If `y=cos^(-1)((2cosx-3sinx)/(sqrt(13)))`, then `(dy)/(dx)`, is

Options

A. zero
B. constant=1
C. `"constant "ne1`
D. none of these

Answer

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

### Step 1: Rewrite the function
We start with the given function:
\[ y = \cos^{-1}\left(\frac{2\cos x - 3\sin x}{\sqrt{13}}\right) \]

### Step 2: Use the cosine identity
We can express this in terms of cosine:
\[ y = \cos^{-1}\left(\frac{2}{\sqrt{13}} \cos x - \frac{3}{\sqrt{13}} \sin x\right) \]
Let \(A = \frac{2}{\sqrt{13}}\) and \(B = \frac{3}{\sqrt{13}}\).

### Step 3: Recognize the cosine of a sum
Using the cosine of a sum identity:
\[ \cos(a + b) = \cos a \cos b - \sin a \sin b \]
we can interpret:
\[ y = \cos^{-1}(\cos(x + a)) \]
where \(a\) is such that:
\[ \cos a = \frac{2}{\sqrt{13}}, \quad \sin a = \frac{3}{\sqrt{13}} \]

### Step 4: Find \(a\)
From the definitions of sine and cosine:
\[ \tan a = \frac{\sin a}{\cos a} = \frac{3/\sqrt{13}}{2/\sqrt{13}} = \frac{3}{2} \]
Thus,
\[ a = \tan^{-1}\left(\frac{3}{2}\right) \]

### Step 5: Substitute back into the equation
Now we can write:
\[ y = x + a \]
where \(a = \tan^{-1}\left(\frac{3}{2}\right)\).

### Step 6: Differentiate
Now we differentiate \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = \frac{d}{dx}(x + a) = 1 + 0 = 1 \]

### Final Answer
Thus, the derivative is:
\[ \frac{dy}{dx} = 1 \]

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