Question

consider the following curve.

$y = \frac{1 + x}{4 + e^{x}}$

find $y(x)$.

$y(x)=$

find an equation of the tangent line to the given curve at the point $(0,\frac{1}{5})$.

$y=$

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = 1 + x$, so $u'=1$; $v = 4+e^{x}$, so $v'=e^{x}$.
$y'=\frac{(1)(4 + e^{x})-(1 + x)e^{x}}{(4 + e^{x})^{2}}=\frac{4 + e^{x}-e^{x}-xe^{x}}{(4 + e^{x})^{2}}=\frac{4-xe^{x}}{(4 + e^{x})^{2}}$

Step2: Find the slope of the tangent line at $x = 0$

Substitute $x = 0$ into $y'(x)$.
$y'(0)=\frac{4-0\times e^{0}}{(4 + e^{0})^{2}}=\frac{4}{(4 + 1)^{2}}=\frac{4}{25}$

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 $(x_{1},y_{1})=(0,\frac{1}{5})$ and $m=\frac{4}{25}$.
$y-\frac{1}{5}=\frac{4}{25}(x - 0)$
$y=\frac{4}{25}x+\frac{1}{5}$

Answer:

$y'(x)=\frac{4-xe^{x}}{(4 + e^{x})^{2}}$
$y=\frac{4}{25}x+\frac{1}{5}$