$f^{\text{'}}(ln2)=$

$($This answer should not contain II

$3.[-/0\mathrm{.}7$ Points$]$

Let $f(x)=\frac{6x-1}{x5}$

Find $f^{\text{'}}(x).$ First set up the limits and simplify the first one.

Using equation $(6)$ from Section $3\mathrm{.}1$

$f^{\text{'}}(x)=\underset{tx}{lim}($simplify; no complex fraction$).$

Using equation $(2)$ from Section $3\mathrm{.}2$

$f^{\text{'}}(x)=\underset{h0}{lim},(do$ not simplify $)$

$f^{\text{'}}(x)=\frac{x}{(x5)}$

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