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July 31, 2014

July 31, 2014

Posted by **dt** on Sunday, December 2, 2012 at 12:08pm.

Can someone help me find the first and second derivatives please!

- calculus quick question -
**Reiny**, Sunday, December 2, 2012 at 12:42pmwrite it as

g(x) = (1 - 4x^2)^(1/2)

g ' (x) = (1/2)(1 - 4x^2)^(-1/2) (-8x)

= -4x(1 - 4x^2)^(-1/2)

now by the product rule

g '' (x)

= (-4x)(-1/2)(1-4x^2)^(-3/2) (-8x)- 4(1 - 4x^2)^(-1/2)

= (-16x^2)(1-4x^2)^(-3/2) - 4(1 - 4x^2)^(-1/2)

= -4(1 - 4x^2)^(-3/2) [ 4x^2 - (1-4x^2) ]

= -4(1-4x^2)^(-3/2) (4x^2 + x - 2)

= -4( 1-4x^2)^(-3/2)

or

= -4/(1 - 4x^2)^(3/2)

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