if $y = \\frac{x - 5}{\\ln x}$, find $y(e^{2})$.
if $y = \\frac{x - 5}{\\ln x}$, find $y(e^{2})$.
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, $u = x - 5$, so $u'=1$, and $v=\ln x$, so $v'=\frac{1}{x}$. Then $y'=\frac{1\times\ln x-(x - 5)\times\frac{1}{x}}{(\ln x)^{2}}=\frac{\ln x-\frac{x - 5}{x}}{(\ln x)^{2}}=\frac{\ln x - 1+\frac{5}{x}}{(\ln x)^{2}}$.
Step2: Substitute $x = e^{2}$
When $x = e^{2}$, $\ln(e^{2})=2$. Substitute into $y'$: [ \begin{align*} y'(e^{2})&=\frac{2-1+\frac{5}{e^{2}}}{2^{2}}\ &=\frac{1+\frac{5}{e^{2}}}{4}\ &=\frac{e^{2}+5}{4e^{2}} \end{align*} ]
Answer:
$\frac{e^{2}+5}{4e^{2}}$