Q. If $y = ae^{x} + be^{- x} + c$ where $a,\ b,\ c$ are parameters, then $y’$ is equal to
A) $y$
B) $y^{'}$
C) 0
D) $y^{''}$
Explanation
(b)
$y = ae^{x} + be^{- x} + c$
On differentiating w.r.t. $x$, we get
$y^{'} = ae^{x} - be^{- x}$
Again differentiating w.r.t. $x$, we get
$y^{''} = ae^{x} + be^{- x}$
Again differentiating w.r.t. $x$, we get
$y^{'''} = ae^{x} - be^{- x} = y'$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.