Question

If \( y = \csc(\cot^{-1} x) \), then prove that \( \sqrt{1 + x^2} \frac{dy}{dx} - x = 0 \).

Hint:

When dealing with inverse trigonometric functions, it is often helpful to represent them as triangle relationships first, simplifying the differentiation process.

Answer 1

Step 1: Express \( \csc(\cot^{-1} x) \) in terms of \( x \).
Let \( \theta = \cot^{-1} x \). Then: \[ \cot \theta = x \quad \Rightarrow \quad \text{Adjacent side} = 1, \quad \text{Opposite side} = x, \quad \text{Hypotenuse} = \sqrt{1 + x^2}. \] From the definition of cosecant: \[ \csc \theta = \frac{\text{Hypotenuse}}{\text{Opposite side}} = \frac{\sqrt{1 + x^2}}{x}. \] Thus, we have: \[ y = \csc(\cot^{-1} x) = \frac{\sqrt{1 + x^2}}{x}. \] Step 2: Differentiate \( y = \frac{\sqrt{1 + x^2}}{x} \).
To differentiate \( y = \frac{\sqrt{1 + x^2}}{x} \), we use the quotient rule: \[ \frac{dy}{dx} = \frac{\frac{d}{dx}(\sqrt{1 + x^2}) \cdot x - \sqrt{1 + x^2} \cdot \frac{d}{dx}(x)}{x^2}. \] The derivative of \( \sqrt{1 + x^2} \) is: \[ \frac{d}{dx} (\sqrt{1 + x^2}) = \frac{x}{\sqrt{1 + x^2}}. \] Substitute into the quotient rule: \[ \frac{dy}{dx} = \frac{\left(\frac{x}{\sqrt{1 + x^2}} \cdot x\right) - \sqrt{1 + x^2}}{x^2}. \] Simplify: \[ \frac{dy}{dx} = \frac{\frac{x^2}{\sqrt{1 + x^2}} - \sqrt{1 + x^2}}{x^2}. \] Now, combine the terms in the numerator under a common denominator: \[ \frac{dy}{dx} = \frac{\frac{x^2 - (1 + x^2)}{\sqrt{1 + x^2}}}{x^2} = \frac{\frac{-1}{\sqrt{1 + x^2}}}{x^2}. \] Simplify further: \[ \frac{dy}{dx} = -\frac{1}{x^2 \sqrt{1 + x^2}}. \] Step 3: Verify the given expression.
Substitute \( \frac{dy}{dx} \) into the expression \( \sqrt{1 + x^2} \frac{dy}{dx} - x \): \[ \sqrt{1 + x^2} \cdot \left(-\frac{1}{x^2 \sqrt{1 + x^2}}\right) - x = -\frac{1}{x^2} - x. \] Simplify the expression: \[ -\frac{1}{x^2} - x + x = 0. \] Step 4: Final Conclusion.
The given expression is verified: \[ \boxed{\sqrt{1 + x^2} \frac{dy}{dx} - x = 0}. \]