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June 16th, 2019, 09:39 AM   #1
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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).
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June 16th, 2019, 11:09 AM   #2
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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)
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June 16th, 2019, 12:45 PM   #3
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Quote:
Originally Posted by Khoxy View Post
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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