If `y=log(cose^(x))," then "dy/dx` = _______.

Answer

To solve the problem, we need to differentiate the function \( y = \log(\cos(e^x)) \) with respect to \( x \). Let's go through the steps systematically.

### Step 1: Differentiate the logarithmic function
We start with the function:
\[ y = \log(\cos(e^x)) \]
Using the chain rule for differentiation, we have:
\[ \frac{dy}{dx} = \frac{1}{\cos(e^x)} \cdot \frac{d}{dx}(\cos(e^x)) \]

### Step 2: Differentiate the cosine function
Next, we need to differentiate \( \cos(e^x) \). Using the chain rule again:
\[ \frac{d}{dx}(\cos(e^x)) = -\sin(e^x) \cdot \frac{d}{dx}(e^x) \]
Since the derivative of \( e^x \) is \( e^x \), we get:
\[ \frac{d}{dx}(\cos(e^x)) = -\sin(e^x) \cdot e^x \]

### Step 3: Substitute back into the derivative
Now we substitute this back into our expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{1}{\cos(e^x)} \cdot (-\sin(e^x) \cdot e^x) \]
This simplifies to:
\[ \frac{dy}{dx} = -\frac{\sin(e^x) \cdot e^x}{\cos(e^x)} \]

### Step 4: Simplify using trigonometric identities
Using the identity \( \tan(x) = \frac{\sin(x)}{\cos(x)} \), we can rewrite our expression:
\[ \frac{dy}{dx} = -e^x \tan(e^x) \]

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