Find the slope of tangent to the curve x^2 + y^2 = -xy+4y+11 at (3,-1)
solution:
d/dx(x^2 +y^2) = d/dx(-xy+4y+11)
2x + 2y(dy/dx) = -x(dy/dx) -y +4(dy/dx)
dy/dx = (-2x-y)/(x+2y-4)
Next step:substitude x=3,y=-1 into dy/dx.After that,the slope of tangent at (3,-1) will be found
I want to ask why dy/dx is the slope of tangent of the equation.
x^2 + y^2 = -xy+4y+11 is an implicit function,which cannot be expressed in terms of x,so why differentiating y with respect to x will give the slope of tangent?