Differential coefficient of `log_(10)x "
" w.r.t "
" log_x 10 ` is

Options

A. `-((log x)^2)/((log 10)^2)`
B. `((log_(10)x)^2)/( (log 10)^2)`
C. `(log_x 10)/( log 10)`
D. `-((log 10)^2)/((log x)^2)`

Answer

To find the differential coefficient of \( \log_{10} x \) with respect to \( \log_x 10 \), we will follow these steps:

### Step 1: Define the Functions
Let:
- \( y = \log_{10} x \)
- \( z = \log_x 10 \)

### Step 2: Use the Change of Base Formula
Using the change of base formula for logarithms, we can express \( y \) and \( z \) as:
\[ y = \frac{\log_e x}{\log_e 10} \]
\[ z = \frac{\log_e 10}{\log_e x} \]

### Step 3: Differentiate \( y \) with respect to \( x \)
To find \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{1}{\log_e 10} \cdot \frac{d}{dx} (\log_e x) = \frac{1}{\log_e 10} \cdot \frac{1}{x} \]
Thus,
\[ \frac{dy}{dx} = \frac{1}{x \log_e 10} \]

### Step 4: Differentiate \( z \) with respect to \( x \)
To find \( \frac{dz}{dx} \):
\[ \frac{dz}{dx} = \frac{d}{dx} \left( \frac{\log_e 10}{\log_e x} \right) \]
Using the quotient rule:
\[ \frac{dz}{dx} = \frac{0 \cdot \log_e x - \log_e 10 \cdot \frac{1}{x}}{(\log_e x)^2} = -\frac{\log_e 10}{x (\log_e x)^2} \]

### Step 5: Use the Chain Rule to Find \( \frac{dy}{dz} \)
Now, we can find \( \frac{dy}{dz} \) using the formula:
\[ \frac{dy}{dz} = \frac{\frac{dy}{dx}}{\frac{dz}{dx}} \]
Substituting the values we found:
\[ \frac{dy}{dz} = \frac{\frac{1}{x \log_e 10}}{-\frac{\log_e 10}{x (\log_e x)^2}} \]

### Step 6: Simplify the Expression
Simplifying the expression:
\[ \frac{dy}{dz} = \frac{1}{x \log_e 10} \cdot \left(-\frac{x (\log_e x)^2}{\log_e 10}\right) \]
The \( x \) cancels out:
\[ \frac{dy}{dz} = -\frac{(\log_e x)^2}{(\log_e 10)^2} \]

### Final Result
Thus, the differential coefficient of \( \log_{10} x \) with respect to \( \log_x 10 \) is:
\[ \frac{dy}{dz} = -\frac{(\log_e x)^2}{(\log_e 10)^2} \]