Question

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

Hint:

Use trigonometric identities and triangle relationships to simplify inverse trigonometric functions before differentiation.

Answer 1

Step 1: Express \( \csc(\cot^{-1} x) \) in terms of \( x \).
Let \( \theta = \cot^{-1} x \). Then: \[ \cot \theta = x \quad \Rightarrow \quad \text{Adjacent side} = 1, \quad \text{Opposite side} = x, \quad \text{Hypotenuse} = \sqrt{1 + x^2}. \] Thus: \[ \csc \theta = \frac{\text{Hypotenuse}}{\text{Opposite side}} = \frac{\sqrt{1 + x^2}}{x}. \] So: \[ y = \csc(\cot^{-1} x) = \frac{\sqrt{1 + x^2}}{x}. \] Step 2: Differentiate \( y = \frac{\sqrt{1 + x^2}}{x} \).
Using 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: \[ \frac{d}{dx} (\sqrt{1 + x^2}) = \frac{1}{2\sqrt{1 + x^2}} \cdot 2x = \frac{x}{\sqrt{1 + x^2}}. \] 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: \[ \frac{dy}{dx} = \frac{\frac{x^2}{\sqrt{1 + x^2}} - \sqrt{1 + x^2}}{x^2}. \] Combine terms 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 \( \sqrt{1 + x^2} \frac{dy}{dx} - x \): \[ \sqrt{1 + x^2} \cdot \left(-\frac{1}{x^2 \sqrt{1 + x^2}}\right) - x = -\frac{1}{x^2} - x. \] Simplify: \[ -\frac{1}{x^2} - x + x = 0. \] Step 4: Conclusion.
The given expression is verified: \[ \boxed{\sqrt{1 + x^2} \frac{dy}{dx} - x = 0}. \]