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 Calculus Calculus Math Forum

 June 16th, 2019, 09:39 AM #1 Newbie   Joined: Apr 2019 From: Malawi Posts: 19 Thanks: 0 Math Focus: Trigonometry and calculus Derivatives Differentiate the function below with respect to x. y = In (√4x^3 + 3x^2 + 2x). May somebody help please.( The whole expression is in square root). June 16th, 2019, 11:09 AM #2 Senior Member   Joined: Mar 2019 From: iran Posts: 318 Thanks: 14 y = ln(√(4x^3 + 3x^2 + 2x)) y' = 1/√(4x^3 + 3x^2 + 2x) × 1/(2√(4x^3 + 3x^2 + 2x)) × (12x^2 + 6x + 2) = (6x^2 + 3x + 1)/(4x^3 + 3x^2 + 2x) Thanks from topsquark June 16th, 2019, 12:45 PM   #3
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Quote:
 Originally Posted by Khoxy Differentiate the function below with respect to x. y = In (√4x^3 + 3x^2 + 2x). May somebody help please.( The whole expression is in square root).
power property of logs ...

$y = \dfrac{1}{2} \ln(4x^3+3x^2+2x)$

simplifying log expressions first makes the derivative easier to determine ...

$\dfrac{dy}{dx} = \dfrac{1}{2} \cdot \dfrac{12x^2+6x+2}{4x^3+3x^2+2x} = \dfrac{6x^2 + 3x + 1}{4x^2+3x^2+2x}$ Tags derivatives 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 lolly Calculus 4 February 11th, 2015 10:11 AM johngalt47 Calculus 2 November 23rd, 2013 11:44 AM texaslonghorn15 Calculus 1 July 1st, 2013 12:38 PM wulfgarpro Algebra 15 May 14th, 2010 03:23 PM Calavan11 Calculus 2 December 11th, 2009 04:42 AM

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