Question

How do you use the chain rule to differentiate `y=((x+1)/(x-2))^5`?

Answer 1

`(dy)/(dx) = (-15(x+1)^4)/(x-2)^6`

Explanation:

`y=((x+1)/(x-2))^5`

Let u=`(x+1)/(x-2)`
so `y=u^5`
and `(dy)/(du) = 5u^4`

then,

`(du)/(dx)=((x-2)(1)-(x+1)(1))/(x-2)^2`

`(du)/(dx)=(x-2-x-1)/(x-2)^2`

`(du)/(dx)=-3/(x-2)^2`

`(dy)/(dx) = (dy)/(du)times(du)/(dx)`

`(dy)/(dx) = 5u^4times-3/(x-2)^2`

`(dy)/(dx) = 5((x+1)/(x-2))^4times-3/(x-2)^2`

`(dy)/(dx) = (-15(x+1)^4)/(x-2)^6`

Answer 2

`dy/dx=-(15(x+1)^4)/(x-2)^6`

Explanation:

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

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

`rArrdy/dx=5((x+1)/(x-2))^4xxd/dx((x+1)/(x-2))`

`"differentiate "(x+1)/(x-2)" using the "color(blue)"quotient rule"`

`"given "f(x)=(g(x))/(h(x))" then"`

`f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larrcolor(blue)"quotient rule"`

`g(x)=x+1rArrg'(x)=1`

`h(x)=x-2rArrh'(x)=1`

`rArrd/dx((x+1)/(x-2))`

`=(x-2-x-1)/(x-2)^2=-3/(x-2)^2`

`rArrdy/dx=5((x+1)/(x-2))^4xx-3/(x-2)^2`

`color(white)(rArrdy/dx)=-(15(x+1)^4)/(x-2)^6`