Differentiate the following w.r.t.x. `log_(10)x+log_(x)10+log_(x)x+log_(10)10`

Answer

To differentiate the expression \( y = \log_{10} x + \log_{x} 10 + \log_{x} x + \log_{10} 10 \) with respect to \( x \), we will follow these steps:

### Step 1: Rewrite the logarithmic expressions
We start with:
\[ y = \log_{10} x + \log_{x} 10 + \log_{x} x + \log_{10} 10 \]
Using the change of base formula, we can rewrite the logarithms:
\[ \log_{10} x = \frac{\log x}{\log 10}, \quad \log_{x} 10 = \frac{\log 10}{\log x}, \quad \log_{x} x = 1, \quad \log_{10} 10 = 1 \]
Substituting these into the equation gives:
\[ y = \frac{\log x}{\log 10} + \frac{\log 10}{\log x} + 1 + 1 \]
This simplifies to:
\[ y = \frac{\log x}{\log 10} + \frac{\log 10}{\log x} + 2 \]

### Step 2: Differentiate with respect to \( x \)
Now we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx} \left( \frac{\log x}{\log 10} \right) + \frac{d}{dx} \left( \frac{\log 10}{\log x} \right) + \frac{d}{dx}(2) \]
The derivative of a constant (2) is 0, so we focus on the first two terms.

1. For the first term:
\[ \frac{d}{dx} \left( \frac{\log x}{\log 10} \right) = \frac{1}{\log 10} \cdot \frac{1}{x} \]

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

Combining these, we have:
\[ \frac{dy}{dx} = \frac{1}{x \log 10} - \frac{\log 10}{x (\log x)^2} \]

### Step 3: Simplify the expression
Combining the two terms gives:
\[ \frac{dy}{dx} = \frac{1}{x \log 10} - \frac{\log 10}{x (\log x)^2} \]
Factoring out \(\frac{1}{x}\):
\[ \frac{dy}{dx} = \frac{1}{x} \left( \frac{1}{\log 10} - \frac{\log 10}{(\log x)^2} \right) \]

### Final Answer
Thus, the derivative of the given expression with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{1}{x} \left( \frac{1}{\log 10} - \frac{\log 10}{(\log x)^2} \right) \]