What is the derivative of `sin("ln"x)+cos("lnx")`
with respect to x at x=e?

Options

A. `(cos1-sin1)/e`
B. `(sin1-cos1)/e`
C. `(cos1+sin1)/e`
D. None of these

Answer

To find the derivative of the function \( y = \sin(\ln x) + \cos(\ln x) \) with respect to \( x \) at \( x = e \), we can follow these steps:

### Step 1: Differentiate the function
We start with the function:
\[ y = \sin(\ln x) + \cos(\ln x) \]

To find the derivative \( \frac{dy}{dx} \), we will use the chain rule. The derivative of \( \sin(u) \) is \( \cos(u) \) and the derivative of \( \cos(u) \) is \( -\sin(u) \), where \( u = \ln x \).

Thus, we differentiate each term:
\[ \frac{dy}{dx} = \frac{d}{dx}[\sin(\ln x)] + \frac{d}{dx}[\cos(\ln x)] \]
Using the chain rule:
\[ \frac{dy}{dx} = \cos(\ln x) \cdot \frac{d}{dx}[\ln x] + (-\sin(\ln x)) \cdot \frac{d}{dx}[\ln x] \]
Since \( \frac{d}{dx}[\ln x] = \frac{1}{x} \), we have:
\[ \frac{dy}{dx} = \cos(\ln x) \cdot \frac{1}{x} - \sin(\ln x) \cdot \frac{1}{x} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{1}{x} \left( \cos(\ln x) - \sin(\ln x) \right) \]

### Step 2: Evaluate the derivative at \( x = e \)
Now, we need to evaluate \( \frac{dy}{dx} \) at \( x = e \):
\[ \frac{dy}{dx} \bigg|_{x=e} = \frac{1}{e} \left( \cos(\ln e) - \sin(\ln e) \right) \]
Since \( \ln e = 1 \), we substitute:
\[ \frac{dy}{dx} \bigg|_{x=e} = \frac{1}{e} \left( \cos(1) - \sin(1) \right) \]

### Final Answer
Thus, the derivative of \( y = \sin(\ln x) + \cos(\ln x) \) with respect to \( x \) at \( x = e \) is:
\[ \frac{1}{e} \left( \cos(1) - \sin(1) \right) \]

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