When you find the derrivative of f(x)=1/[sqrt(x+2)] you get f'(x)=-1/[2(x+2)^3/2], i was just wondering how they got the power of 3/2 in the denominator? because when i do it i just get 2(x+2)

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- Oct 12th 2008, 02:00 PMjohntuana very easy limit
When you find the derrivative of f(x)=1/[sqrt(x+2)] you get f'(x)=-1/[2(x+2)^3/2], i was just wondering how they got the power of 3/2 in the denominator? because when i do it i just get 2(x+2)

- Oct 12th 2008, 02:19 PMskeeter
$\displaystyle f(x) = \frac{1}{\sqrt{x+2}} = (x+2)^{-\frac{1}{2}}$

$\displaystyle f'(x) = -\frac{1}{2}(x+2)^{-\frac{3}{2}} = -\frac{1}{2(x+2)^{\frac{3}{2}}}$