Question

find the second derivative of x^2 - y^2= 1

Answer 1

2x-2yy'=0
solve for y'
2yy'=2x
y'=2x/[2y]=x/y
so
y''=[(x)'(y)-(x)(y)']/y^2
y''=[1y-xy']/y^2
y''=[y-xy']/y^2
but y'=x/y
y''=[y-x(x/y)]/y^2
y''=[y-x^2/y]/y^2 lets get rid of the double fraction
y''=[y^2-x^2]/y^3 after multiply both top and bottom by y
y''=-[x^2-y^2]/y^3 but omg x^2-y^2=1
so
y''=-1/y^3
and done

Answer 2

thank you so much! :)

Answer 3

np