# Use implicit differentiation to find `dy/dx` and then `(d^2y)/(dx^2)` . `a) x^2+y^2=1 ` `b) y^2+2y=2x-1` ``

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hala718 | Certified Educator

`a) x^2 + y^2 = 1`

`==> 2x + 2yy' = 0`

`==> 2yy' = -2x`

`==> y' = (-2x)/(2y) = (-x)/y`

`==> yy' = -x`

`==> y'y' + yy'' = -1`

`==> y'^2 + yy'' = -1`

`==> yy'' = -1 -y'^2`

`==> y'' = -(1+y'^2)/y`

`==> y'' = -(1+(-x/y)^2)/y = -(1+x^2/y^2)/y = -(y^2+x^2)/y^3`

`==> y'' = -1/y - x^2/y^3`

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b)`y^2 + 2y= 2x-1`

`==> 2yy' + 2y' = 2 `

`==> y'(2y+2) = 2`

`==> y' = 2/(2y+2)= 1/(y+1)`

`==> (y+1)y' = 1`

`==> yy' + y' = 1`

`==> y'y' +yy'' + y'' = 0`

`==> yy'' + y'' = -y'^2`

`==> y''(y+1) = -y'^2`

`==> y'' = -(y'^2)/(y+1)`

`==> y'' = -(1/(y+1)^2)/(y+1) = -1/(y+1)^3`

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