consider the following curve.\n\n$y = \\frac{x - 1}{x + 1}$\n\nfind $y(x)$.\n\n$y(x)=$\n\nfind equations of…

consider the following curve.\n\n$y = \\frac{x - 1}{x + 1}$\n\nfind $y(x)$.\n\n$y(x)=$\n\nfind equations of the tangent lines to the curve that are parallel to

consider the following curve.\n\n$y = \\frac{x - 1}{x + 1}$\n\nfind $y(x)$.\n\n$y(x)=$\n\nfind equations of the tangent lines to the curve that are parallel to

Answer

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = x - 1$, $u'=1$, $v=x + 1$, and $v'=1$.

Step2: Calculate the derivative

Substitute $u$, $u'$, $v$, and $v'$ into the quotient - rule formula: [ \begin{align*} y'(x)&=\frac{1\times(x + 1)-(x - 1)\times1}{(x + 1)^{2}}\ &=\frac{x + 1-x + 1}{(x + 1)^{2}}\ &=\frac{2}{(x + 1)^{2}} \end{align*} ]

Answer:

$\frac{2}{(x + 1)^{2}}$