The derivative of `sin^(2)x` with respect to `e^(cosx)` is :

Options

A. `-2cosxe^(-cosx)`
B. `2cosxe^(-cosx)`
C. `2cosxe^(cosx)`
D. `-2cosxe^(cosx)`

Answer

To find the derivative of \( \sin^2 x \) with respect to \( e^{\cos x} \), we will use the chain rule. Here’s a step-by-step solution:

### Step 1: Define the functions
Let:
- \( u = \sin^2 x \)
- \( v = e^{\cos x} \)

We want to find \( \frac{du}{dv} \).

### Step 2: Find \( \frac{du}{dx} \)
Using the chain rule, we differentiate \( u \) with respect to \( x \):
\[ \frac{du}{dx} = \frac{d}{dx}(\sin^2 x) = 2 \sin x \cdot \cos x \]
This is derived from the power rule and the fact that the derivative of \( \sin x \) is \( \cos x \).

### Step 3: Find \( \frac{dv}{dx} \)
Now, we differentiate \( v \) with respect to \( x \):
\[ \frac{dv}{dx} = \frac{d}{dx}(e^{\cos x}) = e^{\cos x} \cdot (-\sin x) \]
This uses the chain rule, where the derivative of \( e^x \) is \( e^x \) and we multiply by the derivative of \( \cos x \), which is \( -\sin x \).

### Step 4: Use the chain rule to find \( \frac{du}{dv} \)
Using the relationship \( \frac{du}{dv} = \frac{du}{dx} \cdot \frac{dx}{dv} \), we can express \( \frac{du}{dv} \) as:
\[ \frac{du}{dv} = \frac{du}{dx} \cdot \frac{1}{\frac{dv}{dx}} = \frac{2 \sin x \cos x}{e^{\cos x} (-\sin x)} \]

### Step 5: Simplify the expression
Now, we can simplify:
\[ \frac{du}{dv} = \frac{2 \sin x \cos x}{-e^{\cos x} \sin x} = \frac{-2 \cos x}{e^{\cos x}} \]
Here, the \( \sin x \) terms cancel out.

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