If `y=sin^(-1){(5x+12 sqrt(1-x^(2)))/(13)},` find `(dy)/(dx).`

Options

A. `(1)/(sqrt(1-x^(2)))`
B. `(-1)/(sqrt(1-x^(2)))`
C. `(3)/(sqrt(1-x^(2)))`
D. `(1)/(sqrt(1+x^(2)))`

Answer

To find \(\frac{dy}{dx}\) for the function \(y = \sin^{-1}\left(\frac{5x + 12\sqrt{1 - x^2}}{13}\right)\), we will follow these steps:

### Step 1: Rewrite the expression
We start with the given function:
\[ y = \sin^{-1}\left(\frac{5x + 12\sqrt{1 - x^2}}{13}\right) \]

### Step 2: Use the substitution
Let \(x = \cos \theta\). Therefore, \(\theta = \cos^{-1}(x)\). This gives us:
\[ \sqrt{1 - x^2} = \sin \theta \]
Substituting this into the equation for \(y\):
\[ y = \sin^{-1}\left(\frac{5\cos \theta + 12\sin \theta}{13}\right) \]

### Step 3: Recognize the sine addition formula
The expression inside the inverse sine can be recognized as:
\[ \frac{5\cos \theta + 12\sin \theta}{13} = \sin(\alpha + \theta) \]
where \(\sin \alpha = \frac{5}{13}\) and \(\cos \alpha = \frac{12}{13}\). Thus, we can write:
\[ y = \sin^{-1}(\sin(\alpha + \theta)) \]

### Step 4: Simplify using the property of inverse sine
Since \(\sin^{-1}(\sin(\theta)) = \theta\) for \(\theta\) in the range of \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), we have:
\[ y = \alpha + \theta \]
Substituting back for \(\theta\):
\[ y = \alpha + \cos^{-1}(x) \]

### Step 5: Differentiate with respect to \(x\)
Now we differentiate both sides with respect to \(x\):
\[ \frac{dy}{dx} = \frac{d}{dx}(\alpha) + \frac{d}{dx}(\cos^{-1}(x)) \]
Since \(\alpha\) is a constant, its derivative is \(0\). The derivative of \(\cos^{-1}(x)\) is:
\[ \frac{d}{dx}(\cos^{-1}(x)) = -\frac{1}{\sqrt{1 - x^2}} \]
Thus, we have:
\[ \frac{dy}{dx} = 0 - \frac{1}{\sqrt{1 - x^2}} = -\frac{1}{\sqrt{1 - x^2}} \]

### Final Result
The final result is:
\[ \frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2}} \]

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