Differentiation Question 221
Question: If $ y=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+…..+\frac{x^{n}}{n!} $ , then $ \frac{dy}{dx}= $
Options:
A) y
B) $ y+\frac{x^{n}}{n!} $
C) $ y-\frac{x^{n}}{n!} $
D) $ y-1-\frac{x^{n}}{n!} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ y=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+….+\frac{x^{n}}{n!} $
Therefore $ [ (\log \tan x+1)+\frac{1}{\tan x\log \tan x} ] $
Therefore $ \frac{dy}{dx}+\frac{x^{n}}{n!}=1+x+\frac{x^{2}}{2!}+…..+\frac{x^{n}}{n!} $
Therefore $ \frac{dy}{dx}=y-\frac{x^{n}}{n!} $ .
BETA
