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anonymous
 one year ago
Using the difference quotient, find f'(x) from 2sqrtx
anonymous
 one year ago
Using the difference quotient, find f'(x) from 2sqrtx

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I know the answer is \[\frac{ 1 }{ \sqrt{x} }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But I get only as far as \[\frac{ 2x + 2h + 2x}{ 2\sqrt{x} + 2\sqrt{x}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so difference quotient: \(\dfrac{f(x+h)f(x)}{h}\) I would find what \(f(x+h)\) is and state what \(f(x)\) is separately, then just plug it into the formula. \[f(x) = 2\sqrt{x}\]\[f(x+h) = 2\sqrt{x+h}\]\[\begin{align} f'(x)= \dfrac{2\sqrt{x+h}2\sqrt{x}}{h} \\ &=\frac{2\sqrt{x+h} 2\sqrt{x}}{h} \cdot \frac{2\sqrt{x+h}+2\sqrt{x}}{2\sqrt{x+h}+2\sqrt{x}} \\ &=\frac{4(x+h)4x}{h(2\sqrt{x+h}+2\sqrt{x})} \\&=\frac{4h}{h(2\sqrt{x+h}+2\sqrt{x})} \\ &=\color{red}{\frac{4}{2\sqrt{x+h}+2\sqrt{x}}} \end{align} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes I messed up I saw my mistake you da bestest!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Remember that: \((ab)(a+b) = a^2b^2\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks i follow....:)
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