Q. The derivative of $\left\lbrack \frac{e^{x} + 1}{e^{x}} \right\rbrack$ is equal to
A) 0
B) $\frac{1}{e^{x}}$
C) $- \frac{1}{e^{x}}$
D) $e^{x}$
Explanation
(c)
$\frac{d}{dx}\left\lbrack \frac{e^{x} + 1}{e^{x}} \right\rbrack = \frac{d}{dx}\left\lbrack 1 + e^{- x} \right\rbrack = - e^{- x} = - \frac{1}{e^{x}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.