What is the derivative of `sin^(2)x` with respect to `cos^(2)x`?

Options

A. `tan^(2)x`
B. `cot^(2)x`
C. `-1`
D. 1

Explanation

To find the derivative of \( \sin^2 x \) with respect to \( \cos^2 x \), we will use the chain rule. Let's denote:

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

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

### Step 1: Find \( \frac{du}{dx} \)

Using the chain rule, we differentiate \( u = \sin^2 x \):

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

### Step 2: Find \( \frac{dv}{dx} \)

Now, we differentiate \( v = \cos^2 x \):

\[ \frac{dv}{dx} = 2 \cos x \cdot \frac{d}{dx}(\cos x) = 2 \cos x \cdot (-\sin x) = -2 \cos x \sin x \]

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

Now we can find \( \frac{du}{dv} \) using the formula:

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

Substituting the values we found:

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

### Step 4: Simplify

The \( 2 \sin x \cos x \) in the numerator and the \( -2 \cos x \sin x \) in the denominator cancel each other out:

\[ \frac{du}{dv} = -1 \]

### Final Answer

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

\[ \frac{du}{dv} = -1 \]

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