Differentiation Question 40
Question: If $ y=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+…+\frac{x^{n}}{n!} $ , then $ \frac{dy}{dx} $ is equal to
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:
[c] $ y=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+…+\frac{x^{n}}{n!} $ Or $ \frac{dy}{dx}=0+1+x+\frac{x^{2}}{2!}+…+\frac{{x^{n-1}}}{(n-1)!} $ Or $ \frac{dy}{dx}+\frac{x^{n}}{n!}=1+x+\frac{x^{2}}{2!}+…+\frac{x^{n}}{n!} $ Or $ \frac{dy}{dx}=y-\frac{x^{n}}{n!} $
BETA
