Question

What is the instantaneous rate of change of `f(x)=x/(-x-8)` at `x=4`?

Answer 1

`-1/18`

Explanation:

`"the instantaneous rate of change at x = 4"`

`"is the value of the derivative at x = 4"`

`"differentiate 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)=xrArrg'(x)=1`

`h(x)=-x-8rArrh'(x)=-1`

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

`rArrf'(4)=-8/(144)=-1/18`

Answer 2

`\`

`"Answer is:" \qquad \quad - 1/18.`

Explanation:

`\`

`"The required quantity is the derivative of" \ \ f(x) \ \ "evaluated at"`
`\qquad \qquad x = 4.`

`"So, let's calculate the derivative of" \ \ f(x):`

1. `"Original Function:" \qquad f(x) = x / {-x-8}`
2. `"Rewrite a little:" \qquad f(x) = - x / {x+8}`
3. `"Quotient Rule:" \qquad f'(x) = - { (x+8) [ x ]' - x [x+8]' }/ [ (x+8)^2 ]`
   `\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = - { (x+8) [ 1 ] - x [1+0] }/ [ (x+8)^2 ]`
   `\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = - { (x+8) - x }/ [ (x+8)^2 ]`
   `\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = - { 8 }/ [ (x+8)^2 ].`
4. `"Summary:" \qquad \qquad \qquad \quad \ f'(x) = - { 8 }/ [ (x+8)^2 ].`
5. `"Evaluate" \ \ f'(x) \ \ "at" \ \ x = 4,`
   `\qquad \quad \quad "to get the instanteous rate of change at" \ \ x=4:`
   `f'(4) = - { 8 }/ [ (4+8)^2 ] \ = - \ 8/12^2 \ = - \ 8/ ( 12 \cdot 12 )`
   `\qquad\quad \quad \ = - \color{red}{ cancel {8} }/ ( color{red}{ cancel {2} } \cdot 6 \cdot color{red}{ cancel {4} } \cdot 3 ) = \ \ - 1/18.`
6. `"Summarize:" \qquad \qquad \qquad \quad \quad \ f'(4) = - 1/18.`

`:. "instantaneous rate of change of"\ \ f(x) \ \ "at" \ (x=4) \ = - 1/18.`