Differentiation Question 249
Question: $ \frac{d}{dx}[ \frac{e^{ax}}{\sin (bx+c)} ]= $
[AI CBSE 1983]
Options:
A) $ \frac{e^{ax}[a\sin (bx+c)+b\cos (bx+c)]}{{{\sin }^{2}}(bx+c)} $
B) $ \frac{e^{ax}[a\sin (bx+c)-b\cos (bx+c)]}{\sin (bx+c)} $
C) $ \frac{e^{ax}[a\sin (bx+c)-b\cos (bx+c)]}{{{\sin }^{2}}(bx+c)} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \frac{d}{dx}( \frac{e^{ax}}{\sin (bx+c)} ) $
$ =\frac{ae^{ax}\sin (bx+c)-be^{ax}\cos (bx+c)}{{{{\sin (bx+c)}}^{2}}} $
$ =\frac{e^{ax}[a\sin (bx+c)-b\cos (bx+c)]}{{{\sin }^{2}}(bx+c)} $ .
BETA
