How do you find the derivative of #f(x)=(x-1)(x-2)(x-3)#?

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Answer 1 · 2016-08-11T10:40:43

#f^'(x)=(x-3)(3x-5)#

#f(x)=(x-1)(x-2)(x-3)#

#"let k(x),l(x),m(x) be equal (x-1),(x-2),(x-3);"#

#k(x)=(x-1)" ; "k^'(x)=1#
#l(x)=(x-2)" ; "l^'(x)=1#
#m(x)=(x-3)" ; "m^'(x)=1#

#f^'(x)=k^'(x)*l(x)*m(x)+l^'(x)*k(x)*m(x)+m^'(x)*k(x)*l(x)#

#f^'(x) = 1*(x-2) * (x-3)+ 1 * (x-1) * (x-3)+1*(x-2)*(x-3)#

#(x-2)*(x-3)=x^2-3x-2x+6=x^2-5x+6#

#(x-1)*(x-3)=x^2-3x-x+3=x^2-4x+3#

#f^'(x)=x^2-5x+6+x^2-4x+3+x^2-5x+6#

#f^'(x)=3x^2-14x+15#

#f^'(x)=(x-3)(3x-5)#