`dy/dx(cosec^(-1)x)` = __________.

Answer

To find the derivative \( \frac{dy}{dx} \) of \( y = \csc^{-1}(x) \), we can follow these steps:

### Step 1: Set \( y = \csc^{-1}(x) \)
Let \( y = \csc^{-1}(x) \). This implies that \( \csc(y) = x \).

### Step 2: Differentiate both sides
Differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(\csc(y)) = \frac{d}{dx}(x) \]
Using the chain rule, we have:
\[ -\csc(y) \cot(y) \frac{dy}{dx} = 1 \]

### Step 3: Solve for \( \frac{dy}{dx} \)
Rearranging the equation gives:
\[ \frac{dy}{dx} = -\frac{1}{\csc(y) \cot(y)} \]

### Step 4: Express \( \csc(y) \) and \( \cot(y) \) in terms of \( x \)
Since \( \csc(y) = x \), we can substitute:
\[ \frac{dy}{dx} = -\frac{1}{x \cot(y)} \]

### Step 5: Find \( \cot(y) \)
To find \( \cot(y) \), we can use the identity:
\[ \cot(y) = \frac{\cos(y)}{\sin(y)} \]
From the definition of cosecant, we know:
\[ \sin(y) = \frac{1}{\csc(y)} = \frac{1}{x} \]
Using the Pythagorean identity, we can find \( \cos(y) \):
\[ \cos^2(y) + \sin^2(y) = 1 \implies \cos^2(y) = 1 - \sin^2(y) = 1 - \left(\frac{1}{x}\right)^2 = \frac{x^2 - 1}{x^2} \]
Thus,
\[ \cos(y) = \sqrt{\frac{x^2 - 1}{x^2}} = \frac{\sqrt{x^2 - 1}}{x} \]
Now substituting into \( \cot(y) \):
\[ \cot(y) = \frac{\cos(y)}{\sin(y)} = \frac{\frac{\sqrt{x^2 - 1}}{x}}{\frac{1}{x}} = \sqrt{x^2 - 1} \]

### Step 6: Substitute \( \cot(y) \) back into \( \frac{dy}{dx} \)
Now substituting back, we have:
\[ \frac{dy}{dx} = -\frac{1}{x \sqrt{x^2 - 1}} \]

### Final Answer
Thus, the derivative of \( y = \csc^{-1}(x) \) is:
\[ \frac{dy}{dx} = -\frac{1}{x \sqrt{x^2 - 1}} \]

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