Differentiation Question 278
Question: $ \frac{d}{dx}[e^{ax}\cos (bx+c)] $ =
[AISSE 1989]
Options:
A) $ e^{ax}[a\cos (bx+c)-b\sin (bx+c)] $
B) $ e^{ax}[a\sin (bx+c)-b\cos (bx+c)] $
C) $ e^{ax}[\cos (bx+c)-\sin (bx+c)] $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{d}{dx}[e^{ax}\cos (bx+c)] $ = $ \frac{dx}{dt}=-2\sin t+2\sin 2t $
= $ e^{ax}[a\cos (bx+c)-b\sin (bx+c)] $ .
BETA
