If ` y= log_(10) x + log_x 10 + log_x x- log _(10) 10 `, then `(dy)/(dx) =….`

Answer

To solve the problem \( y = \log_{10} x + \log_x 10 + \log_x x - \log_{10} 10 \) and find \( \frac{dy}{dx} \), we will follow these steps:

### Step 1: Simplify the Expression for \( y \)

Using the change of base formula, we can rewrite the logarithmic terms:
- \( \log_{10} x = \frac{\log_e x}{\log_e 10} \)
- \( \log_x 10 = \frac{\log_e 10}{\log_e x} \)
- \( \log_x x = 1 \) (since the logarithm of a number to its own base is always 1)
- \( \log_{10} 10 = 1 \)

Substituting these into the equation for \( y \):
\[ y = \frac{\log_e x}{\log_e 10} + \frac{\log_e 10}{\log_e x} + 1 - 1 \]
This simplifies to:
\[ y = \frac{\log_e x}{\log_e 10} + \frac{\log_e 10}{\log_e x} \]

### Step 2: Differentiate \( y \)

Now we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\log_e x}{\log_e 10}\right) + \frac{d}{dx}\left(\frac{\log_e 10}{\log_e x}\right) \]

The first term is a constant multiplied by \( \log_e x \):
\[ \frac{d}{dx}\left(\frac{\log_e x}{\log_e 10}\right) = \frac{1}{\log_e 10} \cdot \frac{1}{x} \]

For the second term, we apply the quotient rule:
\[ \frac{d}{dx}\left(\frac{\log_e 10}{\log_e x}\right) = \log_e 10 \cdot \frac{d}{dx}\left(\frac{1}{\log_e x}\right) \]
Using the chain rule:
\[ \frac{d}{dx}\left(\frac{1}{\log_e x}\right) = -\frac{1}{(\log_e x)^2} \cdot \frac{1}{x} \]
Thus,
\[ \frac{d}{dx}\left(\frac{\log_e 10}{\log_e x}\right) = -\frac{\log_e 10}{(\log_e x)^2} \cdot \frac{1}{x} \]

### Step 3: Combine the Results

Putting it all together:
\[ \frac{dy}{dx} = \frac{1}{x \log_e 10} - \frac{\log_e 10}{x (\log_e x)^2} \]
We can factor out \( \frac{1}{x} \):
\[ \frac{dy}{dx} = \frac{1}{x} \left( \frac{1}{\log_e 10} - \frac{\log_e 10}{(\log_e x)^2} \right) \]

### Final Result

Thus, the final expression for \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{1}{x} \left( \frac{1}{\log_e 10} - \frac{\log_e 10}{(\log_e x)^2} \right) \]

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