Hi there!

can i also get some with this one? #10 and #14. show your work. i'll do the rest : (

Hello from MrBillDoesMath!

#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

how is that middle school

step-by-step explanation:

step-by-step explanation:

i think you should think about the question and not look online.