Differentiation Question 237
Question: If $ y=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+…..\infty , $ then $ \frac{dy}{dx}= $
[Karnataka CET 1999]
Options:
A) y
B) $ y-1 $
C) $ y+1 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ y=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+……\infty $
Therefore $ y=e^{x} $
Differentiating with respect to x, we get $ \frac{dy}{dx}=e^{x}=y $ .
BETA
