Question

In Problem 10, find $\mathbf{Y}\mathbf{"}$at the origin.

Short answer

The second derivative of the equation $x{e}^y+y{e}^x=0$

Is${\left(\frac{d^{2}y}{d{x}^2}\right)}_{\left(0,0\right)}=4$.

Step-by-step solution

Explanation of Solution

The provided equation or curve is$x{e}^y+y{e}^x=0.$

Step 2: Implicit Differentiation

To find the slopes of tangents to curves that aren't functions, we can utilize implicit differentiation (they fail the vertical line test). Parts ofyare assumed to be functions that satisfy the supplied equation, but yis not a function ofx.

Calculation

Consider the equation of the curve below:

$x{e}^y+y{e}^x=0$

Differentiate the provided above equation of curve with respect to x,

$$\begin{gathered} x{e}^y\frac{dy}{dx}+e^x+y{e}^x+e^x\frac{dy}{dx}=0 \\ \left(x{e}^y+e^x\right)\frac{dy}{dx}+e^x+y{e}^x=0 \end{gathered}$$

Hence,

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

Again, differentiate the provided above equation of a curve,

$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=-\left(e^x+y{e}^x\right) \\ =\frac{\left\{\begin{array}{c}\left(x{e}^y+e^x\right)\left(e^y\left(-\left({\displaystyle \frac{e^y+y{e}^x}{x{e}^y+e^x}}\right)\right)\right)+y{e}^x+e^x\left(-\left({\displaystyle \frac{e^y+y{e}^x}{x{e}^y+e^x}}\right)\right)\\ -\left(e^y+y{e}^x\right)\left(x{e}^y\left(-\left({\displaystyle \frac{e^y+y{e}^x}{x{e}^y+e^x}}\right)\right)+e^y+e^x\right)\end{array}\right\}}{x{e}^y+e^x} \end{gathered}$$

At the point (0,0),

${\left(\frac{d^{2}y}{d{x}^2}\right)}_{\left(0,0\right)}=4$