How do you find the second derivative of #x^2+y^2=1#?

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Answer 1 · 2018-03-09T05:27:00

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

differentiating #x^2+y^2=1#, we get

#2x+2y(dy)/(dx)=0# i.e. #(dy)/(dx)=-x/y#

and differentiating it further

#2+2[(dy)/(dx)(dy)/(dx)+y*(d^2y)/(dx^2)}=0#

or #2+2[(dy)/(dx)]^2+2y(d^2y)/(dx^2)=0#

or #2+2x^2/y^2+2y(d^2y)/(dx^2)=0#

or #2y(d^2y)/(dx^2)=-2-2x^2/y^2#

or #(d^2y)/(dx^2)=-2/(2y)-2x^2/y^2*1/(2y)#

= #-1/y-x^2/y^3#