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Solve the following:

If $g(x) = \dfrac{2x}{x+1}$, find: $$\dfrac{g(x+h)-g(x)}{h}$$

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What have you tried? Have you been able to determine $g(x+h)$? –  Michael Albanese Jan 26 '13 at 10:17
    
Is it $\displaystyle \lim_{h \rightarrow 0} \frac{g(x+h)-g(x)}{h}$ –  juantheron Jan 26 '13 at 10:25
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1 Answer

If $$g(x)=\frac{2(x)}{(x)+1},g(x+h)=\frac{2(x+h)}{(x+h)+1}$$

So, $$g(x+h)-g(x)=\frac{2(x+h)}{x+h+1}-\frac{2x}{x+1}=\frac{2(x+h)(x+1)-2x(x+h+1)}{(x+h+1)(x+1)}=\frac {2h}{(x+h+1)(x+1)}$$

So, $$\frac{g(x+h)-g(x)}h=\frac2{(x+h+1)(x+1)}$$

If you know, $g'(x)=\lim_{h\to0}\frac{g(x+h)-g(x)}h,$

$$g'(x)=\lim_{h\to0}\frac2{(x+h+1)(x+1)}=\frac2{(x+1)^2}$$

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