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)^(2))/((log 10)^(2))`
D. `- ((log 10)^(2))/((log x)^(2))`

Answer

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

### Step 1: Define the variables
Let:
\[ u = \log_{10} x \]
\[ v = \log_x 10 \]

### Step 2: Change the base of logarithms
Using the change of base formula, we can express \( u \) and \( v \) in terms of natural logarithms:
\[ u = \frac{\ln x}{\ln 10} \]
\[ v = \frac{\ln 10}{\ln x} \]

### Step 3: Find \( \frac{du}{dx} \)
Now, we differentiate \( u \) with respect to \( x \):
\[ \frac{du}{dx} = \frac{1}{\ln 10} \cdot \frac{1}{x} = \frac{1}{x \ln 10} \]

### Step 4: Find \( \frac{dv}{dx} \)
Next, we differentiate \( v \) with respect to \( x \):
\[ \frac{dv}{dx} = -\frac{\ln 10}{(\ln x)^2} \cdot \frac{1}{x} = -\frac{\ln 10}{x (\ln x)^2} \]

### Step 5: Find \( \frac{du}{dv} \)
Now, we can find \( \frac{du}{dv} \) using the chain rule:
\[ \frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{\frac{1}{x \ln 10}}{-\frac{\ln 10}{x (\ln x)^2}} \]

### Step 6: Simplify \( \frac{du}{dv} \)
Simplifying the expression:
\[ \frac{du}{dv} = \frac{1}{x \ln 10} \cdot \left(-\frac{x (\ln x)^2}{\ln 10}\right) = -\frac{(\ln x)^2}{(\ln 10)^2} \]

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