Find `f'(x) if f(x) =log(cose^x)`

Answer

To find the derivative \( f'(x) \) of the function \( f(x) = \log(\cos(e^x)) \), we will use the chain rule and the properties of logarithmic and trigonometric functions. Here’s the step-by-step solution:

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

### Step 2: Apply the chain rule
Using the chain rule, we differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx}[\log(\cos(e^x))] \]
The derivative of \( \log(u) \) where \( u = \cos(e^x) \) is:
\[ f'(x) = \frac{1}{\cos(e^x)} \cdot \frac{d}{dx}[\cos(e^x)] \]

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

### Step 4: Substitute back into the derivative
Now substituting back into our expression for \( f'(x) \):
\[ f'(x) = \frac{1}{\cos(e^x)} \cdot (-\sin(e^x) \cdot e^x) \]
This simplifies to:
\[ f'(x) = -\frac{e^x \sin(e^x)}{\cos(e^x)} \]

### Step 5: Simplify using trigonometric identity
Using the identity \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \):
\[ f'(x) = -e^x \tan(e^x) \]

### Final Answer
Thus, the derivative \( f'(x) \) is:
\[ f'(x) = -e^x \tan(e^x) \]

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