June 16th, 2019, 09:39 AM  #1 
Newbie Joined: Apr 2019 From: Malawi Posts: 13 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: 317 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) 
June 16th, 2019, 12:45 PM  #3  
Math Team Joined: Jul 2011 From: Texas Posts: 2,980 Thanks: 1573  Quote:
$y = \dfrac{1}{2} \ln(4x^3+3x^2+2x)$ simplifying log expressions first makes the derivative easier to determine ... $\dfrac{dy}{dx} = \dfrac{1}{2} \cdot \dfrac{12x^2+6x+2}{4x^3+3x^2+2x} = \dfrac{6x^2 + 3x + 1}{4x^2+3x^2+2x}$  

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