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

Options

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

Answer

To find the derivative of \( \sin^2 x \) with respect to \( \cos^2 x \), we can use the chain rule and implicit differentiation. Here’s a step-by-step solution:

### Step 1: Define the Functions
Let:
- \( u = \sin^2 x \)
- \( v = \cos^2 x \)

### Step 2: Differentiate \( u \) with respect to \( x \)
Using the chain rule:
\[ \frac{du}{dx} = \frac{d}{dx}(\sin^2 x) = 2\sin x \cdot \frac{d}{dx}(\sin x) = 2\sin x \cdot \cos x \]
This simplifies to:
\[ \frac{du}{dx} = 2\sin x \cos x = \sin(2x) \]

### Step 3: Differentiate \( v \) with respect to \( x \)
Again using the chain rule:
\[ \frac{dv}{dx} = \frac{d}{dx}(\cos^2 x) = 2\cos x \cdot \frac{d}{dx}(\cos x) = 2\cos x \cdot (-\sin x) \]
This simplifies to:
\[ \frac{dv}{dx} = -2\cos x \sin x = -\sin(2x) \]

### Step 4: Find the Derivative \( \frac{du}{dv} \)
Using the derivatives found in Steps 2 and 3:
\[ \frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{\sin(2x)}{-\sin(2x)} = -1 \]

### Final Result
Thus, the derivative of \( \sin^2 x \) with respect to \( \cos^2 x \) is:
\[ \frac{du}{dv} = -1 \]

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