The differential coefficient of `f(x)=log(logx)` with respect to x is

Options

A. `(x)/(logx)`
B. `(logx)/(x)`
C. `(xlogx)^(-1)`
D. `xlogx`

Answer

To find the differential coefficient of the function \( f(x) = \log(\log x) \) with respect to \( x \), we will use the chain rule of differentiation. Here are the steps to solve the problem:

### Step 1: Identify the function
We have the function:
\[ f(x) = \log(\log x) \]

### Step 2: Differentiate using the chain rule
To differentiate \( f(x) \), we apply the chain rule. The chain rule states that if you have a composite function \( f(g(x)) \), then the derivative is given by:
\[ f'(x) = f'(g(x)) \cdot g'(x) \]
In our case, let \( g(x) = \log x \). Thus, we can rewrite \( f(x) \) as:
\[ f(x) = \log(g(x)) \]

### Step 3: Differentiate the outer function
The derivative of \( \log(u) \) with respect to \( u \) is:
\[ \frac{d}{du} \log(u) = \frac{1}{u} \]
So, applying this to our function:
\[ f'(g(x)) = \frac{1}{g(x)} = \frac{1}{\log x} \]

### Step 4: Differentiate the inner function
Next, we differentiate \( g(x) = \log x \):
\[ g'(x) = \frac{1}{x} \]

### Step 5: Apply the chain rule
Now, we combine the derivatives using the chain rule:
\[ f'(x) = f'(g(x)) \cdot g'(x) = \frac{1}{\log x} \cdot \frac{1}{x} \]

### Step 6: Simplify the expression
Thus, we have:
\[ f'(x) = \frac{1}{x \log x} \]

### Final Result
The differential coefficient of \( f(x) = \log(\log x) \) with respect to \( x \) is:
\[ f'(x) = \frac{1}{x \log x} \]

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