Q224 / 270 · NDA 2012 (Phase 2) · Mathematics

If $y=\ln \left(e^{m x}+e^{-m x}\right)$, then what is $\frac{d y}{d x}$ at $x=0$ equal to?

  • A
    -1
  • Correct answer
    B
    0
  • C
    1
  • D
    2

Solution

$y=\log _e\left(e^{m x}+e^{-m x}\right)$ $\frac{d y}{d x}=\frac{1}{\left(e^{m x}+e^{-m x}\right)} \cdot\left(e^{m x} \cdot m+e^{-m x}(-m)\right)$ $=\frac{1}{\left(e^{m x}+e^{-m x}\right)}\left(e^{m x}-e^{-m x}\right)$ $\left(\frac{d y}{d x}\right)_{x=0}=0$

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

Relevant Questions