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

Options

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

Answer

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

### Step 1: Identify the outer and inner functions
The outer function is \( \log(u) \) where \( u = \cos(e^x) \). The inner function is \( \cos(v) \) where \( v = e^x \).

### Step 2: Differentiate the outer function
The derivative of \( \log(u) \) with respect to \( u \) is:
\[ \frac{d}{du} \log(u) = \frac{1}{u} \]
Thus, we have:
\[ \frac{dy}{du} = \frac{1}{\cos(e^x)} \]

### Step 3: Differentiate the inner function \( \cos(v) \)
Now we differentiate \( \cos(v) \) with respect to \( v \):
\[ \frac{d}{dv} \cos(v) = -\sin(v) \]
Substituting \( v = e^x \), we get:
\[ \frac{du}{dv} = -\sin(e^x) \]

### Step 4: Differentiate \( e^x \)
Next, we differentiate \( e^x \) with respect to \( x \):
\[ \frac{dv}{dx} = e^x \]

### Step 5: Apply the chain rule
Now, we apply the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{1}{\cos(e^x)} \cdot (-\sin(e^x)) \cdot e^x \]

### Step 6: Simplify the expression
This simplifies to:
\[ \frac{dy}{dx} = -\frac{e^x \sin(e^x)}{\cos(e^x)} \]
Using the identity \( \tan(v) = \frac{\sin(v)}{\cos(v)} \), we can rewrite this as:
\[ \frac{dy}{dx} = -e^x \tan(e^x) \]

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