If `y=log_(10)x+log_(x)10+log_(x)x+log_(10)` 10 then what is `((dy)/(dx))_(x=10)` equal to ?

Options

A. 10
B. 2
C. 1
D. 0

Answer

To solve the problem, we need to differentiate the given function \( y \) and evaluate \( \frac{dy}{dx} \) at \( x = 10 \).

### Step-by-Step Solution:

1. **Write the function**:
\[ y = \log_{10} x + \log_x 10 + \log_x x + \log_{10} 10 \]

2. **Simplify the terms**:
- We know that \( \log_{10} 10 = 1 \).
- The term \( \log_x x = 1 \) (since any log of a number to its own base is 1).
- Therefore, we can simplify \( y \):
\[ y = \log_{10} x + \log_x 10 + 1 + 1 = \log_{10} x + \log_x 10 + 2 \]

3. **Use the change of base formula**:
- Recall that \( \log_x 10 = \frac{1}{\log_{10} x} \).
- Substitute this into the equation:
\[ y = \log_{10} x + \frac{1}{\log_{10} x} + 2 \]

4. **Differentiate \( y \) with respect to \( x \)**:
- Let \( u = \log_{10} x \). Then \( y = u + \frac{1}{u} + 2 \).
- Differentiate \( y \):
\[ \frac{dy}{dx} = \frac{du}{dx} + \frac{d}{dx}\left(\frac{1}{u}\right) \]
- We know \( \frac{du}{dx} = \frac{1}{x \ln 10} \) and \( \frac{d}{dx}\left(\frac{1}{u}\right) = -\frac{1}{u^2} \cdot \frac{du}{dx} \):
\[ \frac{dy}{dx} = \frac{1}{x \ln 10} - \frac{1}{u^2} \cdot \frac{1}{x \ln 10} \]
- Substitute \( u = \log_{10} x \):
\[ \frac{dy}{dx} = \frac{1}{x \ln 10} - \frac{1}{(\log_{10} x)^2} \cdot \frac{1}{x \ln 10} \]
- Combine the terms:
\[ \frac{dy}{dx} = \frac{1}{x \ln 10} \left(1 - \frac{1}{(\log_{10} x)^2}\right) \]

5. **Evaluate at \( x = 10 \)**:
- At \( x = 10 \), we have \( \log_{10} 10 = 1 \):
\[ \frac{dy}{dx} \bigg|_{x=10} = \frac{1}{10 \ln 10} \left(1 - \frac{1}{1^2}\right) = \frac{1}{10 \ln 10} (1 - 1) = \frac{1}{10 \ln 10} \cdot 0 = 0 \]

### Final Answer:
\[ \frac{dy}{dx} \bigg|_{x=10} = 0 \]