`cos(a sin ^(-1)"1/x)`

Answer

To differentiate the function \( y = \cos(a \sin^{-1}(\frac{1}{x})) \) with respect to \( x \), we will follow these steps:

### Step 1: Identify the function
We start with the function:
\[ y = \cos(a \sin^{-1}(\frac{1}{x})) \]

### Step 2: Differentiate using the chain rule
To differentiate \( y \) with respect to \( x \), we apply the chain rule. The derivative of \( \cos(u) \) is \( -\sin(u) \cdot \frac{du}{dx} \), where \( u = a \sin^{-1}(\frac{1}{x}) \).

Thus, we have:
\[ \frac{dy}{dx} = -\sin(a \sin^{-1}(\frac{1}{x})) \cdot \frac{d}{dx}(a \sin^{-1}(\frac{1}{x})) \]

### Step 3: Differentiate \( a \sin^{-1}(\frac{1}{x}) \)
Next, we need to differentiate \( a \sin^{-1}(\frac{1}{x}) \). The derivative of \( \sin^{-1}(v) \) is \( \frac{1}{\sqrt{1 - v^2}} \cdot \frac{dv}{dx} \). Here, \( v = \frac{1}{x} \).

First, we find \( \frac{dv}{dx} \):
\[ v = \frac{1}{x} \implies \frac{dv}{dx} = -\frac{1}{x^2} \]

Now, we can differentiate \( \sin^{-1}(\frac{1}{x}) \):
\[ \frac{d}{dx}(\sin^{-1}(\frac{1}{x})) = \frac{1}{\sqrt{1 - (\frac{1}{x})^2}} \cdot \left(-\frac{1}{x^2}\right) \]
\[ = -\frac{1}{x^2 \sqrt{1 - \frac{1}{x^2}}} \]
\[ = -\frac{1}{x^2 \sqrt{\frac{x^2 - 1}{x^2}}} = -\frac{1}{\sqrt{x^2 - 1}} \cdot \frac{1}{x^2} \]

### Step 4: Combine the derivatives
Now substituting back into the derivative of \( y \):
\[ \frac{dy}{dx} = -\sin(a \sin^{-1}(\frac{1}{x})) \cdot a \left(-\frac{1}{\sqrt{x^2 - 1}} \cdot \frac{1}{x^2}\right) \]
\[ = a \sin(a \sin^{-1}(\frac{1}{x})) \cdot \frac{1}{x^2 \sqrt{x^2 - 1}} \]

### Step 5: Final expression
Thus, the final expression for the derivative is:
\[ \frac{dy}{dx} = \frac{a \sin(a \sin^{-1}(\frac{1}{x}))}{x^2 \sqrt{x^2 - 1}} \]