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November 30th, 2017, 07:00 PM   #1
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Second derivative help

I figured out the first derivative, and I know you have to use the quotient rule to get the second, but I get lost somewhere during the quotient rule.
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November 30th, 2017, 07:13 PM   #2
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$y = x \sqrt{3-x}$

$y^\prime = \sqrt{3-x} - \dfrac{x}{2\sqrt{3-x}}$

$y^{\prime \prime} = -\dfrac{1}{2\sqrt{3-x}} -

\dfrac 1 2 \left(\dfrac{(1)(\sqrt{3-x})-(x)\left(\dfrac{-1}{2\sqrt{3-x}}\right)}{3-x}\right)$

and I leave simplifying the algebra to you
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December 9th, 2017, 05:45 AM   #3
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If you don't like to use the quotient rule, or as a check, you can write the denominator with a negative exponent and use the product rule.

For example, we can write $\displaystyle \frac{x}{2\sqrt{3- x}}$ as $\frac{1}{2}x(3- x)^{-1/2}$.

By the product rule, the derivative is $\frac{1}{2}\left((3- x)^{-1/2}+ x(3- x)^{-3/2}(-1)\right)$.

If you want to put the answer back into the same form as the problem was given write those negative exponents as fractions with common denominator $(3- x)^{3/2}$: $\frac{(3- x)}{2(3- x)^{3/2}}- \frac{x}{2(3- x)^{3/2}}= \frac{3- 2x}{2(3- x)^{3/2}}= \frac{3- 2x}{2(3- x)\sqrt{3- x}}$
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December 25th, 2017, 07:47 AM   #4
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Here is a proposed solution to your problem
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