Question

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

Answer 1

`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`