If `f(x)=x^(3)+3x+4` and `g` is the inverse function of `f(x)`, then the value of `(d)/(dx)((g(x))/(g(g(x))))` at x = 4 equals

Options

A. `(-1)/(6)`
B. 6
C. `(-1)/(3)`
D. non-existent

Explanation

To solve the problem step by step, we need to find the value of \(\frac{d}{dx}\left(\frac{g(x)}{g(g(x))}\right)\) at \(x = 4\), where \(g\) is the inverse function of \(f(x) = x^3 + 3x + 4\).

### Step 1: Differentiate the function
We will use the quotient rule for differentiation. The quotient rule states that if \(u(x)\) and \(v(x)\) are functions of \(x\), then:

\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

In our case, let \(u = g(x)\) and \(v = g(g(x))\). Thus, we have:

\[ \frac{d}{dx}\left(\frac{g(x)}{g(g(x))}\right) = \frac{g(g(x)) g'(x) - g(x) g'(g(x)) g'(x)}{(g(g(x)))^2} \]

### Step 2: Evaluate at \(x = 4\)
We need to evaluate this derivative at \(x = 4\):

\[ \frac{d}{dx}\left(\frac{g(x)}{g(g(x))}\right) \bigg|_{x=4} = \frac{g(g(4)) g'(4) - g(4) g'(g(4)) g'(4)}{(g(g(4)))^2} \]

### Step 3: Find \(g(4)\)
Since \(g\) is the inverse of \(f\), we need to find \(g(4)\). This means we need to find \(x\) such that \(f(x) = 4\):

\[ f(x) = x^3 + 3x + 4 \]
Setting \(f(x) = 4\):

\[ x^3 + 3x + 4 = 4 \implies x^3 + 3x = 0 \implies x(x^2 + 3) = 0 \]

This gives us \(x = 0\) (since \(x^2 + 3\) has no real roots). Therefore, \(g(4) = 0\).

### Step 4: Find \(g(g(4))\)
Now we need to find \(g(g(4))\):

\[ g(4) = 0 \implies g(g(4)) = g(0) \]

Next, we need to find \(g(0)\). We need \(f(x) = 0\):

\[ f(x) = x^3 + 3x + 4 = 0 \]

By testing \(x = -1\):

\[ f(-1) = (-1)^3 + 3(-1) + 4 = -1 - 3 + 4 = 0 \]

Thus, \(g(0) = -1\).

### Step 5: Find \(g'(4)\)
Now we need to find \(g'(4)\). By the inverse function theorem:

\[ g'(x) = \frac{1}{f'(g(x))} \]

We need \(f'(x)\):

\[ f'(x) = 3x^2 + 3 \]

Now evaluate \(f'(g(4)) = f'(0)\):

\[ f'(0) = 3(0)^2 + 3 = 3 \]

Thus, \(g'(4) = \frac{1}{f'(0)} = \frac{1}{3}\).

### Step 6: Substitute values into the derivative
Now we substitute back into our derivative:

\[ \frac{d}{dx}\left(\frac{g(x)}{g(g(x))}\right) \bigg|_{x=4} = \frac{g(g(4)) g'(4) - g(4) g'(g(4)) g'(4)}{(g(g(4)))^2} \]

Substituting the values we found:

\[ = \frac{(-1) \cdot \frac{1}{3} - 0 \cdot g'(-1) \cdot \frac{1}{3}}{(-1)^2} \]

This simplifies to:

\[ = \frac{-\frac{1}{3}}{1} = -\frac{1}{3} \]

### Final Answer
Thus, the value of \(\frac{d}{dx}\left(\frac{g(x)}{g(g(x))}\right)\) at \(x = 4\) is:

\[ \boxed{-\frac{1}{3}} \]