Step 1 Recall the Chain Rule in terms of the functions $$ h(x) $$ and $$ g(x) $$, which states that if $$ g $$ is differentiable at $$ x $$ and $$ h $$ is differentiable at $$ g(x) $$, then the composite function $$ F=h $$ o $$ g $$ defined by $$ F(x)=h(g(x)) $$ is differentiable at $$ x $$ and $$ F^{\prime} $$ is given by the following. $$ F^{\prime}(x)=h^{\prime}(g(x)) \cdot g^{\prime}(x) $$ We are given the function $$ f(x)=\sin (8 \ln (x)) $$ To write this is the form $$ F(x)=h(g(x)) $$ we can let $$ g(x)=8 \ln (x) $$ and $$ h(x)= $$ SUBMI SUB (YOU CANNOT COME BACK) Need Help? Read It SUBMIT ANSWER SKI

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To express the function $$ f(x) = \sin(8 \ln(x)) $$ in the form $$ F(x) = h(g(x)) $$, we need to identify the inner function $$ g(x) $$ and the outer function $$ h(x) $$.

From the problem, we have already defined:
- $$ g(x) = 8 \ln(x) $$

Now, we need to determine $$ h(x) $$. Since $$ f(x) $$ is the sine of $$ g(x) $$, we can define $$ h(x) $$ as follows:
- $$ h(x) = \sin(x) $$

Thus, we can write:
- $$ F(x) = h(g(x)) = h(8 \ln(x)) = \sin(8 \ln(x)) $$

In summary:
- $$ g(x) = 8 \ln(x) $$
- $$ h(x) = \sin(x) $$

Now, we can apply the Chain Rule to find the derivative $$ F'(x) $$.

1. First, we need to find $$ h'(x) $$ and $$ g'(x) $$:
   - $$ h'(x) = \cos(x) $$
   - $$ g'(x) = \frac{8}{x} $$

2. Now, we can apply the Chain Rule:
   $$
   F'(x) = h'(g(x)) \cdot g'(x)
   $$

Substituting the values we found:
$$
F'(x) = h'(8 \ln(x)) \cdot g'(x) = \cos(8 \ln(x)) \cdot \frac{8}{x}
$$

Thus, the derivative of $$ f(x) = \sin(8 \ln(x)) $$ is:
$$
F'(x) = \frac{8 \cos(8 \ln(x))}{x}
$$
To express $$ f(x) = \sin(8 \ln(x)) $$ as $$ F(x) = h(g(x)) $$, let: - $$ g(x) = 8 \ln(x) $$ - $$ h(x) = \sin(x) $$ Applying the Chain Rule: $$ F'(x) = \cos(8 \ln(x)) \cdot \frac{8}{x} $$ So, the derivative is $$ \frac{8 \cos(8 \ln(x))}{x} $$.

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