The derivative of `cos^(3)x ` w.r.t. `sin^(3)x ` is

Options

A. `-cot x `
B. cotx
C. tan x
D. `-tan x `

Answer

To find the derivative of \( \cos^3 x \) with respect to \( \sin^3 x \), we can use the chain rule. Let's denote:

- \( u = \cos^3 x \)
- \( v = \sin^3 x \)

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

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

Using the chain rule, we differentiate \( u \) with respect to \( x \):

\[ u = \cos^3 x \]

Using the power rule and the chain rule:

\[ \frac{du}{dx} = 3 \cos^2 x \cdot \frac{d}{dx}(\cos x) = 3 \cos^2 x \cdot (-\sin x) = -3 \sin x \cos^2 x \]

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

Now we differentiate \( v \) with respect to \( x \):

\[ v = \sin^3 x \]

Again using the power rule and the chain rule:

\[ \frac{dv}{dx} = 3 \sin^2 x \cdot \frac{d}{dx}(\sin x) = 3 \sin^2 x \cdot \cos x \]

### 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 derivatives we found:

\[ \frac{du}{dv} = \frac{-3 \sin x \cos^2 x}{3 \sin^2 x \cos x} \]

### Step 4: Simplify the expression

We can simplify this expression:

\[ \frac{du}{dv} = \frac{-3 \sin x \cos^2 x}{3 \sin^2 x \cos x} = \frac{-\cancel{3} \sin x \cos^2 x}{\cancel{3} \sin^2 x \cos x} \]

Canceling out the common terms:

\[ = \frac{-\cos x}{\sin x} = -\cot x \]

### Final Answer

Thus, the derivative of \( \cos^3 x \) with respect to \( \sin^3 x \) is:

\[ \frac{du}{dv} = -\cot x \]

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