Question

Find d^2y/dx^2 at the following point using implicit differentiation:

x^4 + y^4 = 16, (0,2)

Answer 1

\[x^4+y^4=16\]\[4x^3+4y^3y'=0\quad\Rightarrow\quad y'=-\frac{x^3}{y^3}\]\[y''=-\frac{3x^2y^3-3x^3y^2y'}{y^6}=\frac{3x^2(xy'-y)}{y^4}\]\[y''(0,2)=0\]