What is the instantaneous rate of change of f(x)=x/(-x-8)f(x)=xx8 at x=4 x=4?

2 Answers
Feb 9, 2018

-1/18118

Explanation:

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

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

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

"given "f(x)=(g(x))/(h(x))" then"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

Feb 9, 2018

\

"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.