Question

Using implicit differentiation, sqrt(x+y) = 1 + x^2y^2

Answer 1

\[d/dx[(x+y)^{1/2}]=d/dx[1+x^2y^2]\]
\[1/2(x+y)^{-1/2}(1+dy/dx)=2xy^2+2x^2y dy/dx\]
\[1/2(x+y)^{-1/2}+1/2(x+y)^{-1/2}dy/dx=2xy^2+2x^2y dy/dx\]
\[[1/2(x+y)^{-1/2}-2x^2y]dy/dx=2xy^2-1/2(x+y)^{-1/2}\]
\[dy/dx=[2xy^2-1/2(x+y)^{-1/2}]/[1/2(x+y)^{-1/2}-2x^2y]\]
\[dy/dx=(4x^2(x+y)^{1/2}-1)/(2(x+y)^{1/2})\]
\[\div (1-4x^2y(x+y)^{1/2})/(2(x+y)^{1/2})\]
\[dy/dx=[4xy^2(x+y)^{1/2}-1]/[1-4x^2y(x+y)^{1/2}]\]

Answer 2

nadeem how did you use latex , so pretty

Answer 3

lol.... I used a similar program to type up my labs for physics so I know most of the shortcuts by memory

Answer 4

Ah, beautiful =) Wish I knew all the shortcuts..