Question 733544
{{{(1/(x + h)^2 - 1/x^2)/h}}} =

{{{(1/(x + h)^2 - 1/x^2)}}}{{{"÷"}}}{{{h/1}}} =

Get LCD

{{{(1*x^2/((x + h)^2x^2) - (x+h)^2/x^2(x+h)^2)))}}}{{{"÷"}}}{{{h/1}}} =


{{{(x^2-(x+h)^2)/(x + h)^2x^2}}}{{{"÷"}}}{{{h/1}}} =

Invert the second fraction and change division to multiplication:

{{{(x^2-(x+h)^2)/(x + h)^2x^2)}}}{{{""*""}}}{{{1/h}}} =

{{{(x^2-(x+h)(x+h))/(x + h)^2x^2}}}{{{""*""}}}{{{1/h}}} =

{{{(x^2-(x^2+2hx+h^2))/(x + h)^2x^2}}}{{{""*""}}}{{{1/h}}} =
 
{{{(x^2-x^2-2hx-h^2)/(x + h)^2x^2}}}{{{""*""}}}{{{1/h}}} =

{{{(cross(x^2)-cross(x^2)-2hx-h^2)/(x + h)^2x^2}}}{{{""*""}}}{{{1/h}}} =

{{{(-2hx-h^2)/(x + h)^2x^2}}}{{{""*""}}}{{{1/h}}} =

{{{(-h(2x+h))/(x + h)^2x^2}}}{{{""*""}}}{{{1/h}}} =

{{{-(h(2x+h))/(x + h)^2x^2}}}{{{""*""}}}{{{1/h}}} =

{{{-(cross(h)(2x+h))/(x + h)^2x^2}}}{{{""*""}}}{{{1/cross(h)}}} =

{{{-(2x+h)/(x + h)^2x^2}}}

Edwin</pre>