November 30th, 2017, 07:00 PM  #1 
Newbie Joined: Nov 2017 From: California Posts: 1 Thanks: 0  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.

November 30th, 2017, 07:13 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,494 Thanks: 1369 
$y = x \sqrt{3x}$ $y^\prime = \sqrt{3x}  \dfrac{x}{2\sqrt{3x}}$ $y^{\prime \prime} = \dfrac{1}{2\sqrt{3x}}  \dfrac 1 2 \left(\dfrac{(1)(\sqrt{3x})(x)\left(\dfrac{1}{2\sqrt{3x}}\right)}{3x}\right)$ and I leave simplifying the algebra to you 
December 9th, 2017, 05:45 AM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
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 $(3 x)^{3/2}$: $\frac{(3 x)}{2(3 x)^{3/2}} \frac{x}{2(3 x)^{3/2}}= \frac{3 2x}{2(3 x)^{3/2}}= \frac{3 2x}{2(3 x)\sqrt{3 x}}$ 
December 25th, 2017, 07:47 AM  #4 
Newbie Joined: Dec 2017 From: Netherlands Posts: 21 Thanks: 0 
Here is a proposed solution to your problem


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