Question

How do you differentiate `f(x)=4sqrt((x^2+1)/(x^2-1))`?

Answer 1

`f'(x)=(-8x)/{(x^2-1)^(3/2)(x^2+1)^(1/2)}.`

Explanation:

`y=f(x)=4sqrt{(x^2+1)/(x^2-1)}=4{(x^2+1)/(x^2-1)}^(1/2).`

Taking ln of both sides,
`lny=ln4+1/2{ln(x^2+1)-ln(x^2-1)}`

Diff. both sides w.r.t x,
`d/dxlny=d/dxln4+1/2{d/dxln(x^2+1)-d/dxln(x^2-1)}.`
`d/dylny*dy/dx=0+1/2{1/(x^2+1)*d/dx(x^2+1)-1/(x^2-1)*d/dx(x^2-1)}`
`1/y*dy/dx=1/2[1/(x^2+1)*2x-1/(x^2-1)*2x}=1/2*2x[{x^2-1-x^2-1]/{(x^2-1)(x^2+1)]=-(2x)/{(x^2-1)(x^2+1)}`

Therefore, `dy/dx=f'(x)=y*(-2x)/{(x^2-1)(x^2+1)}=4sqrt{(x^2+1)/(x^2-1)}*(-2x)/{(x^2-1)(x^2+1)}=(-8x)/{(x^2-1)^(3/2)(x^2+1)^(1/2)}.`