How do you differentiate #f(x)=x/(x-1)^2+x^2-4/(1-1x)# using the sum rule?

1 Answer
Jan 17, 2018

#f'(x)=(2x)/(x-1)^3+(x-1)^2+2x-4/(1-x)^2#

Explanation:

#"differentiate using the "color(blue)"chain/product rules"#

#"given "y=f(g(x))"then"#

#dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"#

#"given "y=g(x)h(x)" then"#

#dy/dx=g(x)h'(x)+h(x)g'(x)larrcolor(blue)"product rule"#

#"differentiate "x/(x-1)^2" using chain/product rules"#

#x/(x-1)^2=x(x-1)^-2#

#d/dx(x(x-1)^2)#

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

#=-2x(x-1)^-3+(x-1)^2=(2x)/(x-1)^3+(x-1)^2#

#"differentiate "4/(1-x)" using the chain rule"#

#4/(1-x)=4(1-x)^-1#

#d/dx(4(1-x)^-1)=-4(1-x)^-2(-1)=4/(1-x)^2#

#"returning to the original question"#

#f'(x)=(2x)/(x-1)^3+(x-1)^2+2x-4/(1-x)^2#