According to a web page on the internet, the derivative of a quotient is the quotient of the derivatives: in other words,

(fg)′(x)=(f′(x))(g′(x))\displaystyle {\left(\frac{{f}}{{g}}\right)}'{\left({x}\right)}={\left({f}'{\left({x}\right)}\right)}{\left({g}'{\left({x}\right)}\right)}

Let f(x)=9x+2\displaystyle {f{{\left({x}\right)}}}={9}{x}+{2}, g(x)=5x\displaystyle {g{{\left({x}\right)}}}={5}{x}.

f′(x)=\displaystyle {f}'{\left({x}\right)}=  

g′(x)=\displaystyle {g}'{\left({x}\right)}=  

f′(x)g′(x)=\displaystyle \frac{{{f}'{\left({x}\right)}}}{{{g}'{\left({x}\right)}}}=  

(fg)(x)\displaystyle {\left(\frac{{f}}{{g}}\right)}{\left({x}\right)} =  

(fg)′(x)\displaystyle {\left(\frac{{f}}{{g}}\right)}'{\left({x}\right)} =  

According to the web page, (fg)′(x)=f′(x)g′(x)\displaystyle {\left(\frac{{f}}{{g}}\right)}'{\left({x}\right)}=\frac{{{f}'{\left({x}\right)}}}{{{g}'{\left({x}\right)}}}. Based on your work above (check all that apply):

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