# How do you use implicit differentiation to find the slope of the curve given 1/(x+1)+1/(y+1)=1 at (1,1)?

Nov 13, 2016

The curve has a slope of $- 1$.

#### Explanation:

$\frac{0 \left(x + 1\right) - 1 \times 1}{x + 1} ^ 2 + \frac{0 \left(y + 1\right) - 1 \times 1 \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{y + 1} ^ 2 = 0$

$- \frac{1}{x + 1} ^ 2 - \frac{1 \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{y + 1} ^ 2 = 0$

Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$.

$- \frac{{\left(y + 1\right)}^{2} + {\left(x + 1\right)}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}}{{\left(x + 1\right)}^{2} {\left(y + 1\right)}^{2}} = 0$

$- {\left(y + 1\right)}^{2} - {\left(x + 1\right)}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$- {\left(x + 1\right)}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = {\left(y + 1\right)}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {\left(y + 1\right)}^{2} / {\left(x + 1\right)}^{2}$

The slope, or rate of change of the curve is

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {\left(1 + 1\right)}^{2} / {\left(1 + 1\right)}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{4}{4}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 1$

Nov 13, 2016

$- 1$

#### Explanation:

$\frac{1}{x + 1} + \frac{1}{y + 1} = \frac{x + y + 2}{x y + x + y + 1} = 1$

then

$x + y + 2 = x y + x + y + 1$ simplifying

$x y = 1$ now

$\frac{d}{\mathrm{dx}} \left(x y - 1\right) = y + x y ' = 0$ so

$y ' = - \frac{1}{x} ^ 2$ so $y ' \left(1\right) = - 1$

NOTE. What is de difference between $\frac{1}{x + 1} + \frac{1}{y + 1} = 1$ and $x y = 1$?

Attached two graphics. The first is associated to $\frac{1}{x + 1} + \frac{1}{y + 1} = 1$

and the second is associated to $x y = 1$

This second graphic is clean, without the singularities in $x = - 1$ and $y = - 1$. If the desired results are asked apart from the singularities, this second option offers a terse environment to the problem.