Question 1204134
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Through a bit of trial-and-error, it appears the given equation should be {{{x^2+y^2 = 2y}}}.
Please be more careful when entering the equation.


The 1st derivative would be: {{{dy/dx = x/(1-y)}}}
The scratch work is shown in the next section.


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{{{x^2+y^2 = 2y}}}


{{{expr(d/(dx))( x^2 + y^2 ) = expr(d/(dx))(2y)}}}


{{{expr(d/(dx))( x^2 ) + expr(d/(dx))( y^2 ) = expr(d/(dx))(2y)}}}


{{{2x + 2y*expr((dy)/(dx)) = 2*expr((dy)/(dx))}}}


{{{2x = 2*expr((dy)/(dx)) -  2y*expr((dy)/(dx))}}}


{{{2x = (2 -  2y)*expr((dy)/(dx))}}}


{{{(dy)/(dx) = (2x)/(2-2y)}}}


{{{(dy)/(dx) = (2x)/(2(1-y))}}}


{{{(dy)/(dx) = x/(1-y)}}}
This will be used later.


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We then apply another round of implicit differentiation with respect to x.


{{{(dy)/(dx) = x/(1-y)}}}


{{{(d^2y)/(dx^2) = expr(d/(dx))(x/(1-y))}}}


{{{(d^2y)/(dx^2) = ( expr(d/(dx))(x)*(1-y) - (x)*expr(d/(dx))(1-y) )/( (1-y)^2 )}}} Quotient Rule


{{{(d^2y)/(dx^2) = ( 1*(1-y) + x*expr((dy)/(dx)) )/( (1-y)^2 )}}} 


{{{(d^2y)/(dx^2) = ( 1-y + x*expr((dy)/(dx)) )/( (1-y)^2 )}}} 




From here we plug in the first derivative

{{{(d^2y)/(dx^2) = ( 1-y + x*expr((dy)/(dx)) )/( (1-y)^2 )}}} 


{{{(d^2y)/(dx^2) = ( 1-y + x*(x/(1-y)) )/( (1-y)^2 )}}} 


{{{(d^2y)/(dx^2) = ( ((1-y)^2)/(1-y) + (x^2)/(1-y))/( (1-y)^2 )}}}

 
{{{(d^2y)/(dx^2) = ( ((1-y)^2+x^2)/(1-y))/( (1-y)^2 )}}} 


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


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


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

 
{{{(d^2y)/(dx^2) = (1-2y+highlight(x^2+y^2))/( (1-y)^3 )}}} 


{{{(d^2y)/(dx^2) = (1-2y+highlight(2y))/( (1-y)^3 )}}} Use the equation x^2+y^2 = 2y


{{{(d^2y)/(dx^2) = 1/( (1-y)^3 )}}} 



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Here's another way to do it 


{{{(dy)/(dx) = x/(1-y)}}}


{{{expr((dy)/(dx))*(1-y) = x}}}


{{{expr((d^2y)/(dx^2))*(1-y) + expr((dy)/(dx))*expr(d/(dx))(1-y) = expr(d/(dx))(x)}}} Product Rule


{{{expr((d^2y)/(dx^2))*(1-y) + ((dy)/(dx))*(-(dy)/(dx))= 1}}}


{{{expr((d^2y)/(dx^2))*(1-y) - ((dy)/(dx))^2 = 1}}}


{{{expr((d^2y)/(dx^2))*(1-y) - (x/(1-y))^2 = 1}}} Plug in dy/dx = x/(1-y)


{{{expr((d^2y)/(dx^2))*(1-y) = 1 + (x/(1-y))^2}}}


{{{expr((d^2y)/(dx^2))*(1-y) = ((1-y)^2)/((1-y)^2) + (x^2)/((1-y)^2)}}}


{{{expr((d^2y)/(dx^2))*(1-y) = ((1-y)^2+x^2)/((1-y)^2))}}}


{{{(d^2y)/(dx^2) = (((1-y)^2+x^2)/((1-y)^2))*(1/(1-y))}}}


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


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


{{{(d^2y)/(dx^2) = (1-2y+2y)/((1-y)^3))}}} Replace x^2+y^2 with 2y


{{{(d^2y)/(dx^2) = 1/((1-y)^3))}}}
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