If `f(x)=e^(sin(log cos x))` and `g(x)=log cos x`, then what is the derivative of f(x) with respect to g(x) ?

Options

A. `f(x)cos[g(x)]`
B. `f(x)sin[g(x)]`
C. `g(x)cos[f(x)]`
D. `g(x)sin[f(x)]`

Answer

To find the derivative of \( f(x) = e^{\sin(\log(\cos x))} \) with respect to \( g(x) = \log(\cos x) \), we will use the chain rule. Here’s the step-by-step solution:

### Step 1: Differentiate \( f(x) \) with respect to \( x \)

We start by differentiating \( f(x) \):

\[ f(x) = e^{\sin(\log(\cos x))} \]

Using the chain rule, we differentiate \( f(x) \):

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

### Step 2: Differentiate \( \sin(\log(\cos x)) \)

Now, we need to differentiate \( \sin(\log(\cos x)) \):

\[ \frac{d}{dx}[\sin(\log(\cos x))] = \cos(\log(\cos x)) \cdot \frac{d}{dx}[\log(\cos x)] \]

### Step 3: Differentiate \( \log(\cos x) \)

Next, we differentiate \( \log(\cos x) \):

\[ \frac{d}{dx}[\log(\cos x)] = \frac{1}{\cos x} \cdot \frac{d}{dx}[\cos x] = \frac{1}{\cos x} \cdot (-\sin x) = -\frac{\sin x}{\cos x} = -\tan x \]

### Step 4: Combine the derivatives

Now we can substitute this back into our expression for \( \frac{df}{dx} \):

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

This simplifies to:

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

### Step 5: Differentiate \( g(x) \) with respect to \( x \)

Now we differentiate \( g(x) \):

\[ g(x) = \log(\cos x) \]

Using the result from Step 3:

\[ \frac{dg}{dx} = -\tan x \]

### Step 6: Apply the chain rule to find \( \frac{df}{dg} \)

Now we can find \( \frac{df}{dg} \) using the chain rule:

\[ \frac{df}{dg} = \frac{df/dx}{dg/dx} = \frac{-\tan x \cdot \cos(\log(\cos x)) \cdot e^{\sin(\log(\cos x))}}{-\tan x} \]

The \( -\tan x \) cancels out:

\[ \frac{df}{dg} = \cos(\log(\cos x)) \cdot e^{\sin(\log(\cos x))} \]

### Final Answer

Thus, the derivative of \( f(x) \) with respect to \( g(x) \) is:

\[ \frac{df}{dg} = e^{\sin(\log(\cos x))} \cdot \cos(\log(\cos x)) \]