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December 19, 2014

December 19, 2014

Posted by **cynthia** on Tuesday, February 28, 2012 at 1:57am.

2. f(x)=√4x^2+3x+4, find f'(x).

i got stuck.

f(x)=√5x+6=(5x+6)^1/2

f'(x)= 1/2(5x+6)^-3/2*(5).

- Calculus AP need help! -
**Steve**, Tuesday, February 28, 2012 at 11:09amAlmost. How did you get from ^1/2 to ^-3/2? 1/2 - 1 = -1/2.

f = (5x+6)^(1/2)

use the power rule and chain rule:

f' = 1/2 * (5x+6)^(-1/2) * 5 = 5/(2√(5x+6))

f = √(4x^2+3x+4)^(1/2)

f' = 1/2 * (4x^2+3x+4)^-1/2) * (8x+3) = (8x+3)/(2√(4x^2+3x+4))

- Calculus AP need help! -
**cynthia**, Tuesday, February 28, 2012 at 2:23pmoops.. i thought 1/2 to -3/2. now i understand. thanks

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