Differentiate ` sin ^(-1) ((2 cos x + 3 sin x )/(sqrt(13))) ` w.r.to x

Answer

To differentiate the function \( y = \sin^{-1} \left( \frac{2 \cos x + 3 \sin x}{\sqrt{13}} \right) \) with respect to \( x \), we will follow these steps:

### Step 1: Set up the equation
Let
\[ y = \sin^{-1} \left( \frac{2 \cos x + 3 \sin x}{\sqrt{13}} \right) \]

### Step 2: Use the formula for differentiation of inverse sine
The derivative of \( \sin^{-1}(u) \) with respect to \( x \) is given by:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} \]
where \( u = \frac{2 \cos x + 3 \sin x}{\sqrt{13}} \).

### Step 3: Differentiate \( u \)
First, we need to find \( \frac{du}{dx} \):
\[ u = \frac{2 \cos x + 3 \sin x}{\sqrt{13}} \]
Differentiating \( u \):
\[ \frac{du}{dx} = \frac{1}{\sqrt{13}} \cdot \left( -2 \sin x + 3 \cos x \right) \]

### Step 4: Calculate \( u^2 \)
Next, we need to calculate \( u^2 \):
\[ u^2 = \left( \frac{2 \cos x + 3 \sin x}{\sqrt{13}} \right)^2 = \frac{(2 \cos x + 3 \sin x)^2}{13} \]
Expanding the numerator:
\[ (2 \cos x + 3 \sin x)^2 = 4 \cos^2 x + 12 \cos x \sin x + 9 \sin^2 x \]
Thus,
\[ u^2 = \frac{4 \cos^2 x + 12 \cos x \sin x + 9 \sin^2 x}{13} \]

### Step 5: Substitute into the derivative formula
Now we can substitute \( u \) and \( \frac{du}{dx} \) into the derivative formula:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - \frac{4 \cos^2 x + 12 \cos x \sin x + 9 \sin^2 x}{13}}} \cdot \frac{1}{\sqrt{13}} \cdot (-2 \sin x + 3 \cos x) \]

### Step 6: Simplify the expression
Now simplify the expression inside the square root:
\[ 1 - u^2 = 1 - \frac{4 \cos^2 x + 12 \cos x \sin x + 9 \sin^2 x}{13} = \frac{13 - (4 \cos^2 x + 12 \cos x \sin x + 9 \sin^2 x)}{13} \]
This can be simplified further, but we can leave it in this form for now.

### Final Result
Thus, the derivative \( \frac{dy}{dx} \) can be expressed as:
\[ \frac{dy}{dx} = \frac{(-2 \sin x + 3 \cos x)}{\sqrt{13(13 - (4 \cos^2 x + 12 \cos x \sin x + 9 \sin^2 x))}} \]