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

Question

Christopher Agresta · Accepted Answer

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

Charles Anctil · Answer

#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)#

Charles Arocha · Answer

undefined To implicitly differentiate $1 - xy = x - y$, follow these steps:

1. Differentiate each term with respect to $x$.
2. Apply the chain rule where necessary.
3. Solve for $\frac{dy}{dx}$ after differentiation.

The steps are as follows:

$$
\begin{align*}
\frac{d}{dx}(1 - xy) &= \frac{d}{dx}(x - y) \
0 - \frac{d}{dx}(xy) &= 1 - \frac{dy}{dx} \
- y \frac{dx}{dx} - x \frac{dy}{dx} &= 1 - \frac{dy}{dx} \
- y - x \frac{dy}{dx} &= 1 - \frac{dy}{dx} \
-x \frac{dy}{dx} + \frac{dy}{dx} &= 1 + y \
\left(-x + 1\right) \frac{dy}{dx} &= 1 + y \
\frac{dy}{dx} &= \frac{1 + y}{1 - x}
\end{align*}
$$

So, the implicit derivative of $1 - xy = x - y$ with respect to $x$ is $\frac{1 + y}{1 - x}$.