My Math Forum Diff Eqn Product

 Calculus Calculus Math Forum

 September 13th, 2013, 11:54 PM #1 Member   Joined: Nov 2011 Posts: 72 Thanks: 0 Diff Eqn Product Differenciate Y = Ln((X^2)(Sq root of 1-X^2)) to get (X/2) -( X/(1-X^2))
 September 14th, 2013, 05:29 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,950 Thanks: 1141 Math Focus: Elementary mathematics and beyond Re: Diff Eqn Product $\frac{d}{dx}\ln$$x^2\sqrt{1\,-\,x^2}$$\,=\,\frac{2\,-\,3x^2}{x$$1\,-\,x^2$$}$
September 14th, 2013, 08:27 AM   #3
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

Re: Diff Eqn Product

Hello, hatchelhoff!

$\text{You have a typo . . . }
\;\;\text{It should be }\frac{2}{x}\,\text{ instead of }\frac{x}{2}.$

Quote:
 $\text{Differentiate: }\:y\:=\:\ln\left(x^2\sqrt{1\,-\,x^2}\right)\,\text{ to get: }\:y#39; \:=\:\frac{2}{x}\,-\,\frac{x}{1-x^2}$

$\text{W\!e have: }\;y \;=\;\ln(x^2) \,+\,\ln(1\,-\,x^2)^{\frac{1}{2}}$

[color=beige]. . . . . . . . .[/color]$y \;=\;2\cdot\ln(x)\,+\,\frac{1}{2}\cdot\ln(1\,-\,x^2}$

$\text{Hence: }\;y' \;=\;2\cdot\frac{1}{x}\,+\,\frac{1}{2}\cdot\frac{-2x}{1\,-\,x^2}$

[color=beige]. . . . . . [/color]$y' \;=\;\frac{2}{x}\,-\,\frac{x}{1\,-\,x^2}$

 September 14th, 2013, 11:27 AM #4 Member   Joined: Nov 2011 Posts: 72 Thanks: 0 Re: Diff Eqn Product Great thanks soroban

 Tags diff, eqn, product

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Niko Bellic Calculus 2 July 8th, 2013 10:01 AM otaniyul Linear Algebra 0 October 30th, 2009 06:40 PM amaciose Calculus 5 October 12th, 2008 04:10 AM keeps14soccer Calculus 2 September 14th, 2008 04:56 AM kaanmm Calculus 4 June 7th, 2008 06:46 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top