CALCULUS HELP URGENT!!
posted by Anonymous .
find y' if arctan(xy)=6+x^(4)y
here is one way:
xy = tan(6 + x^4y)
xdy/dx + y = sec^2 (6 + x^4y) (x^4 dy/dx + 4x^3y_
= x^4 dy/dx sec^2 (6+x^4y) + 4x^3y sec^2(6+x^4y)
dy/dx(x - x^4sec^2 (6+x^4y) ) = 4x^3 sec^2 (6+x^4y) - y
dy/dx = (4x^3 sec^2 (6+x^4y) - y)/(x - x^4sec^2 (6+x^4y) )
check my algebra, I should have written it out on paper first.
Or, more directly,
1/(1+x^2y^2) (y+xy') = 4x^3y+x^4y'
y+xy' = 4x^3y (1+x^2y^2) + x^4(1+x^2y^2)y'
y' = (4x^3y + 4x^5y^3-y)/(x-x^4-x^6y^2)
This can also be obtained (after restoring the y that was accidentally dropped) from Reiny's result by recalling the sec^2 = 1+tan^2 = 1+x^2y^2