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December 27th, 2015, 11:02 AM  #1 
Newbie Joined: Dec 2015 From: PL Posts: 1 Thanks: 0  Derivative from definition
Hello i have a little problem with this Deriverative from definition do i correctly arranged it? cause i have some problem with with the result it comes 0 on numerator and good value on denominator 
December 27th, 2015, 11:45 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,649 Thanks: 2630 Math Focus: Mainly analysis and algebra 
$$\begin{aligned} \lim _{x \to x_0}{{1 \over \sqrt x}  {1 \over \sqrt x_0} \over x x_0} &= \lim_{h \to 0} {{1 \over \sqrt{x_0+h}}{1 \over \sqrt x_0} \over h} \\ &= \lim_{h \to 0} \frac1h \cdot {\sqrt x_0  \sqrt{x_0+h} \over \sqrt{x_0+h}\sqrt x_0} \\ &= \lim_{h \to 0} \frac1h \cdot {\sqrt x_0  \sqrt{x_0+h} \over \sqrt{x_0+h}\sqrt x_0} \cdot {\sqrt x_0 + \sqrt{x_0+h} \over \sqrt x_0 + \sqrt{x_0+h} } \\ &= \lim_{h \to 0} \frac1h \cdot {x_0  (x_0+h) \over \sqrt{x_0+h}\sqrt x_0 \left(\sqrt x_0 + \sqrt{x_0+h} \right) } \\ &= \lim_{h \to 0} \frac1h \cdot {h \over \sqrt{x_0+h}\sqrt x_0 \left(\sqrt x_0 + \sqrt{x_0+h} \right) } \\ &= \lim_{h \to 0} {1 \over \sqrt{x_0+h}\sqrt x_0 \left(\sqrt x_0 + \sqrt{x_0+h} \right) } \\ &= {1 \over \sqrt x_0\sqrt x_0 \left(\sqrt x_0 + \sqrt x_0 \right) } \\ &=  {1 \over x_0 \cdot 2\sqrt x_0 } \\ &= {1 \over 2 x_0^\frac32} \end{aligned}$$ 
January 23rd, 2016, 06:25 AM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902  I don't know what you mean by a "good value on denominator". If you take the limit of numerator and denominator separately, you should get 0 for both. Instead "rationalize the numerator" by multiplying both numerator and denominator by $\displaystyle \frac{\sqrt{x_0}+ \sqrt{x_0+ h}}{\sqrt{x_0}+ \sqrt{x_0+ h}}$ as v8archie did. (His "h" is $\displaystyle x x_0$.)


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definition, derivative, deriverative 
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