Find `(dy)/(dx)` for the function: `y=a^((sin^(-1)x)^2`

Explanation

To find \(\frac{dy}{dx}\) for the function \(y = a^{(\sin^{-1} x)^2}\), we will use the chain rule and the properties of logarithmic differentiation. Let's go through the steps systematically.

### Step 1: Identify the function
We have:
\[ y = a^{(\sin^{-1} x)^2} \]

### Step 2: Differentiate using the chain rule
To differentiate \(y\) with respect to \(x\), we can use the formula for the derivative of an exponential function:
\[ \frac{d}{dx}(a^u) = a^u \ln(a) \frac{du}{dx} \]
where \(u = (\sin^{-1} x)^2\).

### Step 3: Differentiate the exponent
Now, we need to differentiate \(u = (\sin^{-1} x)^2\) using the chain rule:
\[ \frac{du}{dx} = 2(\sin^{-1} x) \cdot \frac{d}{dx}(\sin^{-1} x) \]
We know that:
\[ \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}} \]
Thus,
\[ \frac{du}{dx} = 2(\sin^{-1} x) \cdot \frac{1}{\sqrt{1 - x^2}} \]

### Step 4: Substitute back into the derivative
Now we can substitute \(u\) and \(\frac{du}{dx}\) back into the derivative of \(y\):
\[ \frac{dy}{dx} = a^{(\sin^{-1} x)^2} \ln(a) \cdot \frac{du}{dx} \]
Substituting \(\frac{du}{dx}\):
\[ \frac{dy}{dx} = a^{(\sin^{-1} x)^2} \ln(a) \cdot 2(\sin^{-1} x) \cdot \frac{1}{\sqrt{1 - x^2}} \]

### Step 5: Final expression
Combining everything, we get:
\[ \frac{dy}{dx} = 2a^{(\sin^{-1} x)^2} \ln(a) \cdot \frac{\sin^{-1} x}{\sqrt{1 - x^2}} \]

This is the required derivative \(\frac{dy}{dx}\).

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