If `y = cos^(-1) ((2cos x - 3 sin x)/(sqrt(13))), " then" (dy)/(dx) ` is equal to

Options

A. 1
B. 0
C. constant ` (ne 1)`
D. None of these

Explanation

To find the derivative \( \frac{dy}{dx} \) for the function \( y = \cos^{-1} \left( \frac{2 \cos x - 3 \sin x}{\sqrt{13}} \right) \), we will follow these steps:

### Step 1: Differentiate using the chain rule
We start by applying the chain rule for differentiation. The derivative of \( \cos^{-1}(u) \) is given by:
\[ \frac{d}{dx} \left( \cos^{-1}(u) \right) = -\frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} \]
where \( u = \frac{2 \cos x - 3 \sin x}{\sqrt{13}} \).

### Step 2: Find \( u \) and its derivative \( \frac{du}{dx} \)
Now, we need to differentiate \( u \):
\[ u = \frac{2 \cos x - 3 \sin x}{\sqrt{13}} \]
To find \( \frac{du}{dx} \), we differentiate the numerator:
\[ \frac{du}{dx} = \frac{d}{dx}(2 \cos x - 3 \sin x) \cdot \frac{1}{\sqrt{13}} \]
Calculating the derivative of the numerator:
\[ \frac{d}{dx}(2 \cos x) = -2 \sin x \]
\[ \frac{d}{dx}(-3 \sin x) = -3 \cos x \]
Thus,
\[ \frac{du}{dx} = \frac{-2 \sin x - 3 \cos x}{\sqrt{13}} \]

### Step 3: Substitute \( u \) and \( \frac{du}{dx} \) into the derivative formula
Now substituting \( u \) and \( \frac{du}{dx} \) into the derivative of \( y \):
\[ \frac{dy}{dx} = -\frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} \]
We need to calculate \( 1 - u^2 \):
\[ u^2 = \left( \frac{2 \cos x - 3 \sin x}{\sqrt{13}} \right)^2 = \frac{(2 \cos x - 3 \sin x)^2}{13} \]
Calculating \( (2 \cos x - 3 \sin x)^2 \):
\[ (2 \cos x - 3 \sin x)^2 = 4 \cos^2 x - 12 \cos x \sin x + 9 \sin^2 x \]
Thus,
\[ 1 - u^2 = 1 - \frac{4 \cos^2 x - 12 \cos x \sin x + 9 \sin^2 x}{13} \]
This can be simplified to:
\[ 1 - u^2 = \frac{13 - (4 \cos^2 x - 12 \cos x \sin x + 9 \sin^2 x)}{13} \]
\[ = \frac{13 - 4 \cos^2 x - 9 \sin^2 x + 12 \cos x \sin x}{13} \]

### Step 4: Final expression for \( \frac{dy}{dx} \)
Now substituting \( u \) and \( \frac{du}{dx} \) back into the derivative:
\[ \frac{dy}{dx} = -\frac{1}{\sqrt{1 - \frac{(2 \cos x - 3 \sin x)^2}{13}}} \cdot \frac{-2 \sin x - 3 \cos x}{\sqrt{13}} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{2 \sin x + 3 \cos x}{\sqrt{13} \sqrt{1 - \frac{(2 \cos x - 3 \sin x)^2}{13}}} \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{2 \sin x + 3 \cos x}{\sqrt{13 - (2 \cos x - 3 \sin x)^2}} \]