`d/(dx)[log ((cos(e^(x))))]=`

Options

A. `cos(e^(x-1))`
B. `e^-xcos(e^(x))`
C. `e^xsin(e^(x))`
D. `-e^xtan(e^(x))`

Answer

To solve the problem \( \frac{d}{dx} \left[ \log(\cos(e^x)) \right] \), we will use the chain rule and properties of logarithms. Here is a step-by-step solution:

### Step 1: Apply the Chain Rule
We start by applying the chain rule to differentiate the logarithmic function. The derivative of \( \log(u) \) is \( \frac{1}{u} \cdot \frac{du}{dx} \), where \( u = \cos(e^x) \).

\[ \frac{d}{dx} \left[ \log(\cos(e^x)) \right] = \frac{1}{\cos(e^x)} \cdot \frac{d}{dx}[\cos(e^x)] \]

**Hint:** Remember that the derivative of \( \log(u) \) involves the derivative of \( u \).

### Step 2: Differentiate \( \cos(e^x) \)
Next, we need to differentiate \( \cos(e^x) \). Using the chain rule again, we have:

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

The derivative of \( e^x \) is \( e^x \), so we can substitute that in:

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

**Hint:** When differentiating \( \cos(u) \), remember that it becomes \( -\sin(u) \), and apply the chain rule for \( u = e^x \).

### Step 3: Substitute Back
Now we substitute this back into our expression from Step 1:

\[ \frac{d}{dx} \left[ \log(\cos(e^x)) \right] = \frac{1}{\cos(e^x)} \cdot \left(-\sin(e^x) \cdot e^x\right) \]

This simplifies to:

\[ \frac{d}{dx} \left[ \log(\cos(e^x)) \right] = -\frac{\sin(e^x) \cdot e^x}{\cos(e^x)} \]

### Step 4: Simplify the Expression
We can simplify the expression further using the identity \( \tan(u) = \frac{\sin(u)}{\cos(u)} \):

\[ -\frac{\sin(e^x)}{\cos(e^x)} \cdot e^x = -e^x \tan(e^x) \]

### Final Answer
Thus, the derivative is:

\[ \frac{d}{dx} \left[ \log(\cos(e^x)) \right] = -e^x \tan(e^x) \]