How do I find the deriviative of $\sqrt{\ln(x)}$, by definition? I got stuck thinking of a solution... Would be glad of getting help/hints.
Thanks in advance!
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What do you mean by definition? Also, the expression is not defined for $x \in (0, 1)$. – Michael Albanese Jan 05 '13 at 11:45
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By definition: $$f'(x)=\lim_{x->x_0}{f(x)-f(x_0)\over x-x_0}$$ – pie Jan 05 '13 at 11:54
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2I thought so. When I first learnt calculus, calculating this limit was referred to as finding the derivative by first principles. – Michael Albanese Jan 05 '13 at 11:56
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For $x\in (1,+\infty)$, $$\lim_{h\to 0}\frac{\sqrt{\ln(x+h)}-\sqrt{\ln x}}{h}=\lim_{h\to 0}\frac{\ln(x+h)-\ln x}{h}\frac{1}{\sqrt{\ln(x+h)}+\sqrt{\ln x}}=\\ \lim_{h\to 0}\frac{1}{\sqrt{\ln(x+h)}+\sqrt{\ln x}}\cdot \lim_{h\to 0}\frac{\ln(x+h)-\ln x}{h}=\\\frac{1}{2\sqrt{\ln(x)}}\lim_{h\to 0}\frac{\ln(x+h)-\ln x}{h}$$ The last limit is the derivative of $\ln x$. Check differentiability at $x=1$ separately.
Nameless
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Ah... the alternative definition. Thanks! But I don't understand how you went from the second phase to the third... – pie Jan 05 '13 at 11:57
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@pie You can easily do the same with the "standard" definition as well. I used the continuity of $\ln$ – Nameless Jan 05 '13 at 12:00
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But I don't understand how you went from the second phase to the third. I guess it's the same idea in the other definition. – pie Jan 05 '13 at 12:04
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