using the implicit differentiation, find y' (or dy/dx) of y^4 - 3xy + x = y^3
4y^3 - 3y - 3x + 1 = 3y^2
dy/dx(4y^3 - 3y - 3x) + 1 = 3y^2
dy/dx(4y^3 - 3y - 3x) = 3y^2 - 1
dy/dx = (3y^2 - 1)/(4y^3 - 3y - 3x)
what did i do wrong? =/
the answer is supposed to be (3y - 1)/(4y^3 - 3y^2 - 3x)
Hello, Jeph,
have a look here: http://www.mathhelpforum.com/math-he...rial-latex.pdf
EB