If #(dy)/(dx)=sqrt(y^2+1)#, then #(d^2y)/(d^2x)# is what?

Question

I took the derivative of this function and got #y/(sqrt(y^2+1))#. However, the correct answer is #sqrt(2y)#. How?

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Answer 1 · 2018-05-01T06:09:01

#(d^2y)/(dx^2)=y#
(not #sqrt2y#)

As #(dy)/(dx)=sqrt(y^2+1)#

#(d^2y)/(dx^2)=1/(2sqrt(y^2+1)) * 2y * (dy)/(dx)#

= #1/(2sqrt(y^2+1)) * 2y * sqrt(y^2+1)#

= #y#

Answer 2 · 2018-05-01T06:23:57

#=> (d^2y)/(dx^2) = y #

#=> ((dy)/(dx))^2 = y^2 + 1 #

#=> 2 (dy)/(dx) (d^2y)/(dx^2) = 2y (dy)/(dx) #

#=> cancel(2 (dy)/(dx)) (d^2y)/(dx^2) = cancel(2 (dy)/(dx))y #

#=> (d^2y)/(dx^2) = y #