Differentiate the following w.r.t. x :
`log(cose^(x))`

Answer

To differentiate the function \( y = \log(\cos(e^x)) \) with respect to \( x \), we will apply the chain rule and the properties of logarithmic differentiation. Here’s a step-by-step solution:

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

### Step 2: Differentiate using the chain rule
Using the chain rule, the derivative \( \frac{dy}{dx} \) can be expressed as:
\[ \frac{dy}{dx} = \frac{1}{\cos(e^x)} \cdot \frac{d}{dx}(\cos(e^x)) \]

### Step 3: Differentiate \( \cos(e^x) \)
Now, we need to differentiate \( \cos(e^x) \). Again, we will apply the chain rule:
\[ \frac{d}{dx}(\cos(e^x)) = -\sin(e^x) \cdot \frac{d}{dx}(e^x) \]
Since the derivative of \( e^x \) is \( e^x \), we have:
\[ \frac{d}{dx}(\cos(e^x)) = -\sin(e^x) \cdot e^x \]

### Step 4: Substitute back into the derivative
Now we can 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:
\[ \frac{dy}{dx} = -\frac{e^x \sin(e^x)}{\cos(e^x)} \]

### Step 5: Use the identity for tangent
We can express this in terms of tangent:
\[ \frac{dy}{dx} = -e^x \tan(e^x) \]

### Final Result
Thus, the derivative of \( y = \log(\cos(e^x)) \) with respect to \( x \) is:
\[ \frac{dy}{dx} = -e^x \tan(e^x) \]

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