# What is the slope of the tangent line of e^x-1/(x+y)^2= C , where C is an arbitrary constant, at (3,1)?

##### 1 Answer
May 6, 2018

$- 32 {e}^{3} - 1$

#### Explanation:

do implicit differentiation with respect to x:

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

${e}^{x} + \frac{2}{x + y} ^ 3 \cdot \frac{d}{\mathrm{dx}} \left(x + y\right) = 0$

${e}^{x} + \frac{2}{x + y} ^ 3 \cdot \left(1 + \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

isolate $\frac{\mathrm{dy}}{\mathrm{dx}}$:

$\frac{2}{x + y} ^ 3 \cdot \left(1 + \frac{\mathrm{dy}}{\mathrm{dx}}\right) = - {e}^{x}$

$\left(1 + \frac{\mathrm{dy}}{\mathrm{dx}}\right) = - {e}^{x} \cdot {\left(x + y\right)}^{3} / 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {e}^{x} \cdot {\left(x + y\right)}^{3} / 2 - 1$

plug in point $\left(3 , 1\right)$ to find slope of tangent line:

$\frac{\mathrm{dy}}{\mathrm{dx}}$ at $\left(3 , 1\right) = - {e}^{3} \cdot {\left(3 + 1\right)}^{3} / 2 - 1$

$\frac{\mathrm{dy}}{\mathrm{dx}}$ at $\left(3 , 1\right) = - 32 {e}^{3} - 1$