The differentiation of `sin^3 x + cos ^(3) x` with respect to `sin x + cos x` is

Options

A. `sin x cos x`
B. `-3 sin x cos x`
C. `-sin x cos x`
D. `3sin ^(2) x cos ^(2) x`

Answer

To solve the problem of differentiating \( \sin^3 x + \cos^3 x \) with respect to \( \sin x + \cos x \), we will follow these steps:

### Step 1: Define the functions
Let:
- \( p = \sin^3 x + \cos^3 x \)
- \( q = \sin x + \cos x \)

### Step 2: Differentiate \( p \) with respect to \( x \)
Using the chain rule and the power rule, we differentiate \( p \):
\[ \frac{dp}{dx} = \frac{d}{dx}(\sin^3 x) + \frac{d}{dx}(\cos^3 x) \]
Using the formula \( \frac{d}{dx}(x^n) = n x^{n-1} \frac{dx}{dx} \):
- For \( \sin^3 x \):
\[ \frac{d}{dx}(\sin^3 x) = 3 \sin^2 x \cdot \cos x \]
- For \( \cos^3 x \):
\[ \frac{d}{dx}(\cos^3 x) = 3 \cos^2 x \cdot (-\sin x) = -3 \cos^2 x \sin x \]
Combining these results:
\[ \frac{dp}{dx} = 3 \sin^2 x \cos x - 3 \cos^2 x \sin x \]
Factoring out the common term:
\[ \frac{dp}{dx} = 3 \sin x \cos x (\sin x - \cos x) \]

### Step 3: Differentiate \( q \) with respect to \( x \)
Now we differentiate \( q \):
\[ \frac{dq}{dx} = \frac{d}{dx}(\sin x + \cos x) = \cos x - \sin x \]

### Step 4: Find \( \frac{dp}{dq} \)
We want to find \( \frac{dp}{dq} \):
\[ \frac{dp}{dq} = \frac{dp/dx}{dq/dx} \]
Substituting the derivatives we found:
\[ \frac{dp}{dq} = \frac{3 \sin x \cos x (\sin x - \cos x)}{\cos x - \sin x} \]

### Step 5: Simplify \( \frac{dp}{dq} \)
Notice that \( \cos x - \sin x = -(\sin x - \cos x) \):
\[ \frac{dp}{dq} = \frac{3 \sin x \cos x (\sin x - \cos x)}{- (\sin x - \cos x)} \]
The \( \sin x - \cos x \) terms cancel out:
\[ \frac{dp}{dq} = -3 \sin x \cos x \]

### Final Answer
Thus, the differentiation of \( \sin^3 x + \cos^3 x \) with respect to \( \sin x + \cos x \) is:
\[ \frac{dp}{dq} = -3 \sin x \cos x \]