problem \frac{(x+h)^2-x^2}{h} \frac{(x+h)(x+h)-x^2}{h} What more can I do from here?
S spacewater Junior Member Joined Jul 10, 2009 Messages 67 Jul 13, 2009 #1 problem (x+h)2−x2h\displaystyle \frac{(x+h)^2-x^2}{h}h(x+h)2−x2 (x+h)(x+h)−x2h\displaystyle \frac{(x+h)(x+h)-x^2}{h}h(x+h)(x+h)−x2 What more can I do from here?
problem (x+h)2−x2h\displaystyle \frac{(x+h)^2-x^2}{h}h(x+h)2−x2 (x+h)(x+h)−x2h\displaystyle \frac{(x+h)(x+h)-x^2}{h}h(x+h)(x+h)−x2 What more can I do from here?
D DrMike Full Member Joined Mar 31, 2009 Messages 252 Jul 13, 2009 #2 You could multiply out the brackets to get (x2+2xh+h2)\displaystyle (x^2+2xh+h^2)(x2+2xh+h2) Can you see what to do after that?
You could multiply out the brackets to get (x2+2xh+h2)\displaystyle (x^2+2xh+h^2)(x2+2xh+h2) Can you see what to do after that?
A Aladdin Full Member Joined Mar 27, 2009 Messages 553 Jul 13, 2009 #3 (x+h)2−x2h\displaystyle \frac{(x+h)^2-x^2}{h}h(x+h)2−x2 a^2 - b^2 = (a-b)(a+b) (x+h−x)(x+h+x)h\displaystyle \frac{(x+h-x)(x+h+x)}{h}h(x+h−x)(x+h+x) (h)(2x+h)h\displaystyle \frac{(h)(2x+h)}{h}h(h)(2x+h) Complete ...
(x+h)2−x2h\displaystyle \frac{(x+h)^2-x^2}{h}h(x+h)2−x2 a^2 - b^2 = (a-b)(a+b) (x+h−x)(x+h+x)h\displaystyle \frac{(x+h-x)(x+h+x)}{h}h(x+h−x)(x+h+x) (h)(2x+h)h\displaystyle \frac{(h)(2x+h)}{h}h(h)(2x+h) Complete ...
