It seems to be a simple question, but I couldn't figure it out.
I want to find
$$\lim_{h\to 0} \frac{\frac{1}{\sqrt {x+h}}-\frac{1}{\sqrt {x}}}{h}$$
I don't know how to control the font size.
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$\begingroup$ This is the definition of derivative of $f(x)=\frac 1{\sqrt{x}}$ $\endgroup$– polfosolCommented Nov 27, 2016 at 16:51
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4$\begingroup$ Try multiplying top and bottom by $\sqrt{x}\sqrt{x + h}$ for a more familiar form. Then, multiply by the conjugate to finish it. $\endgroup$– KaynexCommented Nov 27, 2016 at 16:55
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$\begingroup$ @polfosol Yes. I want to simplify this.. $\endgroup$– Ahmed AmirCommented Nov 27, 2016 at 16:56
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3$\begingroup$ Duplicate: Find the derivative of the function $f(x)=\frac 1{\sqrt{x}}$. (Found using Approach0.xyz) $\endgroup$– WorkaholicCommented Nov 27, 2016 at 19:08
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2 Answers
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\begin{align} \lim_{h\to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h} & =\lim_{h\to 0}\frac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x(x+h)}}=\lim_{h\to 0}\frac{(\sqrt{x}-\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}{h\sqrt{x(x+h)}(\sqrt{x}+\sqrt{x+h})} \\ & =\lim_{h\to 0}\frac{x-(x+h)}{h\sqrt{x(x+h)}(\sqrt{x}+\sqrt{x+h})}=-\lim_{h\to 0}\frac{h}{h\sqrt{x(x+h)}(\sqrt{x}+\sqrt{x+h})}\\ &=-\frac{1}{2\sqrt{x^2}\sqrt{x}}=-\frac{1}{2x^{3/2}} \end{align}
answered Nov 27, 2016 at 17:06
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observe that $$\frac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x}\sqrt{x+h}}=\frac{x-x-h}{h\sqrt{x}\sqrt{x+h}(\sqrt{x}+\sqrt{x+h}}$$
answered Nov 27, 2016 at 16:58