Chapter 3.5, Problem 35E

To determine

To find: The value of ${y^{\prime }}^{\prime }$ at the point $x=0$.

Answer to Problem 35E

The second derivative of the equation at $\left(0,1\right)$ is ${y^{\prime }}^{\prime }=\frac{1}{e^2}$.

Explanation of Solution

Given:

The equation $xy+e^y=e$.

Derivative rules:

(1) Chain rule: If $y=f\left(u\right)$ and $u=g\left(x\right)$  are both differentiable function, then $\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$.

(2) Product Rule: If $f_1\left(x\right)$ and $f_2\left(x\right)$ are both differentiable, then $\frac{d}{dx}\left(f_1\left(x\right)f_2\left(x\right)\right)=f_1\left(x\right)\frac{d}{dx}\left(f_2\left(x\right)\right)+f_2\left(x\right)\frac{d}{dx}\left(f_1\left(x\right)\right)$.

Calculation:

The value of y if $x=0$ is computed as follows,

Substitute the value $x=0$ in $xy+e^y=e$,

$\begin{array}{l}\left(0\right)y+e^y=e\\ e^y=e\end{array}$

Take natural logarithmic on both sides,

$\begin{array}{l}\ln e^y=\ln e\\ y=1\left(Q\ln e=1\right)\end{array}$

Thus, the point is $\left(x,y\right)=\left(0,1\right)$.

Obtain the second derivative of the equation at $\left(0,1\right)$.

$xy+e^y=e$

Differentiate implicitly with respect to x,

$\begin{array}{c}\frac{d}{dx}\left(xy+e^y\right)=\frac{d}{dx}\left(e\right)\\ \frac{d}{dx}\left(xy\right)+\frac{d}{dx}\left(e^y\right)=0\end{array}$

Apply the product rule (2) and the chain rule (1),

$\begin{array}{l}\left[x\frac{d}{dx}\left(y\right)+y\frac{d}{dx}\left(x\right)\right]+\frac{d}{dx}\left(e^y\right)=0\\ \left[x\frac{dy}{dx}+y\left(1\right)\right]+\frac{d}{dy}\left(e^y\right)\frac{dy}{dx}=0\\ \left[x{y}^{\prime }+y\right]+e^y{y}^{\prime }=0\end{array}$

$x{y}^{\prime }+y+e^y{y}^{\prime }=0$ (1)

Substitute $\left(0,1\right)$ for $\left(x,y\right)$ in equation (1),

$\begin{array}{l}\left(0\right)y^{\prime }+1+e^1{y}^{\prime }=0\\ 0+1+e{y}^{\prime }=0\\ e{y}^{\prime }=-1\\ y^{\prime }=\frac{-1}{e}\end{array}$

Thus, the derivative $y^{\prime }$ at $\left(0,1\right)$ is $y^{\prime }=\frac{-1}{e}$.

Differentiate the equation (1) implicitly with respect to x,

$\begin{array}{l}\frac{d}{dx}\left(x{y}^{\prime }+y+e^y{y}^{\prime }\right)=\frac{d}{dx}\left(0\right)\\ \frac{d}{dx}\left(x{y}^{\prime }\right)+\frac{d}{dx}\left(y\right)+\frac{d}{dx}\left(e^y{y}^{\prime }\right)=0\end{array}$

Apply the product rule (2) and the chain rule (1),

$\begin{array}{l}\left[x\frac{d}{dx}\left(y^{\prime }\right)+y^{\prime }\frac{d}{dx}\left(x\right)\right]+\frac{d}{dx}\left(y\right)+\left[e^y\frac{d}{dx}\left(y^{\prime }\right)+y^{\prime }\frac{d}{dx}\left(e^y\right)\right]=0\\ \left[x{y^{\prime }}^{\prime }+y^{\prime }\left(1\right)\right]+\frac{dy}{dx}+\left[e^y{y^{\prime }}^{\prime }+y^{\prime }\frac{d}{dy}\left(e^y\right)\frac{dy}{dx}\right]=0\\ x{y^{\prime }}^{\prime }+y^{\prime }\left(1\right)+y^{\prime }+e^y{y^{\prime }}^{\prime }+y^{\prime }{e}^y{y}^{\prime }=0\\ {y^{\prime }}^{\prime }\left(x+e^y\right)+y^{\prime }\left(2+e^y{y}^{\prime }\right)=0\end{array}$

Substitute $\left(0,1\right)$ for $\left(x,y\right)$ and $y^{\prime }=\frac{-1}{e}$,

$\begin{array}{l}{y^{\prime }}^{\prime }\left(0+e^1\right)+\frac{-1}{e}\left(2+e^1\left(\frac{-1}{e}\right)\right)=0\\ {y^{\prime }}^{\prime }e-\frac{1}{e}\left(2-1\right)=0\\ {y^{\prime }}^{\prime }e=\frac{1}{e}\\ {y^{\prime }}^{\prime }=\frac{1}{e^2}\end{array}$

Therefore, the second derivative of the equation at the point $\left(0,1\right)$ is ${y^{\prime }}^{\prime }=\frac{1}{e^2}$.