Question

If \( y = \cos^{-1} x \), find \( \frac{d^2y}{dx^2} \) in terms of \( y \).

• A. \(-\csc^2 y \cot y\)
• B. \(\csc^2 y \cot y\)
• C. \(-\csc y \cot^2 y\)
• D. \(\csc y \cot^2 y\)

Hint:

For second derivatives in inverse trig functions, always convert to \(x = f(y)\) first. It makes chain rule application much cleaner and avoids messy algebra.

Answer 1

The Correct Option is A

Concept: The problem involves differentiating an inverse trigonometric function and expressing the second derivative entirely in terms of \(y\). We use implicit differentiation and the chain rule. Key ideas:
• Implicit differentiation for inverse functions
• Chain rule: \( \frac{d}{dx}[f(y)] = f'(y)\frac{dy}{dx} \)
• Trigonometric derivatives: \( \frac{d}{dy}(\csc y) = -\csc y \cot y \)

Step 1: Converting inverse form to trigonometric form.
\[ y = \cos^{-1} x \Rightarrow x = \cos y \]

Step 2: First derivative.
Differentiate \(x = \cos y\) w.r.t. \(x\): \[ \frac{d}{dx}(x) = \frac{d}{dx}(\cos y) \] \[ 1 = -\sin y \cdot \frac{dy}{dx} \] \[ \frac{dy}{dx} = -\frac{1}{\sin y} = -\csc y \]

Step 3: Second derivative using chain rule.
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(-\csc y) \] Apply chain rule: \[ \frac{d^2y}{dx^2} = \frac{d}{dy}(-\csc y)\cdot \frac{dy}{dx} \] Now, \[ \frac{d}{dy}(-\csc y) = \csc y \cot y \] So, \[ \frac{d^2y}{dx^2} = (\csc y \cot y)\cdot \frac{dy}{dx} \]

Step 4: Substitute \( \frac{dy}{dx} = -\csc y \).
\[ \frac{d^2y}{dx^2} = (\csc y \cot y)(-\csc y) \] \[ \frac{d^2y}{dx^2} = -\csc^2 y \cot y \]