Need to find dy/dx explicitly i.e. in the form dy/dx = g(x,y)?

Question

If: x^2 + 2y^2 -xy + y = 10
and hence determine the gradient of the tangent to the curve represented by the equation at point (0,2)

Answer 1

I'm not sure if a mathematician would like it, but a physicist would just go and differentiate the equation with respect to x, getting
2x+4y (dy/dx) - y - x(dy/dx)+ (dy/dx)=0
sorting it out
(dy/dx)(4y-x+1)=y-2x
(dy/dx)=(y-2x)/(4y-x+1)
(dy/dx)|(0,2)=2/9

Answer 2

Let D = dy/dx.

Let x^2 + 2y^2 -xy + y = 10 ----(1)

Differentiating (1) with respect to x, gives

2x + 4yD -( xD + y) + D = 0.

==> D = dy/dx = y/(x+ 4y +1) ----(2)

Gradient at (0,2) can be obtained by substituting x = 0 and y = 2 in (2), which gives 2/9 = 0.2222...

Answer 3

Differentiate with respect to x:-

(d/dx)(x² + 2y² - xy + y) = 0

2x + 4y (dy/dx) - (y + x dy/dx) + dy/dx = 0

(4y - x + 1)dy/dx = y - 2x

dy/dx = (y - 2x) / (4y - x + 1)

Gradient of tangent at (0,2) = 2/9

Answer 4

that would be 2X +4Y(dy/dx) -y-x(dy/dx)+dy/dx=0 (differentiating wrt x)

simplify and solve with dy/dx on one side of the equation.then substitute the point (0,2) to get the gradient at that point.

Answer 5

x^2+2y^2-xy+y=10
Differentiate d/dx
2x+4y(dy/dx)-y-x(dy/dx)+dy/dx=0
dy/dx=(y-2x)/(4y-x+1)

at point(0,2)
x=0, y =2
dy/dx=(2-2(0))/(4(2)-0+1)
dy/dx=2/9
gradient =2/9

Answer 6

u=x v=y u'=1 v'=dy/dx
d(xy)/dx=xdy/dx+y

2x+4ydy/dx-xdy/dx-y+dy/dx=0
dy/dx(4y-x+1)=y-2x
dy/dx=(y-2x)/(4y-x+1)
dy/dx at (0,2)= 2/(8+1)=2/9