I want to find the second derivative of y in respect to x. y is defined by the equation x^3+y^3=cos[(x^2)*y]?

Question

i'm sure that we will use implicit differentiation; however it gets too messy. I am wondering if there is any shortcut to do this or any simplification while solving it.

Answer 1

It isn't that messy ...

3x^2 + 3y^2 (dy/dx) = - sin(x^2 y) (2x y + x^2 (dy/dx))

Go off to the side and group terms to get
(dy/dx) = u(x,y) / v(x,y).

Then (d^2y/dx^2) = (v (du/dx) - u (dv/dx)) / v^2

The (du/dx) and (dv/dx) terms will contain (dy/dx)'s in them, just replace all of them with the (dy/dx) = u/v that you already know -- or skip the substiutution and just add "... where (dy/dx) = ...". :-)

Dan

Answer 2

Sorry but the equation is not variable separable, so to find the 2nd derivative you have to plow through the mess.
x^3 + y^3 = cos[(x^2)*y]
3x^2dx + 3y^2dy = -sin[(x^2)y] (2xdx + dy)
the differentials ARE variable separable, so you can establish the 1st derivative discretely.
(3y^2 + sin[(x^2)y]) dy = -(3x^2 + sin[(x^2)y]) dx
dy/dx = - (3x^2 + sin[(x^2)y]) / (3y^2 + sin[(x^2)y])

The rest of the fun I leave for you. . .

Answer 3

wow.... wow.... wow.... I'm not trying to be negative or anything but I doubt you are going to get a decent answer to that question on here. Have you tried any math websites? what kind of math is that exactly? you know like algebra stuff like that

Answer 4

F(x,y)=x^3+y^3-cos(xxy)=0
y'= - Fx / Fy = - ( 3xx+sin(xxy)*2xy ) / ( 3yy+sin(xxy)*xx )

İkinci türev için kasamadım valla. Mathematica böyle diyor:
y''=
6*x + 6*y*Dt[y, x]^2 - Cos[x^2*y]*(-2*x*y - x^2*Dt[y, x])*(2*x*y + x^2*Dt[y, x]) + 3*y^2*Dt[y, {x, 2}] -
(-2*y - 4*x*Dt[y, x] - x^2*Dt[y, {x, 2}])*Sin[x^2*y]

burada Dt[y,x] y' ; Dt[y, {x, 2}] y''
Bu ifadede y' i yerine yazıp y'' ları bir tarafa toplayıp çekebilirsin. (kasış!)