How do you find the second derivative of $-ln(x-(x^2+1)^(1/2))$ ?

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Answer 1 · 2017-05-07T19:05:31

$(-x)/(x^2+1)^(3/2)$

$y=-ln(x-(x^2+1)^(1/2))$

To find the first derivative:

$dy/dx=(-1)/(x-(x^2+1)^(1/2))[d/dx(x-(x^2+1)^(1/2))]$

Using the chain rule again:

$dy/dx=(-1)/(x-(x^2+1)^(1/2))(1-1/2(x^2+1)^(-1/2)(2x))$

$dy/dx=(-1)/(x-(x^2+1)^(1/2))(1-x/(x^2+1)^(1/2))$

$dy/dx=(-1)/(x-(x^2+1)^(1/2))(((x^2+1)^(1/2)-x)/(x^2+1)^(1/2))$

$dy/dx=1/((x^2+1)^(1/2)-x)(((x^2+1)^(1/2)-x)/(x^2+1)^(1/2))$

$dy/dx=1/(x^2+1)^(1/2)$

$dy/dx=(x^2+1)^(-1/2)$

Then:

$(d^2y)/dx^2=-1/2(x^2+1)^(-3/2)(2x)$

$(d^2y)/dx^2=-(x^2+1)^(-3/2)(x)$

$(d^2y)/dx^2=(-x)/(x^2+1)^(3/2)$

How do you find the second derivative of -ln(x-(x^2+1)^(1/2))-ln(x-(x^2+1)^(1/2))?