Question

determine the equation of the line tangent to the curve $\frac{x}{e^{y}}+x = 10 + y$ at the point $(5,0)$. attempt 1: 2 attempts remaining. submit answer next item

Explanation:

Step1: Differentiate the given equation implicitly.

Differentiate $\frac{x}{e^{y}}+x = 10 + y$ with respect to $x$. Using the quotient - rule for $\frac{x}{e^{y}}$ (where if $u = x$ and $v=e^{y}$, then $\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^{2}}$) and the fact that $\frac{d}{dx}(e^{y})=e^{y}\frac{dy}{dx}$, $\frac{d}{dx}(x) = 1$, $\frac{d}{dx}(y)=\frac{dy}{dx}$.
We have $\frac{e^{y}-x\cdot e^{y}\frac{dy}{dx}}{(e^{y})^{2}}+1=\frac{dy}{dx}$.

Step2: Substitute the point $(5,0)$ into the derivative equation.

When $x = 5$ and $y = 0$, $e^{y}=e^{0}=1$.
The equation $\frac{e^{y}-x\cdot e^{y}\frac{dy}{dx}}{(e^{y})^{2}}+1=\frac{dy}{dx}$ becomes $\frac{1 - 5\cdot1\cdot\frac{dy}{dx}}{1}+1=\frac{dy}{dx}$.
Expand: $1-5\frac{dy}{dx}+1=\frac{dy}{dx}$.
Rearrange terms: $1 + 1=\frac{dy}{dx}+5\frac{dy}{dx}$.
$2 = 6\frac{dy}{dx}$, so $\frac{dy}{dx}=\frac{1}{3}$.

Step3: Use the point - slope form of a line.

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(5,0)$ and $m=\frac{1}{3}$.
$y-0=\frac{1}{3}(x - 5)$.
$y=\frac{1}{3}x-\frac{5}{3}$.

Answer:

$y=\frac{1}{3}x-\frac{5}{3}$