How do you implicitly differentiation 1-xy = x-y?

2 Answers
Apr 26, 2015

You have to use the implicit function theorem (aka the Dini's theorem)

The implicit equation is F(x,y)=1-x-xy+y=0, F is a smooth function (it's a polynomial), so we can use Dini's theorem

Notation: F_x={dF}/dx

Dini's theorem says that in every point (x,y) such that F_y!=0, we have a neighbourhood where y=f(x) with f smooth and f'=-F_x/F_y

So, F_y=-x+1 => \forall (x,y) != (1,y)
f'(x)=-(-y-1)/(-x+1)=(1+f(x))/(1-x)

Now we invert, F_x= -y-1 => forall (x,y)!=(x,-1)
x=g(y) and g'(y)=-(1-x)/(-y-1)=(1-g(y))/(1+y)

So in (1,-1) the function is not differentiable, elsewhere is differentiable (it's C^1).

Notice that we don't have an "explicit" formula for the derivatives

Apr 26, 2015

1-xy=x-y

When we differentiate, we'll need the product rule for the second term on the left.
I use the order: (fg)' = f'g+fg' for this term we'll have f=-x and g=y

d/(dx)(1-xy) = d/(dx)(x-y)

0 +d/dx(-xy) = 1 - dy/dx

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

-y -x dy/dx = 1 - dy/dx

So

dy/dx - x dy/dx = 1+y

And

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