Question

How do you find the derivative of `F(x) = sqrt( (x-8)/(x^2-2) )`?

Answer 1

`F(x)=sqrt((x-8)/(x^2-2))`
`F'(x)=1/(2sqrt((x-8)/(x^2-2)))(((x^2-2)d/dx(x-8)-(x-8)d/dx(x^2-2))/(x^2-2)^2)`

`F'(x)=1/(2sqrt((x-8)/(x^2-2)))(((x^2-2)(1)-(x-8)(2x))/(x^2-2)^2)`

`F'(x)=sqrt(x^2-2)/(2sqrt((x-8)))((x^2-2-2x^2+16x)/(x^2-2)^2)`

`F'(x)=sqrt(x^2-2)/(2sqrt((x-8)))((-x^2+16x-2)/(x^2-2)^2)`

Answer 2

`F'(x)=-((x^2-16x+2))/{2(x-8)^(1/2)(x^2-2)^(3/2)}.`

Explanation:

Let, `y=F(x)=sqrt{(x-8)/(x^2-2)}={(x-8)/(x^2-2)}^(1/2).`

`:. lny=1/2ln((x-8)/(x^2-2))=1/2{ln(x-8)-ln(x^2-2)}.`

`:. d/dx{lny}=1/2d/dx{ln(x-8)-ln(x^2-2)}.`

By the Chain Rule, then, we have,

`:. d/dy(lny)*dy/dx`

`=1/2[1/(x-8)*d/dx(x-8)-1/(x^2-2)*d/dx(x^2-2)}.`

`:. 1/y*dy/dx=1/2{1/(x-8)*1-1/(x^2-2)*2x},`

`=1/2{{(x^2-2)-2x(x-8)}/{(x-8)(x^2-2)}},`

`=1/2{(-x^2+16x-2)/{(x-8)(x^2-2)}}.`

`rArr dy/dx=-(y(x^2-16x+2))/{2(x-8)(x^2-2)}.`

But, `y={(x-8)/(x^2-2)}^(1/2).`

`:. dy/dx=-{(x-8)/(x^2-2)}^(1/2)*(x^2-16x+2)/{2(x-8)(x^2-2)}, i.e.,`

`dy/dx=F'(x)=-((x^2-16x+2))/{2(x-8)^(1/2)(x^2-2)^(3/2)}.`

Enjoy Mayhs.!