(1/√(3(x+h)) - 1/√(3x))/h.....help, I keep trying to solve and am striking out, tried this:
F friendsofstcorn New member Joined Sep 2, 2011 Messages 2 Sep 2, 2011 #1 (1/√(3(x+h)) - 1/√(3x))/h.....help, I keep trying to solve and am striking out, tried this:
tkhunny Moderator Staff member Joined Apr 12, 2005 Messages 11,339 Sep 2, 2011 #2 Looks rather like you're just making a mess. Why not try to simplify your life a bit. First, incorporate that 'h' in the denominator right away. No need to maintain it as a giant, horrible complicated fraction.
Looks rather like you're just making a mess. Why not try to simplify your life a bit. First, incorporate that 'h' in the denominator right away. No need to maintain it as a giant, horrible complicated fraction.
F friendsofstcorn New member Joined Sep 2, 2011 Messages 2 Sep 2, 2011 #3 i get it to -1/√(3x^2(x+h)) + √(3x(x+h)^2) but can't simplify denominator......suggestions....?
mmm4444bot Super Moderator Joined Oct 6, 2005 Messages 10,962 Sep 2, 2011 #4 Your scan is not easy for me to read. Also, the right side of your image is cut-off. It looks like you somehow first simplified: 13(x+h)−13x\displaystyle \frac{1}{\sqrt{3(x+h)}} - \frac{1}{\sqrt{3x}}3(x+h)1−3x1 to: x−x+h3x(x+h)\displaystyle \frac{\sqrt{x} - \sqrt{x+h}}{\sqrt{3x(x+h)}}3x(x+h)x−x+h. I got the following result, instead: 3[x−x+h]3x(x+h)\displaystyle \frac{\sqrt{3}[\sqrt{x} - \sqrt{x+h}]}{3\sqrt{x(x+h)}}3x(x+h)3[x−x+h] Your idea is good to multiply the numerator and denominator by the conjugate of the numerator. Give it another try.
Your scan is not easy for me to read. Also, the right side of your image is cut-off. It looks like you somehow first simplified: 13(x+h)−13x\displaystyle \frac{1}{\sqrt{3(x+h)}} - \frac{1}{\sqrt{3x}}3(x+h)1−3x1 to: x−x+h3x(x+h)\displaystyle \frac{\sqrt{x} - \sqrt{x+h}}{\sqrt{3x(x+h)}}3x(x+h)x−x+h. I got the following result, instead: 3[x−x+h]3x(x+h)\displaystyle \frac{\sqrt{3}[\sqrt{x} - \sqrt{x+h}]}{3\sqrt{x(x+h)}}3x(x+h)3[x−x+h] Your idea is good to multiply the numerator and denominator by the conjugate of the numerator. Give it another try.