`log_(3)x+3log_(e)x+2tanx`

Answer

To differentiate the expression \( \log_{3}x + 3\log_{e}x + 2\tan x \), we will follow these steps:

### Step-by-Step Solution:

1. **Rewrite the logarithm:**
We start with the expression:
\[ y = \log_{3}x + 3\log_{e}x + 2\tan x \]
Using the change of base formula for logarithms, we can rewrite \( \log_{3}x \) as:
\[ \log_{3}x = \frac{\log_{e}x}{\log_{e}3} \]
Thus, the expression becomes:
\[ y = \frac{\log_{e}x}{\log_{e}3} + 3\log_{e}x + 2\tan x \]

2. **Combine the logarithmic terms:**
We can factor out \( \log_{e}x \):
\[ y = \left(\frac{1}{\log_{e}3} + 3\right) \log_{e}x + 2\tan x \]

3. **Differentiate the expression:**
Now we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \left(\frac{1}{\log_{e}3} + 3\right) \cdot \frac{1}{x} + 2 \cdot \sec^2 x \]
Here, we used the derivative of \( \log_{e}x \) which is \( \frac{1}{x} \) and the derivative of \( \tan x \) which is \( \sec^2 x \).

4. **Final expression:**
Thus, the derivative of the given expression is:
\[ \frac{dy}{dx} = \left(\frac{1}{\log_{e}3} + 3\right) \cdot \frac{1}{x} + 2 \sec^2 x \]