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GIVEN:xf(x)g(x)f '(x)g '(x)2721/7-532-13815A. DERIVATIVEOF 2f(x) = 2f '(x)So, 2 f '(2) =2*(1/7)=2/7 ,at x=2 { from table we see that f'(x)=1/7 at x=2}B.DERIVATIVE OF f(x)+g(x) =f '(x)+g '(x)=f '(3)+g '(3)=381+5{from table we see that f '(x)=381and g '(x)=5 at x=3}DEIVATIVE OF f (x)+g (x) = 381+5,at x=3C.DERIVATIVE OF f(x)*g(x) =f '(x)g(x)+g '(x)f(x)=f '(3)g(3)+g '(3)f(3)=()(-1)+(5)(2){from table we see that f '(x)=381,g(x)=-1 ,f(x)=2 and g '(x)=5  at x=3}DERIVATIVE OF f (x)*g (x)= -3pi+10 ,at x=3D. DERIVATIVEOF  f(x)//g(x)={ f '(x)g(x)-g '(x) f(x)} /g 2(x)= {f '(2)g(2)-g '(2)f(2) }/g 2(2)={(1/7)*2-(-5)*7}/(2)2 {from table we see thatf(2)=7,f '(2)=1/7,g(2)=-5 and g'(2)=-5}=247/28DERIVATIVE OF f (x)/g (x) = 247/28 ,at x=2   E. BY APPLYINGCHAIN RULEDERIVATIVE OF f(g(x)) = f '(g(x))*g '(x)=f '(g(2))*g '(2)=f '(2) *g '(2) { from table we see tha ... See the full answer