Q227 / 270 · NDA 2022 (Phase 2) · Mathematics

If $y=\frac{x \sqrt{x^2-16}}{2}-8 \ln \left|x+\sqrt{x^2-16}\right|$, then what is $\frac{d y}{d x}$ equal to?

  • A
    $x \sqrt{x^2-16}$
  • B
    $x-\sqrt{x^2-16}$
  • Correct answer
    C
    $\sqrt{x^2-16}$
  • D
    $4 \sqrt{x^2-16}$

Solution

We have $\begin{aligned} & \int \sqrt{x^2-a^2} d x=\frac{x}{2} \sqrt{x^2-a^2}-\frac{a^2}{2} \ln \left|x+\sqrt{x^2-a^2}\right| \\ & \therefore \frac{d y}{d x}=\sqrt{x^2-16} \end{aligned}$

Hence, the correct answer is option: 'C'.

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