Derivative of `e^(x) cosx` w.r.t. `e^(-x)`sin x is

Options

A. cot x
B. `-cot x`
C. `e^(2x)`
D. `-e^(2x)`

Answer

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

### Step 1: Define the functions
Let:
- \( y = e^{x} \cos x \)
- \( z = e^{-x} \sin x \)

### Step 2: Find \( \frac{dy}{dx} \)
We need to differentiate \( y \) with respect to \( x \).

Using the product rule:
\[ \frac{dy}{dx} = \frac{d}{dx}(e^{x}) \cdot \cos x + e^{x} \cdot \frac{d}{dx}(\cos x) \]
Calculating the derivatives:
- \( \frac{d}{dx}(e^{x}) = e^{x} \)
- \( \frac{d}{dx}(\cos x) = -\sin x \)

Thus,
\[ \frac{dy}{dx} = e^{x} \cos x - e^{x} \sin x \]
Factoring out \( e^{x} \):
\[ \frac{dy}{dx} = e^{x} (\cos x - \sin x) \]

### Step 3: Find \( \frac{dz}{dx} \)
Now, differentiate \( z \) with respect to \( x \).

Using the product rule:
\[ \frac{dz}{dx} = \frac{d}{dx}(e^{-x}) \cdot \sin x + e^{-x} \cdot \frac{d}{dx}(\sin x) \]
Calculating the derivatives:
- \( \frac{d}{dx}(e^{-x}) = -e^{-x} \)
- \( \frac{d}{dx}(\sin x) = \cos x \)

Thus,
\[ \frac{dz}{dx} = -e^{-x} \sin x + e^{-x} \cos x \]
Factoring out \( e^{-x} \):
\[ \frac{dz}{dx} = e^{-x} (\cos x - \sin x) \]

### Step 4: Find \( \frac{dy}{dz} \)
Using the chain rule:
\[ \frac{dy}{dz} = \frac{dy/dx}{dz/dx} \]
Substituting the expressions we found:
\[ \frac{dy}{dz} = \frac{e^{x} (\cos x - \sin x)}{e^{-x} (\cos x - \sin x)} \]

### Step 5: Simplify
The \( (\cos x - \sin x) \) terms cancel out (assuming \( \cos x - \sin x \neq 0 \)):
\[ \frac{dy}{dz} = \frac{e^{x}}{e^{-x}} = e^{x + x} = e^{2x} \]

### Final Answer
Thus, the derivative of \( e^{x} \cos x \) with respect to \( e^{-x} \sin x \) is:
\[ \frac{dy}{dz} = e^{2x} \]

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