Let $$ f(x)=\frac{x}{x+\frac{6}{x}} $$. Find $$ f^{\prime}(x) $$. (Hint: it may help to simplify first.) $$ f^{\prime}(x)=\square $$

Question

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To find the derivative $$ f'(x) $$ of the function $$ f(x) = \frac{x}{x + \frac{6}{x}} $$, we can start by simplifying the function.

First, simplify the denominator:

$$
x + \frac{6}{x} = \frac{x^2 + 6}{x}
$$

Now, substituting this back into the function $$ f(x) $$:

$$
f(x) = \frac{x}{\frac{x^2 + 6}{x}} = \frac{x \cdot x}{x^2 + 6} = \frac{x^2}{x^2 + 6}
$$

Now we have:

$$
f(x) = \frac{x^2}{x^2 + 6}
$$

Next, we will use the quotient rule to find the derivative $$ f'(x) $$. The quotient rule states that if $$ f(x) = \frac{g(x)}{h(x)} $$, then:

$$
f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}
$$

In our case, let $$ g(x) = x^2 $$ and $$ h(x) = x^2 + 6 $$.

Now, we compute $$ g'(x) $$ and $$ h'(x) $$:

$$
g'(x) = 2x
$$
$$
h'(x) = 2x
$$

Now we can apply the quotient rule:

$$
f'(x) = \frac{(2x)(x^2 + 6) - (x^2)(2x)}{(x^2 + 6)^2}
$$

Simplifying the numerator:

$$
f'(x) = \frac{2x(x^2 + 6) - 2x^3}{(x^2 + 6)^2}
$$

Distributing $$ 2x $$:

$$
f'(x) = \frac{2x^3 + 12x - 2x^3}{(x^2 + 6)^2}
$$

The $$ 2x^3 $$ terms cancel out:

$$
f'(x) = \frac{12x}{(x^2 + 6)^2}
$$

Thus, the derivative $$ f'(x) $$ is:

$$
\boxed{\frac{12x}{(x^2 + 6)^2}}
$$
$$ f^{\prime}(x)=\frac{12x}{(x^2 + 6)^2} $$

Let . Find . (Hint: it may help to simplify first.)