What is \frac{d^{2}y}{dx^{2}} at x=0 equal to?
For the following three (03) items: Let y=f(x)=\frac{x \sin^{-1}x}{\sqrt{1-x^{2}}}+\ln\sqrt{1-x^{2}}.
- A. 0
- B. 0.5
- C. 1 ✓
- D. 1.5
Correct Answer: C. 1
Explanation
From the previous solution, y' = \frac{\sin^{-1}x}{(1-x^2)^{3/2}}. Differentiating again using the product rule on \sin^{-1}x \cdot (1-x^2)^{-3/2}, we obtain y'' = \frac{1}{\sqrt{1-x^2}}(1-x^2)^{-3/2} + \sin^{-1}x[\frac{-3}{2}(1-x^2)^{-5/2}(-2x)]. Substituting x=0, the second term vanishes because \sin^{-1}(0) = 0. The first term becomes \frac{1}{1}(1) = 1. Thus, y''(0) = 1.
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