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)#