$$ \left. \begin{array} { l }{ \frac { d u } { d x } = \frac { ( x + 9 } { ( x + 9 ) ^ { 2 } } ) ( 1 ) - x ( 1 ) } \ { = \frac { } { ( x + 9 ) ^ { 2 } } } \end{array} ight. $$

Question

$$ \left. \begin{array} { l }{ \frac { d u } { d x } = \frac { ( x + 9 } { ( x + 9 ) ^ { 2 } } ) ( 1 ) - x ( 1 ) } \ { = \frac { } { ( x + 9 ) ^ { 2 } } } \end{array} ight. $$

UpStudy AI · Accepted Answer

We start with the derivative computed using the quotient rule. Assume the function is

$$
u(x) = \frac{x}{x+9}.
$$

The quotient rule states that if

$$
u(x)=\frac{f(x)}{g(x)},
$$

then

$$
u'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}.
$$

Here, we set

$$
f(x)=x \quad \text{and} \quad g(x)=x+9.
$$

Therefore,

$$
f'(x)=1 \quad \text{and} \quad g'(x)=1.
$$

Using the quotient rule, we have

$$
\frac{du}{dx}=\frac{(x+9)(1)-x(1)}{(x+9)^2}.
$$

Simplify the numerator:

$$
(x+9)-x=9.
$$

Thus, the derivative simplifies to

$$
\frac{du}{dx}=\frac{9}{(x+9)^2}.
$$
The derivative of $$ u(x) = \frac{x}{x+9} $$ is $$ \frac{9}{(x+9)^2} $$.

\( \left. \begin{array} { l }{ \frac { d u } { d x } = \frac { ( x + 9 } { ( x + 9 ) ^ { 2 } } ) ( 1 ) - x ( 1 ) } \\ { = \frac { } { ( x + 9 ) ^ { 2 } } } \end{array} \right. \)

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