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

Question

Kayla Fitzgerald · Accepted Answer

Explanation:the instantaneous rate of change at x = 4is the value of the derivative at x = 4differentiate using the quotient rulegiven f(x)=g(x)h(x) thenf′(x)=h(x)g′(x)−g(x)h′(x)(h(x))2← quotient ruleg(x)=x⇒g′(x)=1h(x)=−x−8⇒h′(x)=−1⇒f′(x)=−x−8−x(−1)(−x−8)2=−8(−x−8)2⇒f′(4)=−8144=−118

queueryutisteh7n · Accepted Answer

The required quantity is the derivative of f(x) evaluated at
x=4.
So, let's calculate the derivative of f(x):
Original Function: $f (x) = \frac{x}{- x - 8}$
Rewrite a little: $f (x) = - \frac{x}{x + 8}$
Quotient Rule: $f^{'} (x) = - \frac{(x + 8) {[x]}^{'} - x {[x + 8]}^{'}}{{(x + 8)}^{2}}$
$= - \frac{(x + 8) [1] - x [1 + 0]}{{(x + 8)}^{2}}$
$= - \frac{(x + 8) - x}{{(x + 8)}^{2}}$
$= - \frac{8}{{(x + 8)}^{2}}$
Summary: $f^{'} (x) = - \frac{8}{{(x + 8)}^{2}}$
Evaluate f'(x) at x=4,
to get the instanteous rate of change at x=4:
$f^{'} (4) = - \frac{8}{{(4 + 8)}^{2}} = - \frac{8}{12^{2}} = - \frac{8}{12 \cdot 12}$
$= - \frac{8}{2 * 6 * 4 * 3} = - \frac{1}{18}$
Summarize: $f^{'} (4) = - \frac{1}{18}$
$∴$ instantaneous rate of change of $f (x) a t (x = 4) = - \frac{1}{18}$

What is the instantaneous rate of change of

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