Hello from MrBillDoesMath!
Answer:
#10
y' = (2x - 3x^2)/(2y-1)
#14
y' = y^2/x^2
Discussion:
#10
x^2 + y = x^3 + y^2 (*)
Let y' be the first derivative of y with respect to x (i.e. dy/dx). Differentiating (*) gives
2x + y' = 3x^2 + 2yy' => subtract y' from both sides
2x = 3x^2 + 2yy' - y' => factor y' from the rhs
2x = 3x^2 + y' (2y - 1) => subtract 3x^2 from both sides
2x - 3x^2 = y'(2y-1) => divide both sides by (2y-1)
(2x - 3x^2)/(2y-1) = y'
#14
(1/x) + (1/y) = 1 (**)
Differentiating (**) gives
(-1/x^2) + ( -1 y^(-2)y') = 0 => as 1/x = x^(-1) and 1/y = y^(-1)
(-1/x^2) - y^(-2)y' = 0 => add y^(-2)y' to both sides
(-1/x^2) = y^(-2)y' => divide both sides by y^(-2)
(1/x^2)/ (y^(-2)) = y' => 1/(y^(-2)) = y^2
y^2/x^2 = y'
Thank you,
MrB
