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

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

Let f(x)=5x+6\displaystyle {f{{\left({x}\right)}}}={5}{x}+{6}, g(x)=8x+7\displaystyle {g{{\left({x}\right)}}}={8}{x}+{7}.

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

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

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

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

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

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

License