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)#