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 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 ...

$\dfrac{dy}{dx} = \dfrac{1}{2} \cdot \dfrac{12x^2+6x+2}{4x^3+3x^2+2x} = \dfrac{6x^2 + 3x + 1}{4x^2+3x^2+2x}$

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