`(d)/(dx)[sin^(-1)((3cosx+4sinx)/(5))]=`

Options

A. 1
B. 0
C. `-1`
D. `pm1`

Answer

To find the derivative of the function \( y = \sin^{-1}\left(\frac{3\cos x + 4\sin x}{5}\right) \), we will follow these steps:

### Step 1: Define the function
Let:
\[ y = \sin^{-1}\left(\frac{3\cos x + 4\sin x}{5}\right) \]

### Step 2: Differentiate using the chain rule
Using the chain rule, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - \left(\frac{3\cos x + 4\sin x}{5}\right)^2}} \cdot \frac{d}{dx}\left(\frac{3\cos x + 4\sin x}{5}\right) \]

### Step 3: Differentiate the inner function
Now we differentiate the inner function \( \frac{3\cos x + 4\sin x}{5} \):
\[ \frac{d}{dx}\left(\frac{3\cos x + 4\sin x}{5}\right) = \frac{1}{5} \left(-3\sin x + 4\cos x\right) \]

### Step 4: Substitute back into the derivative
Now substitute this back into the derivative:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - \left(\frac{3\cos x + 4\sin x}{5}\right)^2}} \cdot \frac{1}{5} \left(-3\sin x + 4\cos x\right) \]

### Step 5: Simplify the expression
Next, we need to simplify the expression inside the square root:
\[ 1 - \left(\frac{3\cos x + 4\sin x}{5}\right)^2 = 1 - \frac{(3\cos x + 4\sin x)^2}{25} \]
Expanding \( (3\cos x + 4\sin x)^2 \):
\[ (3\cos x + 4\sin x)^2 = 9\cos^2 x + 24\cos x \sin x + 16\sin^2 x \]
Thus,
\[ 1 - \frac{9\cos^2 x + 24\cos x \sin x + 16\sin^2 x}{25} = \frac{25 - (9\cos^2 x + 24\cos x \sin x + 16\sin^2 x)}{25} \]

### Step 6: Final expression for \( \frac{dy}{dx} \)
Now we can write the final expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{-3\sin x + 4\cos x}{5\sqrt{\frac{25 - (9\cos^2 x + 24\cos x \sin x + 16\sin^2 x)}{25}}} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{-3\sin x + 4\cos x}{\sqrt{25 - (9\cos^2 x + 24\cos x \sin x + 16\sin^2 x)}} \]

### Step 7: Evaluate the expression
To find the specific value of \( \frac{dy}{dx} \), we can evaluate it at specific points or simplify further based on the context of the problem.