If `y = log (cos e^(x)), then (dy)/(dx)` is:

Options

A. `cose^(x-1) `
B. `e^(-x) cos e^x`
C. `e^(x) sin e^x`
D. `-e^(x) tan e^x`

Answer

To find the derivative \( \frac{dy}{dx} \) of the function \( y = \log(\cos(e^x)) \), we will use the chain rule. Here’s a step-by-step solution:

### Step 1: Identify the function
We have:
\[ y = \log(\cos(e^x)) \]

### Step 2: Differentiate using the chain rule
To differentiate \( y \), we apply the chain rule. The derivative of \( \log(u) \) is \( \frac{1}{u} \cdot \frac{du}{dx} \) where \( u = \cos(e^x) \).

Thus, we have:
\[ \frac{dy}{dx} = \frac{1}{\cos(e^x)} \cdot \frac{d}{dx}(\cos(e^x)) \]

### Step 3: Differentiate \( \cos(e^x) \)
Now, we need to differentiate \( \cos(e^x) \). Again, we apply the chain rule. The derivative of \( \cos(v) \) is \( -\sin(v) \cdot \frac{dv}{dx} \) where \( v = e^x \).

So,
\[ \frac{d}{dx}(\cos(e^x)) = -\sin(e^x) \cdot \frac{d}{dx}(e^x) \]

### Step 4: Differentiate \( e^x \)
The derivative of \( e^x \) is simply \( e^x \). Therefore,
\[ \frac{d}{dx}(\cos(e^x)) = -\sin(e^x) \cdot e^x \]

### Step 5: Substitute back into the derivative
Now we substitute this back into our expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{1}{\cos(e^x)} \cdot (-\sin(e^x) \cdot e^x) \]

### Step 6: Simplify the expression
This simplifies to:
\[ \frac{dy}{dx} = -\frac{\sin(e^x) \cdot e^x}{\cos(e^x)} \]

### Step 7: Use the identity for tangent
We can rewrite this as:
\[ \frac{dy}{dx} = -e^x \tan(e^x) \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = -e^x \tan(e^x) \]