h(1+x) } -\frac { 1(1+x+h) }{ 1+x(1+x+h) } $$ $$ \frac { -h }{ (1+x+h)(1+x) } \quad *\quad \frac { 1 }{ h } $$

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..

Cherry_Developer
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  • You could divide the numerator and denominator by $h$ and expand the parentheses. – Ragnar Aug 18 '14 at 12:58
  • You're there. "*" is just multiplication, which is normally not indicated explicitly in regular algebra. Remove it and `cancel the $h$'s' (in reality, you're dividing numerator and denominator by $h$). – MGA Aug 18 '14 at 12:59
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    numerator has "-h" – Vikram Aug 18 '14 at 13:01

1 Answers1

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$$\frac { \frac { 1 }{ 1+x+h } -\frac { 1 }{ 1+x } }{ h }\cdot \frac{1/h}{1/h} = \frac{1}{h(1+x+h)} - \frac{1}{h(1+x)} = \frac{(1+x) - (1+x+h)}{h(1+x+h)(1+x)}=\frac{-h}{h(1+x+h)(1+x)}$$

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

amWhy
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  • The textbook answer saids it is:

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

     – Cherry_Developer Aug 18 '14 at 13:13
  • That is the answer. The h's cancel. I assumed, perhaps wrongly, that this was part of a limit problem. I'll edit to take that part out. – amWhy Aug 18 '14 at 13:14
  • So the h's cancel and leave a -1? I am so confused right now.

    I now see the mistake I made that didn't turn the h in the numerator negative.

     – Cherry_Developer Aug 18 '14 at 13:16
  • We have $-h$ in the numerator, and a factor $h$ in the denominator. $$\frac{-h}{h(1+x+h)(1+x)} = \frac{h(-1)}{h(1+x+h)(1+x)} = \frac hh\cdot \frac{-1}{(1+x+h)(1+x)} = 1\cdot \frac{-1}{(1+x+h)(1+x)} = \frac{-1}{(1+x+h)(1+x)}$$ – amWhy Aug 18 '14 at 13:19