Question

How do you find the derivative of `sqrt(x^2-1) / (x^2+1)`?

Answer 1

`-(x(x^2-3))/(sqrt(x^2-1) (x^2+1)^2)`

Explanation:

Using quotient rule:

`((sqrt(x^2-1))'(x^2+1)-(sqrt(x^2-1)*(x^2+1)'))/(x^2+1)^2`

Mind that:
`sqrt(x^2-1)' = ((x^2-1)^(1/2))' = 1/2(x^2-1)^(1/2-1)*(x^2-1)' = 1/2*1/sqrt(x^2-1)(2x) = x/sqrt(x^2-1)`

So the solution is:
#((x^2+1)x/sqrt(x^2-1) -sqrt(x^2-1)(2x))/(x^2+1)^2 = x/(sqrt(x^2-1) (x^2+1)^2)(x^2+1-2x^2+2) = -x(x^2-3)/(sqrt(x^2-1) (x^2+1)^2)#

Answer 2

`(3x-x^3)/(sqrt(x^2-1)(x^2+1)^2`

Explanation:

`"differentiate using the "color(blue)"quotient rule"`

`"Given "y=(g(x))/(h(x))" then"`

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

`g(x)=sqrt(x^2-1)=(x^2-1)^(1/2)`

`"differentiate using the "color(blue)"chain rule"`

`rArrg'(x)=1/2(x^2-1)^(-1/2)xxd/dx(x^2-1)`

`color(white)(rArrg'(x))=x(x^2-1)^(-1/2)`

`h(x)=x^2+1rArrh'(x)=2x`

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

`color(white)rArrdy/dx=(x(x^2-1)^(-1/2)[x^2+1-2(x^2-1)])/(x^2+1)^2`

`color(white)(rArrdy/dx)=(3x-x^3)/(sqrt(x^2-1)(x^2+1)^2)`