Differentiate the following w.r.t. as indicated :
`sin^(2)x" w.r.t. "e^(cosx)`

Answer

To differentiate \( \sin^2 x \) with respect to \( e^{\cos x} \), we can use the chain rule. Let's denote:

- \( u = \sin^2 x \)
- \( v = e^{\cos x} \)

We need to find \( \frac{du}{dv} \), which can be computed using the formula:

\[ \frac{du}{dv} = \frac{du/dx}{dv/dx} \]

### Step 1: Differentiate \( u = \sin^2 x \)

Using the chain rule, we have:

\[ \frac{du}{dx} = 2 \sin x \cdot \cos x \]

### Step 2: Differentiate \( v = e^{\cos x} \)

Using the chain rule again:

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

### Step 3: Compute \( \frac{du}{dv} \)

Now, substituting the derivatives we found:

\[ \frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{2 \sin x \cos x}{e^{\cos x} \cdot (-\sin x)} \]

### Step 4: Simplify \( \frac{du}{dv} \)

Notice that \( \sin x \) in the numerator and denominator can be canceled (assuming \( \sin x \neq 0 \)):

\[ \frac{du}{dv} = \frac{2 \cos x}{-e^{\cos x}} = -\frac{2 \cos x}{e^{\cos x}} \]

### Final Answer

Thus, the derivative of \( \sin^2 x \) with respect to \( e^{\cos x} \) is:

\[ \frac{du}{dv} = -\frac{2 \cos x}{e^{\cos x}} \]

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