the derivative of $ {1\over x} + {1\over y} = 1$ I am finding the derivative of this equation, using the implicit differentiation in term of x.
$$ {1\over x} + {1\over y} = 1$$
Here is what I did.
$$ {1\over x} + {1\over y} = 1$$
$$ x^{-1} + y^{-1} = 1$$
$$$$
$$ D_x  [x^{-1} + y^{-1}] = D_x [1]$$
$$-x^{-2} - y^{-2}\cdot D_x y = 0 $$
$$-y^{-2} \cdot D_x y = x^{-2}$$
$$D_x y = - {y^{2} \over x^2}$$
My derivative is
$$D_x y = - {y^{2} \over x^2}$$
Is this correct?
The answer book says that it is:
$$d_x y = {(y-1) \over (x-1)}$$
Did I do anything wrong? Or is it just the answer book typo? Or can I rewrite this?
 A: Both answers are correct.
$${1\over x} + {1\over y} = 1 \\ \implies y+x=xy \\ \implies y = \frac{x}{x-1}\quad \text{and}\quad x=\frac{y}{y-1}$$
Plug one of those each into your answer to get $$\require{cancel}D_xy = -\frac{y^2}{x^2} = -\frac{y}{x}\frac{y}{x} = -\frac{\color{red}{\cancel {\color{black}y}}}{\color{red}{\cancel {\color{black}x}}}\left(\frac{{\color{red}{\cancel {\color{black}x}}}(y-1)}{{\color{red}{\cancel {\color{black}y}}}(x-1)}\right) = -\frac{y-1}{x-1}$$
Note that you would have gotten this answer immediately if you had implicitly differentiated $y+x=xy$ instead of $\frac1x + \frac 1y = 1$.  So that's likely what the answer key writer did.
A: Your answer is correct. You can rewrite it.
Hint: 
$\frac{1}{x}+\frac{1}{y}=1 \Rightarrow xy=y+x \Rightarrow y=x(y-1)$ and $x=y(x-1).$
Now rewrite $\frac{y^2}{x^2}$ as $\frac{y}{x}\cdot \frac{y}{x}.$
A: Both answers are same(correct) you just have to substitute y as a function of x to see this in your derivative. But the function is undefined when any of y or x is 0.
