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 Calculus Calculus Math Forum

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.
Attached Images IMG_20171130_191614569_HDR.jpg (87.6 KB, 14 views) November 30th, 2017, 07:13 PM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,585 Thanks: 1430 $y = x \sqrt{3-x}$ $y^\prime = \sqrt{3-x} - \dfrac{x}{2\sqrt{3-x}}$ $y^{\prime \prime} = -\dfrac{1}{2\sqrt{3-x}} - \dfrac 1 2 \left(\dfrac{(1)(\sqrt{3-x})-(x)\left(\dfrac{-1}{2\sqrt{3-x}}\right)}{3-x}\right)$ and I leave simplifying the algebra to you Thanks from Jonnykirtland 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
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Here is a proposed solution to your problem
Attached Images Solution.jpg (88.3 KB, 5 views) Tags derivative Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post daltyboy11 Calculus 2 July 10th, 2014 06:57 PM lackofimagination Calculus 1 July 6th, 2014 08:05 PM azerty Calculus 6 April 10th, 2013 10:10 AM supaman5 Linear Algebra 0 November 26th, 2012 10:14 AM rgarcia128 Calculus 4 September 24th, 2011 05:07 PM

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