I need to find $\frac{d}{dx}(\frac{1}{\sqrt{x+2}})$ only using the basic definition of the derivative $f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$, but I'm not having any luck with the algebra. Any hints?
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5Rationalize the denominator – Shailesh Jun 02 '16 at 04:17
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2@Shailesh Actually, rationalise the numerator :) – David Jun 02 '16 at 04:30
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Similar older question: Find derivative of $f(x)=\frac{1}{\sqrt{x+2}}+2x$ by definition – Martin Sleziak Jun 02 '16 at 05:56
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In fact, simply trying Google give you some reasonable results. For example, you could search for derivative 1 square root x or derivative 1 square root x definition. – Martin Sleziak Jun 02 '16 at 06:02
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By elementary algebra (try it!) you can show that $$\frac{f(x+h)-f(x)}h=-\frac1{\sqrt{x+h+2}\,\sqrt{x+2}\,\,(\sqrt{x+h+2}+\sqrt{x+2})}$$ and it is then easy to find the limit as $h\to0$.
David
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$$\frac{d}{dx}\ \frac{1}{\sqrt{x+2}}=\lim_{h\to0}\frac{\left(\frac{1}{\sqrt{x+h+2}}-\frac{1}{\sqrt{x+2}}\right)}{h}$$
$$=\lim_{h\to0}\frac{\left(\frac{1}{\sqrt{x+h+2}}-\frac{1}{\sqrt{x+2}}\right)}{h}\times\frac{\frac{1}{\sqrt{x+h+2}}+\frac{1}{\sqrt{x+2}}}{\frac{1}{\sqrt{x+h+2}}+\frac{1}{\sqrt{x+2}}}$$
$$=\lim_{h\to0}\frac{\left(\frac{1}{x+h+2}-\frac{1}{x+2}\right)}{h\left(\frac{1}{\sqrt{x+h+2}}+\frac{1}{\sqrt{x+2}}\right)}$$
$$=\lim_{h\to0}\frac{\left(\frac{-h}{(x+h+2)(x+2)}\right)}{h\left(\frac{1}{\sqrt{x+h+2}}+\frac{1}{\sqrt{x+2}}\right)}$$
$$=\lim_{h\to0}\frac{\left(\frac{-1}{(x+h+2)(x+2)}\right)}{\left(\frac{1}{\sqrt{x+h+2}}+\frac{1}{\sqrt{x+2}}\right)}$$
$$=\frac{\frac{-1}{(x+2)^2}}{\frac{2}{\sqrt{x+2}}}$$
$$=\frac{-1}{2(x+2)^{\frac{3}{2}}}$$
Ian Miller
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Thanks! I didn't see this as the usual scenario to use a conjugate – peterson.c.joshua Jun 02 '16 at 20:51