Question

Need to find the second derivative of
x^4+y^4=64

Answer 1

\[(x^4)'=4x^3 ; (y^4)=4y^3y' ; (64)'=0\]

Answer 2

\[4x^3+4y^3y'=0\]
solve for y'
\[y'=\frac{-x^3}{y^3}\]
Now apply the quotient rule!

Answer 3

i will let you try

Answer 4

ok give me a moment to type it out or draw it out

Answer 5

yes draw and i will correct errors :)