Question

Implicit Differentiation

Find the 2nd derivative of the following:

1) x=3+√x^2 + y^2

2) (x^2 + y^2)^3 = 8x^2y^2

3) x^2(y-x)^3 = 9

Answer 1

1)

$x=3+\sqrt{x^2+y^2}$

$1=\frac{1}{2\sqrt{x^2+y^2}}(2x+2yy')$

$y'=\frac{\sqrt{x^2+y^2}-x}{y}$

$yy'=\sqrt{x^2+y^2}-x$

$(y')^2+yy''=\frac{x+yy'}{\sqrt{x^2+y^2}}-1$

from two previous equations

$(y')^2+yy''=0$

$y''=-\frac{(y')^2}{y}=-\frac{(\sqrt{x^2+y^2}-x)^2}{y^3}$

2)

$(x^2+y^2)^3=8x^2y^2$

$3(x^2+y^2)^2(2x+2yy')=16xy^2+16x^2yy'$

$6(x^2+y^2)(2x+2yy')^2+3(x^2+y^2)^2(2+2(y')^2+2yy'')=$

$=16y^2+16x*2yy'+32xyy'+16x^2(y')^2+16x^2yy''$

$3x(x^2+y^2)^2-8xy^2=(8x^2y-3y(x^2+y^2)^2)y'$

$y'=\frac{3x(x^2+y^2)^2-8xy^2}{8x^2y-3y(x^2+y^2)^2}$

3)

$x^2(y-x)^3=9$

$2x(y-x)^3+x^2*3(y-x)^2(y'-1)=0$

$y'-1=\frac{-2(y-x)}{3x}=-\frac{2y}{3x}+\frac{2}{3}$

$y'=-\frac{2y}{3x}+\frac{5}{3}$

$y''=-\frac{2y'*3x-2y*3}{9x^2}=\frac{2y-2xy'}{3x^2}=$

$=\frac{2y}{3x^2}-\frac{2}{3x}(-\frac{2y}{3x}+\frac{5}{3})=$

$=\frac{10y}{9x^2}-\frac{10}{9x}=\frac{10(y-x)}{9x^2}$