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March 29, 2015

March 29, 2015

Posted by **sh** on Tuesday, November 10, 2009 at 5:43pm.

y=¡î(x©÷+1)

y'=x(x©÷+1)^(-1/2)

y"=[1(x©÷+1)^(-1/2)]-[(1/2)[(x©÷+1)^(-3/2)](2x)(x)]

y"=[(x©÷+1)^(-1/2)]-[(x©÷)(x©÷+1)^(-3/2)]

Is the second derivative right?

- Calculus -
**sh**, Tuesday, November 10, 2009 at 5:54pmy=root(x^2+1)

y'=x(x^2+1)^(-1/2)

y"=[1(x^2+1)^(-1/2)]-[(1/2)[(x^2+1)^(-3/2)](2x)(x)]

y"=[(x^2+1)^(-1/2)]-[(x^2)(x^2+1)^(-3/2)]

- Calculus -
**Reiny**, Tuesday, November 10, 2009 at 7:04pmI agree

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