Question

If \( y = f(\cosh x) \) and \( f'(x) = \log(x + \sqrt{x^2-1}) \) then \( \frac{d^2 y}{dx^2} = \)

• A. \( \sinh x + x \cosh x \)
• B. \( x \sinh x \)
• C. \( \log(x + \sqrt{x^2+1}) \)
• D. \( \frac{x(2\sqrt{x^2-1}+1)}{\sqrt{x^2-1}(x^2+\sqrt{x^2-1})} \)

Hint:

Recognizing the logarithmic definition of inverse hyperbolic functions can greatly simplify expressions. Specifically, \( \log(x + \sqrt{x^2-1}) \) is the definition of \( \cosh^{-1}(x) \). So \( f'(x) = \cosh^{-1}(x) \), implying \( f'(\cosh x) = x \).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:

We are given a composite function \( y \) and the derivative of the outer function \( f \). We need to apply the chain rule to find the first derivative \( \frac{dy}{dx} \) and then differentiate again to find the second derivative \( \frac{d^2y}{dx^2} \). Properties of hyperbolic functions will be used.
Step 2: Key Formula or Approach:

1. Chain Rule: \( \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \). 2. Hyperbolic Identities:
- \( \frac{d}{dx}(\cosh x) = \sinh x \)
- \( \frac{d}{dx}(\sinh x) = \cosh x \)
- \( \cosh^2 x - \sinh^2 x = 1 \implies \sqrt{\cosh^2 x - 1} = \sinh x \) (for \( x \ge 0 \)). 3. Inverse Hyperbolic Cosine: \( \cosh^{-1} u = \log(u + \sqrt{u^2-1}) \).
Step 3: Detailed Explanation:

Given \( y = f(\cosh x) \). Differentiate with respect to \( x \): \[ \frac{dy}{dx} = f'(\cosh x) \cdot \frac{d}{dx}(\cosh x) \] \[ \frac{dy}{dx} = f'(\cosh x) \cdot \sinh x \] We are given \( f'(u) = \log(u + \sqrt{u^2-1}) \). Substitute \( u = \cosh x \) into the expression for \( f' \): \[ f'(\cosh x) = \log(\cosh x + \sqrt{\cosh^2 x - 1}) \] Using the identity \( \cosh^2 x - 1 = \sinh^2 x \): \[ f'(\cosh x) = \log(\cosh x + \sinh x) \] We know that \( \cosh x + \sinh x = e^x \). \[ f'(\cosh x) = \log(e^x) = x \] So, the first derivative simplifies to: \[ \frac{dy}{dx} = x \cdot \sinh x \] Now, find the second derivative \( \frac{d^2 y}{dx^2} \) using the product rule: \[ \frac{d^2 y}{dx^2} = \frac{d}{dx}(x \sinh x) \] \[ = x \cdot \frac{d}{dx}(\sinh x) + \sinh x \cdot \frac{d}{dx}(x) \] \[ = x \cosh x + \sinh x \cdot 1 \] \[ = \sinh x + x \cosh x \]
Step 4: Final Answer:

The value of \( \frac{d^2 y}{dx^2} \) is \( \sinh x + x \cosh x \).