the derivative of $ {1\over x} + {1\over y} = 1$

Question

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?

Answer 1

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.

Answer 2

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 $\frac{y}{x}\cdot \frac{y}{x}.$

Answer 3

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.