dr|ft Posted April 7, 2005 Share Posted April 7, 2005 find derivate of root (2x - 1) using first principles, Thanks for your help. Link to comment Share on other sites More sharing options...
Dave Posted April 7, 2005 Share Posted April 7, 2005 This is fairly straightforward. You should get, as your limit: [math]\lim_{h \to 0} \frac{\sqrt{2x-1}-\sqrt{2(x+h) - 1}}{h}[/math] Now, by multiplying by the conjugate of the top, we get: [math]\lim_{h \to 0} \frac{-2h}{h(\sqrt{2x-1}+\sqrt{2(x+h) - 1})}[/math] This should give you your answer. Link to comment Share on other sites More sharing options...
VikingF Posted April 10, 2005 Share Posted April 10, 2005 The answer to the problem is: 1/root(2x-1), but when I solve Dave's equation above, I get -1/root(2x-1)....? d/dx(root(f(x))) = f'(x)/2root(f(x)) isn't it? Link to comment Share on other sites More sharing options...
Dave Posted April 10, 2005 Share Posted April 10, 2005 Very sorry, I got my limit the wrong way around. It should be: [math]\lim_{h \to 0} \frac{\sqrt{2(x+h) - 1} - \sqrt{2x-1}}{h}[/math] This gives the correct answer of [math]\frac{1}{\sqrt{2x-1}}[/math]. Link to comment Share on other sites More sharing options...
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