If $f(x)=ln|x^2-11||,$ theil $f^{\text{'}}(x)=$

$($a$)|\frac{2x}{x^2-1}|$

$($b$)\frac{2x}{|x^2-1|}$

$($c$)\frac{2|x|}{x^2-1}$

$($d$)\frac{2x}{x^2-1}$

$9$ Let $g(x)=x^5+3x-2,$ and let $g^{-1}$ denote the inverse function of $g.$ Then $(g^{-1}{)}^{\text{'}}(2)=$

$($a$)\frac{1}{83}$

$($b$)\frac{1}{8}$

$($c$)1$

$($d$)8$

$10.$ If $f(4)=5$ and $f^{\text{'}}(4)=\frac{2}{3},$ then calculate $(f^{-1}{)}^{\text{'}}(5).$

$($a$)\frac{1}{5}$

$($b$)\frac{1}{4}$

$($c$)\frac{2}{3}$

$($d$)\frac{3}{2}$

$11.$ If $\frac{dx}{dz}{x}^{?}x-tanx-cotxcotxtanxy=ln(\frac{x}{2})x=4\frac{1}{8}\frac{1}{4}\frac{1}{2}y=arctan(cosx)\frac{dy}{dx}=\frac{-sinx}{1+co{s}^{2}x}-(se{c}^{-1}(cosx){)}^{2}sinx(se{c}^{-1}(cosx){)}^2\frac{1}{1+co{s}^{2}x}f(x)=e^{3ln(x^2)}{f}^{\text{'}}(x)=e^{3ln(x^2)}6(lnx)e^{3ln(x^2)}5x^{4}6x^{5}sinx=e^y,0,$ what $is\frac{dx}{dz}in$

terms $of{x}^{?}$ terms $ofx?$

$(a)-tanx$

$(b)-cotx$

$(c)cotx$

$(d)tanx$

$12.$ The slope $of$ the line tangent $to$ the graph $ofy=ln(\frac{x}{2})atx=4is$

$(a)\frac{1}{8}$

$(b)\frac{1}{4}$

$(c)\frac{1}{2}$

$(d)1$

$13.Ify=arctan(cosx),$ then $\frac{dy}{dx}=$

$(a)\frac{-sinx}{1+co{s}^{2}x}$

$(b)-(se{c}^{-1}(cosx){)}^{2}sinx$

$(c)(se{c}^{-1}(cosx){)}^2$

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

$14.Iff(x)=e^{3ln(x^2)},$ then $f^{\text{'}}(x)=$

$(a)e^{3ln(x^2)}$

$(b)6(lnx)e^{3ln(x^2)}$

$(c)5x^4$

$(d)6x^5$ Please all ;(