If y=\frac{x\sqrt{x^{2}-16}}{2}-8\ln|x+\sqrt{x^{2}-16}|, then what is \frac{dy}{dx} equal to ?
- A. x\sqrt{x^{2}-16}
- B. x-\sqrt{x^{2}-16}
- C. \sqrt{x^{2}-16} ✓
- D. 4\sqrt{x^{2}-16}
Correct Answer: C. \sqrt{x^{2}-16}
Explanation
We can recognize the given expression as the standard result of the integral \int \sqrt{x^2 - a^2} \,dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\ln|x + \sqrt{x^2 - a^2}|. Comparing this with y, we see that a^2 = 16. Since y is the antiderivative of \sqrt{x^2 - 16}, differentiating y with respect to x will return the original function \sqrt{x^2 - 16}.
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