My Math Forum

My Math Forum (http://mymathforum.com/math-forums.php)
-   Calculus (http://mymathforum.com/calculus/)
-   -   Derivatives (http://mymathforum.com/calculus/346611-derivatives.html)

Khoxy June 16th, 2019 09:39 AM

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).

youngmath June 16th, 2019 11:09 AM

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)

skeeter June 16th, 2019 12:45 PM

Quote:

Originally Posted by Khoxy (Post 610797)
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 ...

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


All times are GMT -8. The time now is 09:13 AM.

Copyright © 2019 My Math Forum. All rights reserved.