Differentiation Question 391
Question: If $ y=1-x+\frac{x^{2}}{2!}-\frac{x^{3}}{3!}+\frac{x^{4}}{4!}- $ ….., then $ \frac{d^{2}y}{dx^{2}}= $
[Karnataka CET 2003]
Options:
A) $ x $
B) $ -x $
C) $ -y $
D) $ y $
Show Answer
Answer:
Correct Answer: D
Solution:
$ y=1-x+\frac{x^{2}}{(2)!}-\frac{x^{2}}{(3)!}+………… $
Therefore $ y={e^{-x}} $
Therefore $ \frac{dy}{dx}={e^{-x}}(-1) $
and $ \frac{d^{2}y}{dx^{2}}=(-1){{e^{-x}}.(-1)} $
$ ={e^{-x}}=y $ .
BETA
