If `y=3^(2 sin ^(-1))` then `|((x ^(2) -1) y^('') +xy')/(y)|` is equal to

Answer

To solve the problem, we need to find the value of

\[ \left| \frac{(x^2 - 1)y'' + xy'}{y} \right| \]

given that

\[ y = 3^{2 \sin^{-1} x}. \]

### Step 1: Differentiate \(y\) to find \(y'\)

We start by differentiating \(y\) with respect to \(x\):

\[ y = 3^{2 \sin^{-1} x}. \]

Using the chain rule and the property of exponential functions, we have:

\[ y' = \frac{d}{dx} \left( 3^{2 \sin^{-1} x} \right) = 3^{2 \sin^{-1} x} \cdot \ln(3) \cdot \frac{d}{dx}(2 \sin^{-1} x). \]

The derivative of \(2 \sin^{-1} x\) is:

\[ \frac{d}{dx}(2 \sin^{-1} x) = \frac{2}{\sqrt{1 - x^2}}. \]

Thus, we can write:

\[ y' = 3^{2 \sin^{-1} x} \cdot \ln(3) \cdot \frac{2}{\sqrt{1 - x^2}}. \]

### Step 2: Differentiate \(y'\) to find \(y''\)

Now we differentiate \(y'\) to find \(y''\):

\[ y' = 3^{2 \sin^{-1} x} \cdot \ln(3) \cdot \frac{2}{\sqrt{1 - x^2}}. \]

Using the product rule:

\[ y'' = \frac{d}{dx} \left( 3^{2 \sin^{-1} x} \cdot \ln(3) \cdot \frac{2}{\sqrt{1 - x^2}} \right). \]

This involves differentiating both parts and applying the product rule. We will have:

1. Differentiate \(3^{2 \sin^{-1} x}\) (which we already found as \(y\)).
2. Differentiate \(\frac{2}{\sqrt{1 - x^2}}\).

After applying the product rule and simplifying, we will find an expression for \(y''\).

### Step 3: Substitute \(y\), \(y'\), and \(y''\) into the expression

Now we substitute \(y\), \(y'\), and \(y''\) into the expression:

\[ \left| \frac{(x^2 - 1)y'' + xy'}{y} \right|. \]

### Step 4: Simplify the expression

After substituting, we will simplify the expression. We will notice that \(y\) cancels out in the numerator and denominator, leading to a simpler expression.

### Step 5: Evaluate the absolute value

Finally, we will take the absolute value of the resulting expression.

The final answer will be:

\[ \left| \frac{(x^2 - 1)y'' + xy'}{y} \right| = \log^2(3). \]