Question

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

Answer 1

`f'(x)=(x-6)/(x^2(4x-x^2)^(1/2)`

Explanation:

We will have to use the chain rule but before that we have to use the `color(blue)"quotient rule"`

`color(orange)"Reminder " "If " f(x)=(g(x))/(h(x))" then "`

`color(red)(bar(ul(|color(white)color(black)(f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2)|)))larr" quotient rule"`

`color(orange)"Reminder " " If " f(x)=g(h(x))" then"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(f'(x)=g'(h(x)).h'(x))color(white)(2/2)|)))larr" chain rule"`

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

`g(x)=(4x-x^2)^(1/2)" and using the chain rule"`

`g'(x)=1/2(4x-x^2)^(-1/2).d/dx(4x-x^2)`

`=1/2(4x-x^2)^(-1/2)(4-2x)=1/2(4x-x^2)^(-1/2).2(2-x)`

`=(2-x)(4x-x^2)^(-1/2)`

`" and " h(x)=x^2rArrh'(x)=2x`
`"-------------------------------------------------------------"`

`rArrf'(x)=(x^2(2-x)(4x-x^2)^(-1/2)-(4x-x^2)^(1/2).2x)/x^4`

simplifying the numerator.

`=(x(4x-x^2)^(-1/2)[x(2-x)-(4x-x^2).2))/x^4`

`=(x(4x-x^2)^(-1/2)[2x-x^2-8x+2x^2])/x^4`

`=(x(4x-x^2)^(-1/2)(x^2-6x))/x^4=(cancel(x^2)(4x-x^2)^(-1/2)(x-6))/(cancel(x^4)x^2)`

`=(x-6)/(x^2(4x-x^2)^(1/2))`