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Using the definition of the derivative and not the power rule, how would I find $f(x) = \frac{1}{2\sqrt{x}}$

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    $\begingroup$ What have you tried? Write out $\frac{f(x)-f(a)}{x-a}$ and simplify, then let $x\to a$. Hint: $x-a=(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})$. $\endgroup$ Oct 10 '15 at 1:23
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Let $h$ be an arbitrary real number, then $$\frac{1}{2}\left(\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}\right) = \frac{1}{2}\left(\frac{\sqrt{x} - \sqrt{x+h}}{\sqrt{x+h}\sqrt{x}}\right) = \frac{1}{2}\left(\frac{(\sqrt{x} - \sqrt{x+h})(\sqrt{x+h}+\sqrt{x})}{\sqrt{x+h}\sqrt{x}(\sqrt{x+h}+\sqrt{x})}\right)$$ so $$\lim_{h\to 0} \frac{1}{2h}\left(\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}\right) = \lim_{h\to 0}\frac{1}{2}\left(\frac{-1}{\sqrt{x+h}\sqrt{x}(\sqrt{x+h}+\sqrt{x})}\right).$$

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  • $\begingroup$ i think the numinator must be -1 $\endgroup$
    – Arashium
    Oct 10 '15 at 1:32
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    $\begingroup$ @Arashium Oh, Thanks. $\endgroup$
    – Hanul Jeon
    Oct 10 '15 at 1:37
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Hint: Compute the rate of variation: $$\frac{\dfrac1{2\sqrt{x'}} -\dfrac1{2\sqrt x}}{x'-x}=\frac{\sqrt x-\sqrt{x'}}{2\sqrt{xx'}{(x'-x)}} =\frac{-1}{2\sqrt{xx'}(\sqrt{x'}+\sqrt x)}$$ and its limit when $x'$ tends to $x$.

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$$\begin{align}\lim_{h\to 0}\frac{1}{h}(f(x+h)-f(x))&=\lim_{h \to 0} \frac{1}{h}\left(\frac{1}{2\sqrt {x+h}}-\frac{1}{2\sqrt {x}}\right)\\ &= \lim_{h \to 0} \frac{1}{2h}\left(\frac{\sqrt {x}-\sqrt {x+h}}{\sqrt {x+h}\sqrt {x}}\right)\\ &=\lim_{h \to 0} \frac{1}{2h}\left(\frac{(\sqrt {x}-\sqrt {x+h})(\sqrt {x}+\sqrt {x+h})}{\sqrt {x+h}\sqrt {x} (\sqrt {x}+\sqrt {x+h})}\right)\\ &=\dots \end{align}$$

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\begin{eqnarray} f(x+h) -f(x) &=& {1 \over 2} ({1 \over \sqrt{x+h}} - {1 \over \sqrt{x}}) \\ &=& {1 \over 2} ({\sqrt{x}-\sqrt{x+h} \over \sqrt{x+h}\sqrt{x}}) \\ &=& {1 \over 2} ({\sqrt{x}-\sqrt{x+h} \over \sqrt{x+h}\sqrt{x}}) ( { \sqrt{x}+\sqrt{x+h} \over \sqrt{x}+\sqrt{x+h} } )\\ &=& {1 \over 2} ( { -h \over \sqrt{x+h}\sqrt{x} (\sqrt{x}+\sqrt{x+h})}) \end{eqnarray}

Now divide by $h$ to get ${f(x+h)-f(x) \over h} = - {1 \over 2} { 1 \over \sqrt{x+h}\sqrt{x} (\sqrt{x}+\sqrt{x+h})} $. Letting $h \to 0$ gives the desired result.

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  • $\begingroup$ Thanks for your help! I wasn't thinking about rationalizing. $\endgroup$ Oct 10 '15 at 1:39
  • $\begingroup$ After a while you will recognise the tricks... $\endgroup$
    – copper.hat
    Oct 10 '15 at 1:40
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Use the definition of the derivative $$\frac{df(x)}{dx} = \lim_{\Delta x\to 0}\frac{f(x+\Delta x) - f(x)}{\Delta x}$$

$$=\frac{d(\frac{1}{2}(x)^{-\frac12})}{dx} = \lim_{\Delta x\to 0}\frac{\frac{1}{2}(x+\Delta x)^{-\frac12} - \frac{1}{2}(x)^{-\frac12}}{\Delta x}$$

$$ = \lim_{\Delta x\to 0}\frac{x^{-\frac12}(1+\frac{\Delta x}{x})^{-\frac12} - (x)^{-\frac12}}{2\Delta x}$$

$$ = \lim_{\Delta x\to 0}\frac{(1+\frac{\Delta x}{x})^{-\frac12} - 1}{2\sqrt{x}\Delta x}$$

$$ = \frac12\lim_{\Delta x\to 0}\frac{(1+\frac{\Delta x}{x})^{-\frac12} - 1}{\sqrt{x}\Delta x}$$

$$ = \frac12\lim_{\Delta x\to 0}\frac{1-\frac{\Delta x}{2x}+0(\frac{\Delta x}{x})^2 - 1}{\sqrt{x}\Delta x}\tag{1}$$

$$ = \frac12\lim_{\Delta x\to 0}\frac{-\frac{\Delta x}{2x}}{\sqrt{x}\Delta x}$$

$$ = \frac12\frac{-1}{2x\sqrt{x}}$$

$$ = \color{blue}{-\frac{1}{4}x^{-\frac32}}$$

Where for $(1)$ I have expanded $(1+\frac{\Delta x}{x})^{-\frac12}$ binomially and $0(\frac{\Delta x}{x})^2$ is 'shorthand' to represent higher order terms which can be neglected since ${\Delta x\to 0}$

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Using the definition of the derivative, i.e

$\frac{\mathrm{d}y}{\mathrm{d}x}=\lim_{h\rightarrow 0}\Big{(}\frac{f(x+h)-f(x)}{(x+h)-x}\Big{)}$

Using $f(x)=\frac{1}{2}x^{-\frac{1}{2}}$, we get...

$\lim_{h\rightarrow 0}\Big{(}\frac{f(x+h)-f(x)}{(x+h)-x}\Big{)}=\lim_{h\rightarrow 0}\Big{(}\frac{\frac{1}{2}(x+h)^{-\frac{1}{2}}-\frac{1}{2}x^{-\frac{1}{2}}}{h}\Big{)}$

Using binomial expansion:
$(x+h)^{-\frac{1}{2}}=(\frac{1}{0!})x^{-\frac{1}{2}} + (\frac{1}{1!})(-\frac{1}{2})x^{-\frac{3}{2}}h + (\frac{1}{2!})(-\frac{1}{2})(-\frac{3}{2})x^{-\frac{5}{2}}h^2 + ...$

Substituting it in you should get:

$\lim_{h\rightarrow 0}\Big{(}\frac{\frac{1}{2}(x+h)^{-\frac{1}{2}}-\frac{1}{2}x^{-\frac{1}{2}}}{h}\Big{)} = \lim_{h\rightarrow 0}\frac{1}{2}\Big{(}(\frac{1}{1!})(-\frac{1}{2})x^{-\frac{3}{2}} + (\frac{1}{2!})(-\frac{1}{2})(-\frac{3}{2})x^{-\frac{5}{2}}h + ...\Big{)}$
$=-\frac{1}{4}x^{-\frac{3}{2}}$

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